Maths new edition

["9.2 Chapter 9 INTRODUCTION The word \u2018statistics\u2019 is derived from the Latin word \u2018status\u2019 which means political state. Political states had to collect information about their citizens to facilitate Governance and plan for the development. Then, in course of time, statistics came to mean a branch of mathematics which deals with the collection, classification and analysis of numerical data. In this chapter, we shall learn about the classification of data, viz., grouped and ungrouped, measures of central tendencies and their properties. DATA The word \u2018data\u2019 means information in the form of numerical figures or a set of given facts. For example, the percentage of marks scored by 10 pupils of a class in a test are: 36, 80, 65, 75, 94, 48, 12, 64, 88 and 98. The set of these figures is the data related to the marks obtained by 10 pupils in a class test. Types of Data Statistics is basically the study of numerical data. It includes methods of collection, classification, presentation, analysis of data and inferences from data. Data as such can be qualitative or quantitative in nature. If one speaks of honesty, beauty, colour, etc., the data is qualitative while height, weight, distance, marks, etc., are quantitative. Data can also be classified as raw data and grouped data. Raw Data Data obtained from direct observation is called raw data. The marks obtained by 10 students in a monthly test is an example of raw data or ungrouped data. In fact, very little can be inferred from this data. So, to make this data clearer and more meaningful, we group it into ordered intervals. Grouped Data To present the data in a more meaningful way, we condense the data into convenient number of classes or groups, generally not exceeding 10 and not less than 5. This helps us in perceiving, at a glance, certain salient features of data. Some Basic Definitions Before getting into the details of tabular representation of data, let us review some basic definitions: Observation Each numerical figure in a data is called an observation. Frequency The number of times a particular observation occurs is called its frequency. Tabulation or Presentation of Data A systematical arrangement of the data in a tabular form is called tabulation or presentation of the data. This grouping results in a table called the frequency table which indicates the number of","Statistics 9.3 scores within each group. Many conclusions about the characteristics of the data, the behaviour of variables etc., can be drawn from this table. The quantitative data that is to be analysed statistically, can be divided into three categories: 1.\t Individual series 2.\t Discrete series 3. \t Continuous series Individual Series Any raw data that is collected, form is an individual series. Examples: 1.\t Runs scored by 6 batsmen in a test match: 13, 89, 56, 24, 48 and 17 2.\t Number of students in different classes in a school: 54, 46, 52, 38, 45, 50, 29, 60, 35, 56 Discrete Series A discrete series is formulated from the raw data by taking the frequency of the observations into consideration. Example:\u2002 Given below is the data showing the number of children in 10 families of a locality: 1, 1, 2, 2, 3, 1, 2, 3, 2, 2 Arranging the data in the ascending order, we get 1, 1, 1, 2, 2, 2, 2, 2, 3, 3. To count, we can use tally marks. We record tally marks in bunches of five, the fifth one crossing the other four diagonally, i.e., |||| . Thus, we may prepare a frequency table as below: Number of Children Tally Marks Number of Families 1 ||| (Frequency) 3 2 |||| 3 || 5 2 Continuous Series When the data contains large number of observations, we put them into different groups, called class intervals such as, 1\u201310, 11\u201320, 21\u201330, etc. Here, 1\u201310 means data whose values lie between 1 and 10, including both 1 and 10. This form is known as an inclusive form. Also, 1 is called the lower limit and 10 is called the upper limit.","9.4 Chapter 9 Example 9.1 Given below are the ages of 40 people in a colony: 33 \u20028 \u20027 28 30 25 6 50 24 44 56 32 27 21 17 62 58 16 14 19 24 31 27 \u20025 12 46 15 42 67 34 \u20024 21 10 40 20 50 48 63 \u20029 21 Taking class intervals 1\u201310, 11\u201320, 21\u201330, 31\u201340, 41\u201350, 51\u201360 and 61\u201370, we construct a frequency distribution table for the above data. Solution First, we write the ages in the ascending order as \u20024 \u20025 \u20026 \u20027 \u20028 \u20029 10 12 14 15 16 17 19 20 21 21 21 24 24 25 27 27 28 30 31 32 33 34 40 42 44 46 48 50 50 56 58 62 63 67 Now, we can prepare the frequency distribution table as below: Class Interval Tally Marks Frequency \u20021\u201310 |||| || \u20027 11\u201320 |||| || \u20027 21\u201330 10 31\u201340 |||| |||| \u20025 41\u201350 \u20026 51\u201360 |||| \u20022 61\u201370 |||| | \u20023 || ||| Class Interval A group into which the raw data is condensed is called a class interval. Each class is bounded by two figures, which are called the class limits. The figure on the LHS is called the lower limit and the figure on the RHS is called the upper limit of the class. Thus 0\u201310 is a class with lower limit being 0 and the upper limit being 10. Class Boundaries In an exclusive form, the lower and upper limits are known as class boundaries or true lower limit and true upper limit of the class respectively. Thus, the boundaries of 15\u201325 which is in exclusive form are 15 and 25. The boundaries in an inclusive form are obtained by subtracting 0.5 to the lower limit and adding 0.5 to the upper limit, if the difference between upper limit of a class and lower limit or the succeeding class is 1. Thus, the boundaries of 15\u201325 which is in inclusive form are 14.5\u201325.5.","Statistics 9.5 Class Size The difference between the true upper limit and the true lower limit is called the class size. Hence, in the above example, the class size = 25 \u2013 15 = 10. Class Mark or Mid-value 1 Class mark = 2 (upper limit + lower limit). Thus, the class mark of 15\u201325 is 1 (25 + 15) = 20 . 2 STATISTICAL GRAPHS The information provided by a numerical frequency distribution is easily understood when represented by diagrams or graphs. The diagrams act as visual aids and leave a lasting impression on the mind. This enables the investigator to make quick conclusions about the distribution. There are different types of graphs or diagrams to represent statistical data. Some of them are: 1.\t Bar chart or bar graph (for unclassified frequency distribution) 2.\t Histogram (for classified frequency distribution) 3.\t Frequency polygon (for classified frequency distribution) 4.\t Frequency curve (for classified frequency distribution) 5.\t Cumulative frequency curves (for classified frequency distribution) \u2009\u2009(i)\t Less than cumulative frequency curve (ii)\t Greater than cumulative frequency curve Bar Graph The important features of bar graphs are: 1.\t Bar graphs are used to represent unclassified frequency distributions. 2.\t F\u0007requency of a value of a variable is represented by a bar (rectangle), whose length (i.e., height) is equal (or proportional) to the frequency. 3.\t T\u0007 he breadth of the bar is arbitrary and the breadth of all the bars are equal. The bars may or may not touch each other. Example 9.2 Represent the following frequency distribution by bar graph: Value of Variable 3 \u20026 9 12 15 Frequency 6 10 4 \u20023 \u20028 Solution Either of the following bar graphs [Fig. 9.1(a) or Fig. 9.2(b)] may be used to represent the above frequency distribution. The first graph takes value of the variable along the X-axis and the frequency along the Y-axis, whereas the second one takes the frequency along the X-axis and the value of the variable on the Y-axis.","9.6 Chapter 9 All the rectangles (bars) should be of same width and uniform spaces should be left between any two consecutive bars. 12 10 15 8 12 69 46 3 2 Frequency Value of variable 0 36 9 12 15 0 2 4 6 8 10 Value of Variable Frequency (a) (b) Figure 9.1 Histograms Classified or grouped data is represented graphically by histograms. A histogram consists of rectangles each of which has its breadth proportional to the size of concerned class interval and its height proportional to the corresponding frequency. In a histogram, two consecutive rectangles have a common side. Hence, in a histogram, we do the following: 1.\t We represent class boundaries along the X-axis. 2.\t Along the Y-axis, we represent class frequencies. 3.\t We construct rectangles with bases along the X-axis and heights along the Y-axis. Example 9.3 Construct a histogram for the frequency distribution below: Class Interval 10\u201320 20\u201330 30\u201340 40\u201350 50\u201360 Frequency 6 3 8 4 7 Solution Frequency 10 Scale: on X-axis,1 cm = 10 units 9 on Y-axis, 1 cm = 1 units Here, the class intervals are continuous. The 8 following histogram is drawn according to the method described in Fig 9.2. 7 6 5 4 3 2 1 0 10 20 30 40 50 60 Clas s es Figure 9.2","Statistics 9.7 Remarks The following points may be noted: 1.\t \u0007A link mark ( ) made on the horizontal axis, between the vertical axis and first vertical rectangle, if there is a gap between 0 and the lower boundary of first class interval. 2.\t We may shade all rectangles. A heading for the histogram may also be given. Important Observations 1.\t \u0007If the class intervals are discontinuous, the distribution has to be changed into continuous intervals and then the histogram has to be drawn. 2.\t B\u0007 ar graphs are used for unclassified frequency distributions, whereas histograms are used for classified frequency distribution. The breadths of rectangles in a bar graph are arbitrary, while those in histogram are determined by class size. Frequency Polygon Frequency polygons are used to represent classified or grouped data graphically. It is a polygon whose vertices are the mid-points of the top sides of the rectangles, forming the histogram of the frequency distribution. To draw a frequency polygon for a given frequency distribution, the mid values of the class intervals are taken on X-axis and the corresponding frequencies on Y-axis and the points are plotted on a graph sheet. These points are joined by straight line segments which form the frequency polygon. Example 9.4 Construct a frequency polygon for the following data: Class Interval 5\u201310 11\u201316 17\u201322 23\u201328 29\u201334 Total Frequency 7 13 8 12 10 50 Solution Here, the class intervals are discontinuous. Hence, first we convert the class intervals to continuous class intervals and then find mid-points of each class interval. We do this by adding 0.5 to each upper limit and subtracting 0.5 from each lower limit. Class Interval Exclusive Mid-value of Class Frequency \u20025\u201310 \u20024.5\u201310.5 \u20027.5 \u20027 11\u201316 10.5\u201316.5 13.5 13 17\u201322 16.5\u201322.5 19.5 \u20028 23\u201328 22.5\u201328.5 25.5 12 29\u201334 28.5\u201334.5 31.5 10 Now, taking the mid-values of class intervals on the X-axis and the corresponding frequencies on the Y-axis, we draw a frequency polygon as shown in the Fig. 9.3.","9.8 Chapter 9 Frequency Scale: on X-axis, 1cm = 5 units on Y-axis, 1 cm = 2 units 16 14 12 10 8 6 4 2 0 5 10 15 20 25 30 35 40 45 50 Mid-value of the classes Figure 9.3 Frequency Curve Frequency curves are used to represent classified or grouped data graphically. As the class-interval in a frequency distribution decreases, the points of the frequency polygon become closer and closer and then the frequency polygon tends to become a frequency curve. So, when the number of scores in the data is sufficiently large and the class intervals become smaller (ultimately tending to zero), the limiting form of frequency polygon becomes frequency curve. Example 9.5 Draw a frequency curve for the data given below: Class Interval 0\u201310 10\u201320 20\u201330 30\u201340 40\u201350 50\u201360 60\u201370 Frequency 3 2 4 6 9 7 10 Solution Following table shows the mid-values of classes and corresponding frequencies for the given data: Class Interval Mid-value Frequency \u20020\u201310 \u20025 3 10\u201320 15 2 20\u201330 25 4 30\u201340 35 6 40\u201350 45 9 50\u201360 55 7 60\u201370 65 10 Now, taking the mid-values of the classes along the X-axis and the corresponding frequencies along the Y-axis, we mark the points obtained from the above table in a graph sheet and join them with a smooth curve, which gives the frequency curve as shown in the Fig. 9.4.","Statistics 9.9 Frequency 16 Scale: on X-axis, 1 cm = 10 units 14 on Y-axis, 1 cm = 2 units 12 10 10 20 30 40 50 60 70 8 Mid-values of the classes 6 Figure 9.4 4 2 0 0 Cumulative Frequency Curves The curves drawn for cumulative frequencies, less than or more than the true limits of the classes of a frequency distribution are called cumulative frequency curves. The curve drawn for the \u2018less than cumulative frequency distribution\u2019 is called the \u2018less than cumulative frequency curve\u2019 and the curve drawn for the \u2018greater than cumulative frequency distribution\u2019 is called the \u2018greater than cumulative frequency curve\u2019. From these curves, we can find the total frequency above or below a particular value of the variable. Example 9.6 For the given distribution, draw the less than and greater than cumulative frequency curves. Class Interval 0\u201310 10\u201320 20\u201330 30\u201340 40\u201350 50\u201360 60\u201370 70\u201380 Frequency 5 3 7 2 14 9 15 10 Solution Less than cumulative frequency distribution: Upper Class Frequency Less than Cumulative Boundary \u20025 Frequency 10 \u20023 \u20025 20 \u20027 \u20028 30 \u20022 15 40 14 17 50 \u20029 31 60 15 40 70 10 55 80 65","9.10 Chapter 9 Greater than cumulative frequency distribution: Lower Class Frequency Greater than Boundary 5 Cumulative Frequency \u20020 3 10 7 65 20 2 60 30 57 40 14 50 50 9 48 60 15 34 70 10 25 10 Cumulative frequency 100 Scale: 1 cm = 10 units 90 80 10 20 30 40 50 60 70 80 90 100 70 Class boundaries 60 Figure 9.5 50 40 30 20 10 0 0 MEASURES OF CENTRAL TENDENCIES FOR UNGROUPED DATA Till now, we have seen that the data collected in statistical enquiry or investigation is in the form of raw data. If the data is very large, the user cannot get much information from such data. For this reason, the data is grouped together to obtain some conclusions. The measure of central tendency is a value which represents the total data, i.e., it is the value in a data around which the values of all the other observations tend to concentrate. The most commonly used measures of central tendencies are: 1.\t Arithmetic mean 2.\t Median 3.\t Mode These measures give an idea about how the data is clustered or concentrated. Arithmetic Mean or Mean (AM) The arithmetic mean (or simply the mean) is the most commonly used measure of central tendency.","Statistics 9.11 Arithmetic Mean for Raw Data Definition\u2002 The arithmetic mean of a statistical data is defined as the quotient obtained when the sum of all the observations or entries is divided by the total number of items. If x1, x2, \u2026, xn are the n items, then n \u2211+ xn xi AM = x1 + x2 + = or briefly \u2211 x . n i =1 n n AM is usually denoted by x . Example 9.7 Find the mean of the first seven natural even numbers. Solution Given data is 2, 4, 6, 8, 10, 12, 14. \u2234 Arithmetic mean ( AM ) = Sum of observations Total number of observations = 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56 = 8. 7 7 Mean of Discrete Series Let x1, x2, x3 \u2026, xn be n observations with respective frequencies f1, f2, \u2026, fn. This can be considered as a special case of raw data where the observation x1 occurs f1 times, x2 occurs f2 times, and so on. \u2234 The mean of the above data = f1x1 + f2x2 + + fnxn . f1 + f2 + + fn n \u2211 fixi It can also be represented by i =1 x = n . \u2211 fi i =1 Weighted Arithmetic Mean When the variables x1, x2, \u2026, xn do not have same importance, and the weights w1, w2, \u2026, wn are given to each of the variables, the weighted arithmetic is given by xw = \u2211 xiwi . \u2211 wi","9.12 Chapter 9 Example 9.8 The population of 50 villages in a state is given below: Population Number of Villages \u20026000 \u20028 \u20027000 10 \u20029000 12 10000 \u20025 11000 \u20027 13000 \u20026 15000 \u20022 Total 50 Find the mean population of the villages of the state. Solution The mean x is given by x = (6000 \u00d7 8) + (7000 \u00d7 10) + (9000 \u00d7 12) + (10000 \u00d7 5) + (11000 \u00d7 7) + (13000 \u00d7 6) + (15000 \u00d7 2) 8 + 10 + 12 + 5 + 7 + 6 + 2 = 461000 50 x = 9220. That is, the mean population of the villages is 9220. Some Important Results about AM n 1.\t The algebraic sum of deviations taken about the mean is zero, i.e., \u2211 (xi \u2212 x ) = 0. 2.\t The value of the mean depends on all the observations. i =1 3.\t The AM of two numbers a and b is a + b . 2 4.\t C\u0007 ombined mean: If x1 and x2 are the arithmetic means of two series with n1 and n2 observations respectively, then the combined mean is: xc = n1x1 + n2x2 n1 + n2 \t The above result can be extended to any number of groups of data. 5.\t \u0007If x is the mean of x1, x2,\u2009\u2026, xn, then the mean of x1 + a, x2 + a, x3 + a, \u2026, xn + a is x + a, for all values of a. 6.\t \u0007If x is the mean of x1, x2,\u2009\u2026, xn, then the mean of ax1, ax2, \u2026, axn is a x \u2009\u2009and that of x1 , x2 , \u2026, xn is x . a a a a 7.\t The mean of the first n natural numbers is \uf8eb x + 1\uf8f6 . \uf8ed\uf8ec 2 \uf8f8\uf8f7","Statistics 9.13 8.\t The mean of the squares of the first n natural numbers = (n + 1)(2n + 1) . 6 9.\t The mean of the cubes of the first n natural numbers = n(n + 1)2 . 4 Median Another measure of the central tendency of a given data is the median. Definition If the values xi in the raw data are arranged either in the increasing or decreasing order of their magnitude, then the middle-most value in this arrangement is called the median. Thus, for the raw (ungrouped) data, the median is computed as follows: 1.\t The values of the observations are arranged in the order of magnitude. 2.\t \u0007The middle-most value is taken as the median. Hence, depending on the number of observations (odd or even), we determine median as follows. (i)\t When the number of observations (n) is odd, then the median is the value of \uf8eb n + 1\uf8f6 th \uf8ec\uf8ed 2 \uf8f7\uf8f8 observation. n (ii)\t If the number of observations (n) is even, then the median is the mean of \uf8eb 2 \uf8f6 th and \uf8ed\uf8ec \uf8f8\uf8f7 n \uf8eb 2 + 1\uf8f7\uf8f6\uf8f8 th observations. \uf8ed\uf8ec Example 9.9 Find the median of the following data: 1, 12, 5, 3, 7, 13, 9, 23, 17, 11 and 6 Solution Arranging the given numbers in the ascending order, we have 1, 3, 5, 6, 7, 9, 11, 12, 13, 17,\u00a023. Here, the middle term is 9 \u2234 Median = 9. Example 9.10 Find the median of the data 11, 5, 3, 13, 16, 9, 18, 10. Solution Arranging the given data in the ascending order, we have 3, 5, 9, 10, 11, 13, 16, 18. As the given number of values is even, we have two middle values, they are \uf8eb n2 \uf8f6\uf8f8\uf8f7 th and \uf8ed\uf8ec \uf8eb n 1\uf8f7\uf8f6\uf8f8 th \uf8ed\uf8ec 2 + observations. Here, they are 10, 11. \u2234 Median of the data = Average of 10 and 11 = 10 + 11 = 10.5. 2","9.14 Chapter 9 Some Important Facts about Median 1.\t The median does not take all the items into consideration. 2.\t The sum of absolute deviations taken about the median is the least. 3.\t The median can be calculated graphically while the mean cannot be. 4.\t The median is not effected by extreme values. 5.\t \u0007The sum of deviations taken about median is less than the sum of absolute deviations taken from any other observation in the data. Mode The third measure of central tendency of a data is the mode. Definition The most frequently found value in the data is called the mode. This is the measure which can be identified in the simplest way. Example 9.11 Find the mode of 1, 0, 2, 1, 0, 2, 3, 1, 3, 1, 0, 4, 2, 4, 1 and 2. Solution Among the observations given, the most frequently found observation is 1. It occurs 5 times. \u2234 Mode = 1. Some Important Facts about Mode 1.\t \u0007For a given data, the mode may or may not exist. In a series of observations, if no item occurs more than once, then the mode is said to be ill-defined. 2.\t If the mode exists for a given data, it may or may not be unique. 3.\t Data having unique mode is uni-modal while data having two modes is bi-modal. Properties of Mode 1.\t It can be calculated graphically. 2.\t It is not effected by extreme values. 3.\t It can be used for open-ended distribution and qualitative data. Empirical Relationship among Mean, Median and Mode For a moderately symmetric data, the above three measures of central tendency can be related by the formula, Mode = 3 Median \u2013 2 Mean. Example 9.12 Find the mode when median is 8 and mean is 10 of a data. Solution Mode = 3 Median \u2013 2 Mean = (3 \u00d7 8) \u2212 (2 \u00d7 10) = 24 \u2212 20 = 4.","Statistics 9.15 Observations 1.\t For a symmetric distribution, Mean = Median = Mode. 2.\t Given any two of the mean, median and mode the third can be calculated. 3.\t This formula is to be applied in the absence of sufficient data. MEASURE OF CENTRAL TENDENCIES FOR GROUPED DATA We studied the measure of central tendencies of ungrouped or raw data. Now we study the measures of central tendencies (mean, median and mode) for grouped data. Mean of Grouped Data If the frequency distribution of \u2018n\u2019 observations of a variable x has k classes, xi is the mid-value and fi is the frequency of ith class, then the mean\u2009x \u2009of grouped data is defined as k fixi \u2211+ fkxk fi x = f1x1 + f2x2 + =\u2211+ fk i =1 f1 + f2 + k (or) simply, i =1 fixi k N where N = fi . i =1 \u2211 \u2211x = In grouped data, it is assumed that the frequency of each class is concentrated at its mid-value. Example 9.13 Calculate the arithmetic mean (AM) of the following data: Salaries of Employees 0\u201310 10\u201320 20\u201330 30\u201340 40\u201350 (in thousand rupees) 4 13 8 9 6 Number of Employees Solution Let us write the tabular form as given below: Salaries of Number of Mid-points of fixi Employees Employees (fi\u2009) Classes (xi) \u200220 \u20025 195 0\u201310 4 200 10\u201320 15 315 20\u201330 13 270 30\u201340 25 40\u201350 8 35 9 45 6 \u2211\u2234 Mean = x = fixi = 1000 = 25. N 40","9.16 Chapter 9 Short-cut Method for Finding the Mean of Grouped Data (Deviation Method) Sometimes, when the frequencies are large in number, the calculation of mean using above formula is cumbersome. This can be simplified if the class interval of each class of grouped data is the same. Under the assumption of equal class intervals, we get the following formula for the mean of grouped data: A+ 1 \uf8ebk fiui \uf8f6 c. \u2211x = N \uf8ec \uf8f7 \u00d7 \uf8ed i=1 \uf8f8 Where,\tA\t= \tassumed mean value from among mid-values \tc\t=\tlength of class interval \tk\t=\tnumber of classes of the frequency distribution k \tN\t=\tSum of frequencies = \u2211 fi i =1 \t ui = xi \u2212 A, i = 1, 2, 3, \u2026, k C \txi\t=\tmid-value of the ith class \t\t\t\u0007ui is called the deviation or difference of the mid-value of the ith class from the assumed value, divided by the class interval. Using this method the previous example can be worked out as follows: Short-cut Method for the Above Example Salaries of Number of Mid-values Deviation Employees Employees (fi) (xi) xi \u2013 A ui = C fiui 0\u201310 \u20024 \u20025 \u22123 \u221212 10\u201320 13 15 \u221226 20\u201330 \u20028 25 \u22122 \u22128 30\u201340 \u20029 35(A) \u20020 40\u201350 \u20026 45 \u22121 \u20026 N = 40 \u03a3fiui = \u221240 \u20020 \u20021 Here, A = 35, N = 40, C = 10, \u03a3fiui = \u221240 \u2234 AM = A + 1 (\u03a3fiui )\u00d7C N \u22c5 = 35 + 1 (\u221240) \u00d710 = 25. 40 Median of Grouped Data Before finding out how to obtain the median of grouped data, we first review what a median class is.","Statistics 9.17 If n is the number of observations, then from the cumulative frequency distribution, the class \uf8ebn\uf8f6 in which \uf8ec\uf8ed 2\uf8f7\uf8f8 th observation lies is called the median class. Formula for calculating median: n \u2212 F 2 f Median (M ) = L + (c) Where,\t L\t=\tLower boundary of the median class, i.e., class in which \uf8eb n\uf8f6 th observation lies \tN\t=\tSum of frequencies \uf8ec\uf8ed 2\uf8f7\uf8f8 \tF\t=\tcumulative frequency of the class just preceding the median class \tf\t=\tfrequency of the median class \tC\t=\tlength of class interval Example 9.14 Given below is the data showing heights of 50 students in a class. Find its median. Height (in cm) 162 164 166 167 168 170 173 175 177 180 1 Number of Students 6 4 5 12 8 3 7 2 2 Solution To find median, we prepare less than cumulative frequency table as given below: Height (in cm) Number of Students Cumulative Frequency (f\u2009) 162 6 \u20026 164 4 10 166 5 15 167 27 168 12 35 170 8 38 173 3 45 175 7 47 177 2 49 180 2 50 1 Here N = 50, which is even. \u2234 Median = N value 50 or 25th observation. 2 2 From the column of cumulative frequency, the value of 25th observation is 167. \u2234 Median = 167 cm. \u2009\u2009Note\u2002 \u2002 In the above example, we do not have any class interval. As there is no class interval, we cannot use the formula.","9.18 Chapter 9 Example 9.15 Find the median of the following data: Class Interval 0\u201310 10\u201320 20\u201330 30\u201340 40\u201350 f 6 8 5 4 7 Solution To find the median, we prepare the following table: Class Interval Frequency Cumulative Frequency 0\u201310 6 6 8 10\u201320 14(F ) 20\u201330 5(\u2009\u2009f\u2009\u2009) 19 30\u201340 4 23 40\u201350 7 30 Total N = 30 Here N = 30 \u21d2 N = 15. 2 This value appears in the class 20\u201330. L = Lower boundary of the median class 20\u201330 = 20 F = 14, f = 5, C = 10 (class length) \uf8ebN \u2212 F \uf8f6 \uf8ed\uf8ec 2 \uf8f8\uf8f7 Median = L + \u00d7C f \u2234 Median = 20 + (15 \u2212 14) \u00d7 10 = 22. 5 Mode of Grouped Data \u22061C \u22061 + \u22062 The formula for determining the mode of grouped data is L1 + . Where, L1 = lower boundary of the modal class (class with highest frequency) \t \u22061 = f \u2212 f1 and \u22062 = f \u2212 f2 where f is the frequency of modal class \tf1\t=\tfrequency of previous class of the modal class \tf2\t=\tfrequency of next class of the modal class Rewriting the formula, Mode = L1 +(f ( f \u2212 f1 )C f2 ) \u2212 f1) + ( f \u2212 Mode = L1 + 2 ( f \u2212 f1 )C ). f \u2212 ( f1 + f2","Statistics 9.19 Example 9.16 The following information gives the marks scored by of students of a class in an examination. Find the mode of the data. Marks 1\u201320 21\u201340 41\u201360 61\u201380 81\u2013100 Number of Students 3 18 23 37 19 Solution Here, the given classes are not continuous. Hence, we first rewrite it as shown below: Marks Adjusted Marks Number of Students 1\u201320 0.5\u201320.5 3 21\u201340 20.5\u201340.5 18 41\u201360 40.5\u201360.5 23(\u2009\u2009f1) 61\u201380 60.5\u201380.5 37(\u2009\u2009f) 81\u2013100 80.5\u2013100.5 19(\u2009\u2009f2) From the above table it can be observed that the maximum frequency occurs in the class interval 61\u201380. \u2234 f = 37, f1 = 23, f2 = 19, L1 = 60.5, C = 20. \u2234 Mode = L1 + ( f ( f \u2212 f1 )C f2 ) \u2212 f1) + ( f \u2212 = 60.5 + (37 \u2212 23)20 = 60.5 + 87.5 = 69.25. 14 + 18 RANGE The difference between the maximum and the minimum values of the given observations is called the range of the data. Given x1, x2, ..., xn (n individual observations) Range = (Maximum value) \u2013 (Minimum Value). Example 9.17 Find the range of 1, 3, 8, 6, 2, 11, 10, 15 and 13. Solution Arranging the given data in the ascending order, We have 1, 2, 3, 6, 8, 10, 11, 13, 15. \u2234 Range = (Maximum value) \u2212 (Minimum value) = 15 \u2013 1 = 14. \u2009\u2009Note\u2002 \u2002 The range of the class interval is the difference of the actual limits of the class.","9.20 Chapter 9 QUARTILES In a given data, the observations that divide the given set of observations into four equal parts are called quartiles. First Quartile or Lower Quartile When the given observations are arranged in ascending order, the observation which lies midway between the lower extreme and the median is called the first quartile or the lower quartile and is denoted as Q1. Third Quartile or Upper Quartile Of the data when the given observations are arranged in ascending order, the observation that lies in midway between the median and the upper extreme observation is called the third quartile or the upper quartile and is denoted by Q3. We can find Q1 and Q3 for an ungrouped data containing n observations as follows. Rearrange the given n observations or items in the ascending order then, lower or first quartile, Q1 is \uf8eb n\uf8f6 th item or observation if n is even and \uf8ebn + 1\uf8f6 th item or observation, when n is odd. \uf8ed\uf8ec 4 \uf8f8\uf8f7 \uf8ed\uf8ec 4 \uf8f8\uf8f7 Example 9.18 Find Q1 for the data 23, 7, 11, 9, 15, 12, 20 and 18. Solution Arranging the given observations in ascending order, we have 7, 9, 11, 12, 15, 18, 20, 23. Here n = 8 (n is even) \u2234 first quartile, Q1 = \uf8eb n \uf8f7\uf8f8\uf8f6th item = \uf8eb 8 \uf8f8\uf8f6\uf8f7th item \uf8ed\uf8ec 4 \uf8ec\uf8ed 4 = second observation of the data, i.e., 9 \u2234Q1 = 9. Example 9.19 Find Q1 for the observations 13, 8, 11, 15, 19, 4 and 10. Solution Arranging the observations in ascending order, we have 4, 8, 10, 11, 13, 15, 19. Here n = 7 (odd) \u2234Q1 = \uf8ebn + 1 \uf8f6 th item, i.e., \uf8eb 7 + 1 \uf8f6 th item = 2nd observation. \uf8ec\uf8ed 4 \uf8f7\uf8f8 \uf8ec\uf8ed 4 \uf8f8\uf8f7 \u2234Q1 = 8.","Statistics 9.21 Example 9.20 The ages of 10 employees in an organization are 26, 23, 27, 33, 39, 43, 41, 36, 42, 25. Find Q1. Solution The given observations when arranged in ascending order, we get 23, 25, 26, 27, 33, 36, 39, 41, 42, 43. Here n = 10 (even) \u2234Q1 = \uf8eb n \uf8f8\uf8f6\uf8f7th observation \uf8ed\uf8ec 4 = \uf8eb 2 1 \uf8f6\uf8f7\uf8f8th observation of the data \uf8ed\uf8ec 2 \u2234Q1 = 2nd observation + 1 (3rd \u2212 2nd ) observation 2 = 25 + 1 (26 \u2212 25) = 25.5 2 \u2234Q1 = 25.5. Third Quartile Q3 = \uf8eb 3n \uf8f6\uf8f8\uf8f7th item, when n is even. \uf8ec\uf8ed 4 = 3 \uf8eb n + 1 \uf8f7\uf8f8\uf8f6th item, when n is odd. \uf8ed\uf8ec 4 Example 9.21 Find Q3 for the data 8, 13, 18, 9, 20, 11. Solution Arranging the data in ascending order, we have 8, 9, 11, 13, 18, 20. Here n = 6 (even) \u2234Q3 = 3 \uf8eb n \uf8f6 th observation = 4 1 th observation \uf8ed\uf8ec 4 \uf8f8\uf8f7 2 \u21d2 Q3 = 4th observation + 1 (5th observation \u2212 4th observation) 2 = 13 + 1 (18 \u2212 13) \u2234Q3 = 15.5. 2 Semi-inter Quartile Range or Quartile Deviation (QD) Quartile deviation, QD = Q3 \u2212 Q1 . 2","9.22 Chapter 9 Example 9.22 13 Find semi-inter quartile range of the following data: 15 X 3 6 \u20027 \u20029 10 11 f 2 9 13 17 10 14 Solution X Frequency (f\u2009) Cumulative Frequency \u20023 \u20022 \u20022 \u20026 \u20029 11 \u20027 13 24 \u20029 17 41 10 10 51 11 14 65 13 15 80 N = 80 Here, N = 80 \u2234Q1 = \uf8eb N \uf8f6\uf8f7\uf8f8th observation = 20th observation \uf8ec\uf8ed 4 \u2234Q1 = 7 (as 20th items lies in the class having 24 as cumulative frequency) \u2234Q3 = 3 \uf8eb N \uf8f8\uf8f6\uf8f7th observation = 60th observation \uf8ec\uf8ed 4 \u2234Q3 \u2261 11 (as 60th item lies in the class having 65 as cumulative frequency) Semi-inter quartile range, (QD) = Q3 \u2212 Q1 = 11 \u2212 7 = 2. 2 2 \u2009\u2009Note\u2002 \u2002 For an individual data, the second quartile Q2 coincides with median. \u21d2 Q2 = Median of the data. Estimation of Median and Quartiles from Ogive 1.\t Prepare the cumulative frequency table with the given data. 2.\t Draw an ogive graph. 3.\t Let, the total number of observations = sum of all frequencies = N. 4.\t Mark the points A, B and C on Y-axis, corresponding to N , N and 3N respectively. 4 2 4 5.\t Mark three points (P, Q and R) on ogive corresponding to 3N , N and N respectively. 4 2 4 6.\t Draw vertical lines from the points R, Q and P to meet X-axis Q1, M and Q3 respectively. 7.\t T\u0007 hen, the abscissas of Q1, M and Q3 gives lower quartile, median and upper quartile respectively.","Statistics 9.23 Example 9.23 The following table shows the distribution of the percentage of marks of a group of students: Percentage of Marks 30\u201340 40\u201350 50\u201360 60\u201370 70\u201380 80\u201390 Number of Students 4 11 3 7 5 2 Solution Class Interval Number of Students (f\u2009) Cumulative Frequency 30\u201340 \u20024 \u20024 40\u201350 11 15 50\u201360 \u20023 18 60\u201370 \u20027 25 70\u201380 \u20025 30 80\u201390 \u20022 32 N = 32 N== 32, N2 1=6 and 34N 24. From the graph, Y Scale: on X-axis, 1 cm = 10 units Ogive on Y-axis, 1 cm = 4 units Lower quartile (Q1) = 44 Cumulative frequency 32 P Upper quartile 28 Q (Q3) = 69 24 C R Median (M) = 20 53 16 B 12 8A 4 0 Q1 M Q3 X 30 40 50 60 70 80 90 Upper boundaries \t Figure 9.6 Estimation of Mode from Histogram 1.\t Draw a histogram to represent the given data. 2.\t F\u0007rom the upper corners of the highest rectangle, draw the line segments to meet the opposite corners of adjacent rectangles, diagonally as shown in the given example. Mark the intersecting point as P. 3.\t Draw PM perpendicular to X-axis, to meet X-axis at M. 4.\t Abscissa of M gives the mode of the data.","9.24 Chapter 9 Example 9.24 Estimate the mode of the following data from the histogram. Class Interval 0\u201310 10\u201320 20\u201330 30\u201340 40\u201350 50\u201360 60\u201370 70\u201380 Frequency 8 10 15 16 20 13 14 12 From the graph, Mode (M) = 44. Y Scale: on X-axis, 1 cm = 10 units 20 AB on Y-axis, 1 cm = 2 units 18 p 16 D Frequency 14 C 12 10 8 6 4 2 0 M X 20 30 40 50 80 90 10 60 70 CI \t Figure 9.7 MEAN DEVIATION (MD) Mean deviation is defined as the average or mean of the deviations taken about mean or median or mode. Hence, it is also called an average deviation. Mean deviation gives the best results when deviations are taken about median. We should take only the positive values of deviations. Mean Deviation (MD) for Ungrouped or Raw Data MD = \u2211|xi \u2212 x| n Where, xi is each observation x is arithmetic mean, or median or mode as specified in the problem n is the number of observations. Example 9.25 Find the mean deviation of the following from median 13, 18, 15, 10, 17, 19 and 21. Solution Writing the given observations in ascending order, we get 10, 13, 15, 17, 18, 19, 21. Given number of observations are 7. \u2234 n = 7 (odd)","Statistics 9.25 \u2234 Median, x = 17 \u2234 Mean Deviation |10 \u2212 17|+ 13 \u2212 17 + 15 \u2212 17 + 17 \u2212 17 + 18 \u2212 17 + 19 \u2212 17 + 21 \u2212 17 =7 7 + 4 + 2 +0 + 1+ 2 + 4 20 = 7 = 7 = 2.857. Mean Deviation for Discrete Data Mean deviation (MD) = \u2211|xi \u2212 x | or \u2211 fD . N N Where, xi is each of the given observations fi is their corresponding frequencies. N = \u03a3fi (sum of the frequencies). D = |xi \u2212 x|. Example 9.26 Find the mean deviation about mean for the following data: x 1 3 7 9 13 14 f 2 5 8 6 \u20023 \u20021 Solution fixi D = |xi - x| fiD \u20022 6.2 12.4 xi fi 15 4.2 21.0 \u20021 2 56 0.2 \u20021.6 \u20023 5 54 1.8 10.8 \u20027 8 39 5.8 17.4 \u20029 6 14 6.8 6.8 13 3 14 1 \u03a3fx = 180 \u03a3fD = 70 N = \u03a3f = 25 x = AM = \u2211 fx \u21d2 x = 180 = 7.2 \u2211 f 25 \u2234 MD = \u2211 fD = 70 = 2.8. 25 N Mean Deviation for Grouped Data \u2211MD = fi |xi \u2212 x | or \u2211 fD N N","9.26 Chapter 9 Where, xi is mid-value of each class fi is the corresponding frequency x is mean or median or mode N is sum of frequencies D = |xi \u2212 x |. \u2002Note\u2002\u2002 The formula for discrete data and grouped data is same except for one change, i.e., xi is to be replaced by mid values of the classes for grouped data. Example 9.27 Find mean deviation for the following data about median: Class Interval 5\u20139 10\u201314 15\u201319 20\u201324 25\u201329 f 5 9 12 8 6 Solution Cumulative Class f x Frequency D = |x - x| fD Interval 5\u20139 \u20025 \u20027 5 10 50 10\u201314 \u20029 12 14F \u20025 45 15\u201319 20\u201324 12f 17 26 \u20020 \u20020 25\u201329 \u20028 22 34 \u20025 40 \t \u20026 27 40 10 60 \t \u03a3fD = 195 N=2 4=20 =20,L 1=4.5, F 1=4, f 1=2,C 5 N \u2212 F 2 f Median = L + \u00d7C = 14.5 + 20 \u2212 14 \u00d7 5 12 = 14.5 + 2.5 = 17 \u2234 MD = \u2211 f |x \u2212 x| or \u2211 fD N N = 195 = 4.875. 40 Some Important Results Based on MD 1.\t MD depends on all observations. 2.\t By default, MD is to be computed about mean. 3.\t MD about median is the least. |a \u2212 b|. 4.\t MD of two numbers a and b is 2","Statistics 9.27 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t If 1\u201310, 11\u201320, 21\u201330, 31\u201340, \u2026, are the classes \t17.\t If lower and upper quartiles of the data are 23 and of a frequency distribution, then the upper bound- 25 respectively, then coefficient of quartile devia- ary of the class 31\u201340 is ______. tion is ______. \t2.\t If 1\u20135, 6\u201310, 11\u201315, \u2026, are classes of a frequency \t18.\t Mode of certain scores is x. If each score is distribution, then the mid-value of the class 11\u201315 decreased by 3, then the mode of the new series is is ______. ______. \t3.\t In a class, seven students got 90 marks each, the 1\t 9.\t If the mean of 2, 4, 6, 8, x and y is 5, then x + y frequency of the observation 90 is ______. is ______. \t4.\t If the lower boundary of a class is 35 and length of \t20.\t Find the mean of the following: the class is 5, then the upper boundary is ______. 1 3 2 1 \t5.\t Range of the scores 27, 35, 47, 36, 25 and x is 23, 10 4 , 9, 4 4 , 8, 2 3 , 12 and 2 3 where x < 25, then x is _____. 2\t 1.\t Find the median of the following data: \t6.\t In a histogram, width of the rectangle represents 2.0025, 2.0205, \u22122.06, \u22122.206 and \u22122.006 _____ of the class and length of the rectangle rep- resents _____ of the class. 2\t 2.\t Find the median of the following data: \t7.\t If the median of the scores 1, 2, x, 4 and 5 (where 23, 92, 43, 34, 54, 48, 82, 14, 62 and 46 1 < 2 < x < 4 < 5) is 3, then the mean of the scores is _____. \t23.\t If the strengths of 10 classes in a school are given as 28, 42, 25, 30, 45, 22, 25, 34, 26 and 36, then find \t8.\t In a less than cumulative frequency distribution, the median strength. frequency and cumulative frequency of a class are 10 and 20 respectively, then the cumulative fre- 2\t 4.\t The mean of 8 observations was found to be 20. PRACTICE QUESTIONS quency of the previous class is _____. Later it was detected that one of the observations was misread as 62. What is the correct observation, \t9.\t Mode of the scores 2, 3, 2, 4, 3, 2, 4, 6 is _____. if the correct mean is 15.5? \t10.\t If median of the sx6coirses__x2_,__x3., x , x and x (where \t25.\t Find the coefficient of range of the scores 25, 30, x > 0) is 6 then 4 5 6 22, 34, 50, 56 and 67. \t11.\t The coefficient of range of the scores 1, 2, 3, 4 and 2\t 6.\t If the median of 33, 28, 20, 25, 34 and x is 29, then find the maximum possible value of x. 5 is ______. 2\t 7.\t If the mode of the scores 3, 4, 3, 5, 4, 6, 6 and x is 4, then find the value of x. \t12.\t If sum of the 20 deviations from the mean is 100, then the mean deviation is ______. 2\t 8.\t If the heights of five students in a class are 132 cm, 158 cm, 150 cm, 145 cm and 155 cm, then find \t13.\t If lower and upper quartiles of the data are x and y their mean height. respectively, then quartile deviation is ______. 1\t 4.\t If mode and median of the certain scores are 2 and \t29.\t Find the median of the following data: 2, then mean is ______. 3 3 , 3 5 , 3 1 , 3 1 , 3 7 and 3 7 1\t 5.\t If the mean of the scores x1, x2, x3, x4, x5 and x6 is 7 8 2 4 9 11 x, then mean of 5x1, 5x2, 5x3, 5x4, 5x5 and 5x6 is ______. 3\t 0.\t Following are the runs scored by 11 members of a cricket team in a test innings. Calculate the quar- 1\t 6.\t If the mean of a, b and c is b, then a + c in terms of tile deviation of the data. b is ______. 20, 22, 30, 32, 39, 41, 42, 60, 62, 65 and 80","9.28 Chapter 9 Short Answer Type Questions \t31.\t If the mean weight of 10 students is 25 kg and the xf xf mean weight of another 10 students is 35 kg, then \u200250 6 \u200280 3 the mean weight of 20 students is ______. \u200260 5 \u200290 3 \u200270 4 100 1 3\t 2.\t Given below are the number of students in 30 class rooms in a school. Construct a frequency distribu- 4\t 0.\t Find the quartile deviation of the following data: tion table for this data with a class interval of 4. 25 30 24 18 20 24 32 35 22 20 xf xf 22 32 40 28 30 25 26 29 34 15 \u20022 \u20024 13 \u20022 38 28 19 16 15 20 24 30 26 18 \u20023 \u20026 17 \u20024 \u20025 \u20028 19 \u20026 3\t 3.\t The average marks of 50 students of a class is 76. \u20027 \u20029 23 \u20026 If the average marks of all boys is 70 and that of 11 10 all girls is 80 in that class, then find the number of boys in the class. \t41.\t Calculate the mean deviation for the following data about median. \t34.\t The mean of 30 observations is 25. If two observa- tions 30 and 60 are misread as 20 and 40, then find 15, 20, 16, 13, 10, 11, 18 the correct mean. \t42.\t Draw a histogram for the following data: 3\t 5.\t The mean expenditure of a person from Monday to Wednesday is `250, and the mean expenditure Class Interval f from Wednesday to Friday is `400. If he spend 0\u20135 2 `300 on Wednesday, find the mean expenditure of 5\u201310 10 the person from Monday to Friday. 8 10\u201315 6 \t36.\t The sum of deviations of n observations about 25 15\u201320 4 is 25 and sum of deviations of the same n obser- 20\u201325 vations about 35 is \u201325. Find the mean of the PRACTICE QUESTIONS observations. \t43.\t Draw the frequency polygon for the following data by drawing a histogram: 3\t 7.\t If the sum of mode and mean of certain observa- tions is 129 and the median of the observations is Class Interval f 63, then find mode and mean. 0\u20135 10 5\u201310 20 \t38.\t The height (in cms) of 20 children of class 9 is 15 given, find the mean. 10\u201315 30 15\u201320 32 Height (in cm) Number of Children 20\u201325 44 120 2 25\u201330 121 4 122 3 4\t 4.\t Draw the less than cumulative frequency curve for 123 2 the following data: 124 5 125 4 Class Interval f 10\u201314 10 \t39.\t Find the median of the following distribution: 15\u201319 \u20028 20\u201324 15 xf xf 25\u201329 20 \u200210 2 \u200230 4 30\u201334 25 \u200220 3 \u200240 5","Statistics 9.29 \t45.\t If the mean of the following data is 26, then find Class Interval f the missing frequency x. 20\u201330 9 30\u201340 5 Class Interval f 40\u201350 6 0\u201310 4 10\u201320 X Essay Type Questions \t46.\t Find the AM of the following data by short-cut Class Interval f method. 20\u201339 20 40\u201359 23 Class Interval f 60\u201379 22 1\u20135 \u20025 80\u201399 13 6\u201310 10 15 4\t 9.\t Calculate the mean deviation about mean for the 11\u201315 10 following data. 16\u201320 5 21\u201325 Class Interval f 10\u201314 \u20025 \t47.\t Find the quartile deviation for the following 15\u201319 \u20026 grouped data. 20\u201324 \u20023 25\u201329 10 Class Interval f 30\u201334 \u20024 PRACTICE QUESTIONS 0\u20134 1 35\u201339 \u20028 5\u20139 3 40\u201344 15 2 45\u201349 10 10\u201314 4 15\u201319 5 \t50.\t Draw the frequency curve for the following data. 20\u201324 3 25\u201329 2 Class Interval f 30\u201334 4 2\u20135 12 35\u201339 3 5\u20138 \u20027 40\u201344 3 8\u201311 \u20028 45\u201349 \u20026 11\u201314 10 \t48.\t Find the mode of the following distribution given 14\u201317 \u20029 below. 17\u201320 Class Interval f 0\u201319 12 CONCEPT APPLICATION Level 1 \t1.\t The mean of first n natural numbers is 5n . Find n. \t2.\t Mean of a certain number of observations is m. 9 If each observation is divided by x(x \u2260 0) and \t\t(a) 5\t\t (b) 4 increased by y, then the mean of new observations is \t\t(c) 9\t\t (d) None of these","9.30 Chapter 9 \t\t(a) mx + y\t\t (b) mx + y \t\t(a) a3\t\t (b) b3 x m + xy \t\t(c) 2a3\t\t (d) 2b3 \t\t(c) x \t\t (d) m + xy \t11.\t If the arithmetic mean of the observations x1, \t3.\t If the difference of mode and median of a data is x2, x3, \u2026, xn is 1, then the arithmetic mean of 24, then the difference of median and mean is x1 , x2 , x3 , , xn (k > 0) is k k k k \t\t(a) 12\t\t (b) 24 \t\t(a) greater than 1.\t (b) less than 1. \t\t(c) 8\t\t (d) 36 \t\t(c) equal to 1.\t (d) None of these \t4.\t The mode of the observations 2x + 3, 3x \u2212 2, 4x + 3, \t12.\t Range of 14, 12, 17, 18, 16 and x is 20. Find x x \u2212 1, 3x \u2212 1, 5x + 2 (x is a positive integer) can be (x\u00a0> 0). \t\t(a) 3\t\t (b) 5 \t\t(a) 2\t\t (b) 28 \t\t(c) 7\t\t (d) 9 \t\t(c) 32\t\t (d) Cannot be determined \t5.\t The median of 21 observations is 18. If two obser- 1\t 3.\t The mean of a set of observation is a. If each vations 15 and 24 are included to the observations, observation is multiplied by b and each product is then the median of the new series is decreased by c, then the mean of new set of obser- vations is ______. \t\t(a) 15\t\t (b) 18 \t\t(c) 24\t\t (d) 16 \t\t(a) a + c \t\t (b) ab \u2013 c b \t6.\t If the quartile deviation of a set of observations is 10 and the third quartile is 35, then the first quar- \t\t(c) a \u2212 c \t\t (d) ab + c tile is b \t\t(a) 24\t\t (b) 30 \t14.\t The mean deviation of first 8 composite numbers \t\t(c) 17\t\t (d) 15 is ______. \t7.\t The upper class limit of inclusive type class interval \t\t(a) 2.9375\t\t (b) 4.83 10\u201320 is ______. \t\t(c) 5.315\t\t (d) 3.5625 PRACTICE QUESTIONS \t\t(a) 10.5\t\t (b) 20 \t15.\t Find the mode of the following discrete series. \t\t(c) 20.5\t\t (d) 17.5 xf xf 15 5 12 \t8.\t The semi-inter quartile range of the observations 24 63 9, 12, 14, 6, 23, 36, 20, 7, 42 and 32 is 36 79 48 8 10 \t\t(a) 12.75\t\t (b) 12.5 \t\t(c) 9.75\t\t (d) 9.5 \t\t(a) 4\t\t (b) 8 \t9.\t Find the mode of the following discrete series. \t\t(c) 5\t\t (d) 7 xf xf 1\t 6.\t The the highest score of certain data exceeds its 15 68 1 37 12 6 lowest score by 16 and coefficient of range is 3 . 53 15 5 Find the sum of the highest score and the lowest score. \t\t(a) 3\t\t (b) 12 \t\t(a) 36\t\t (b) 48 \t\t(c) 8\t\t (d) 6 \t\t(c) 24\t\t (d) 18 1\t 0.\t The mean deviation of a3 + b3 and a3 \u2212 b3 (where 1\t 7.\t Find the mean deviation (approximately) about a and b > 0) is ______. the mode for the following ungrouped data: 20, 25, 30, 18, 15, 40.","Statistics 9.31 \t\t(a) 6.71\t\t (b) 4.52 2\t 5.\t If the mean of x + 2, 2x + 3, 3x + 4 and 4x + 5 is x + 2, then find the value of x. \t\t(c) 7.61\t\t (d) 5.33 n2 \t18.\t The mean of first n odd natural numbers is 81 . \t\t(a) 0\t\t (b) 1 Find n. \t\t(c) \u20131\t\t (d) 2 \t\t(a) 9\t\t (b) 81 2\t 6.\t The range of 15, 14, x, 25, 30, 35 is 23. Find the least possible value of x. \t\t(c) 27\t\t (d) None of these 1\t 9.\t The arithmetic mean of 12 observations is 15. If \t\t(a) 14\t\t (b) 12 two observations 20 and 25 are removed then the arithmetic mean of remaining observations is \t\t(c) 13\t\t (d) 11 2\t 7.\t Find the median of the following data. \t\t(a) 14.5\t\t (b) 13.5 \t\t(c) 12.5\t\t (d) 13 Class Interval f 0\u201310 12 \t20.\t The arithmetic mean and mode of a data is 24 13 and 12 respectively, then the median of the data is 10\u201320 25 ______. 20\u201330 20 30\u201340 10 \t\t(a) 20\t\t (b) 18 40\u201350 \t\t(c) 20\t\t (d) 22 \t\t(a) 25\t\t (b) 23 2\t 1.\t The inter-quartile range of the observations 3, 5, \t\t(c) 24\t\t (d) 26 9, 11, 13, 18, 23, 25, 32 and 39 is \t\t(a) 24\t\t (b) 17 2\t 8.\t In the following table, pass percentage of three schools from the year 2001 to the year 2006 are \t\t(c) 31\t\t (d) 8 given. Which school students\u2019 performance is more consistent? \t22.\t Find the mean deviation from the mode for the following ungrouped data: 2.5, 6.5, 7.3, 12.3, 2001 2002 2003 2004 2005 2006 PRACTICE QUESTIONS 16.2. School X 80 89 79 83 84 65 School Y 92 94 76 75 80 63 \t\t(a) 4.34\t\t (b) 5.57 School Z 93 97 67 63 70 85 \t\t(c) 2.33\t\t (d) 6.72 \t23.\t The mean of the following distribution is 5, then \t\t(a) X\t \t (b) Y find the value of b. xf \t\t(c) Z\t\t (d) X and Y 32 5a \t29.\t The median of the following discrete series is 75 4b xf 35 \t\t(a) 10\t\t (b) 6 62 54 \t\t(c) 8\t\t (d) None of these 86 12 7 \t24.\t The mean deviation of a+b and a\u2212b (where a 76 and b > 0) is ______. 2 2 \t\t(a) b \t\t (b) a \t\t(a) 7\t\t (b) 8 2 2 \t\t(c) 9\t\t (d) 6 \t\t(c) a\t\t (d) b","9.32 Chapter 9 \t30.\t Which of the following does not change for the \t\t(a) Arithmetic mean\t\t (b) Range (d) Quartile deviation observations 23, 50, 27, 2x, 48, 59, 72, 89, 5x, \t\t(c) Median\t\t\t 100, 120, when x lies between 15 and 20? Level 2 \t31.\t If the ratio of mean and median of a certain data is 3\t 7.\t The mean of the following data is 2 : 3, then find the ratio of its mode and mean. \t\t(a) 4 : 3\t\t (b) 7 : 6 Class Interval f \t\t(c) 7 : 8\t\t (d) 5 : 2 10\u201315 5 15\u201320 7 3\t 2.\t If mean of the following distribution is 13, then 20\u201325 3 the value of p is 25\u201330 4 30\u201335 8 x 5 10 12 17 16 20 f 9 \u20023 p \u20028 \u20027 \u20025 \t\t(a) 22 \t\t(b) 23.05 \t\t(a) 6\t\t (b) 7 \t\t(c) 24.05 \t\t(d) 27.05 \t\t(c) 10\t\t (d) 4 3\t 8.\t The median of the following frequency distribu- \t33.\t If the ratio of mode and median of a certain data is tion is 6 : 5, then find the ratio of its mean and median. \t\t(a) 8 : 9\t\t (b) 9 : 10 \t\t(c) 9 : 7\t\t (d) 8 : 11 3\t 4.\t If the arithmetic mean of the following distribu- Class Interval f tion is 8.2, then find the value of p. 0\u201310 \u20025 \u20028 PRACTICE QUESTIONS x 1 3 5 9 11 13 10\u201320 \u20027 f 3 2 7 p \u20024 \u20028 20\u201330 10 30\u201340 20 \t\t(a) 5\t\t (b) 6 40\u201350 \t\t(c) 9\t\t (d) None of these \t\t(a) 35\t \t\t(b) 30 \t35.\t The median of the series 8, 12, 15, 7, x, 19 and 22 \t\t(c) 40\t lies in the interval. \t\t(d) 45 \t\t(a) [12, 15]\t\t (b) [7, 15] 3\t 9.\t Find the quartile deviation of the following dis- crete series. \t\t(c) [15, 17]\t\t (d) [9, 12] x 7 5 4 \u20028 12 10 \t36.\t The mode of the following distribution is f 2 4 6 10 \u20029 \u20027 Class Interval f \t\t(a) 6.5\t 1\u20135 4 \t\t(b) 4.5 6\u201310 7 \t\t(c) 3.5 10 \t\t(d) 2.5 11\u201315 8 16\u201320 6 21\u201325 \t \t (a) 14.5\t\t (b) 16.5 \t\t(c) 10.5\t\t (d) 13.5","Statistics 9.33 Direction for questions 40 and 41: These questions are based on the following data \u00ad(figure). Y-axis 60Cumulative frequency X-axis 55 10 20 30 40 50 60 70 80 90 100 110 120 50 PRACTICE QUESTIONS 45 True limits of the classes 40 35 30 25 20 15 10 5 0 0 \t\tThe given figure represents the percentage of marks on X-axis and the number of students on Y-axis. \t40.\t Find the number of students who scored less than \t\t(a) 4\t\t (b) 3 or equal to 50% of marks. \t\t(c) 3.5\t\t (d) 4.5 \t\t(a) 35\t\t (b) 15 \t \t\t(c) 20\t\t (d) 30 44. Weight (in kg) Number of Students 4\t 1.\t Find the number of students who scored greater 20 8 than or equal to 90% of marks. 22 4 \t\t(a) 47\t\t (b) 45 24 3 \t\t(c) 5\t\t (d) 10 25 7 30 5 \t42.\t In a class of 15 students, on an average, each stu- \t\tFind the mean deviation (approximately) about dent got 12 books. If exactly two students received the median for the above data. same number of books, and the average of books received by remaining students be an integer, then \t\t(a) 2.5\t\t (b) 1.5 which of the following could be the number of books received by each of the two students who \t\t(c) 3\t\t (d) 0.5 received same number of books? \t45.\t Find the mean deviation (approximately) about the mean for the following. \t\t(a) 11\t\t (b) 15 Class Interval f 0\u20135 \u20023 \t\t(c) 20\t\t (d) 25 5\u20130 \u20024 \u20028 \t43.\t Find the quartile deviation of the following dis- 10\u201315 10 crete series. 15\u201320 \u20025 20\u201325 x 3 5 6 8 10 12 f 7 2 3 4 \u20025 \u20026","9.34 Chapter 9 \t\t(a) 5\t\t (b) 4 \t\t\t\t The total weight of 20 girls = 20 \u00d7 35 = 700 kg \t\t(c) 6\t\t (d) 3 \t\t (B) The mean weight of the c=lass 1=74050 37 7 kg 9 4\t 6.\t If the average mark of 15 students is 60 and the \t\t (C) T\u0007 he total weight of 45 students = 1000 kg + average mark of another 10 students is 70, then find the average mark of 25 students. 700 kg = 1700 kg \t\tThe following are the steps involved in solving the \t\t (a) ABC\t\t (b) ACB above problem. Arrange them in sequential order. \t\t(c) BCA\t\t (d) CBA \t\t(A) Average marks of 25 stud=ents 1=62050 64 4\t 8.\t If p < q < 2p; the median and mean of p, q and 2p \t\t(B) The total marks of 15 students = 15 \u00d7 60 = 900 are 36 and 31 respectively, then find the mean of p and q. \t\t\t The total marks of 10 students = 10 \u00d7 70 = 700 \t\t (a) 21.5\t\t (b) 23 \t\t(C) T\u0007 he total marks of 25 students = 900 + 700 \t\t(c) 27.5\t\t (d) 24 = 1600 \t49.\t If x < y < 2x; the median and the mean of x, y and \t\t(a) BCA\t\t (b) BAC 2x are 27 and 33 respectively, then find the mean of x and y. \t\t(c) CBA\t\t (d) CAB \t\t (a) 23.5\t\t (b) 24 \t47.\t In a class of 25 boys and 20 girls, the mean weight of the boys is 40 kg and the mean weight of the \t\t(c) 23\t\t (d) 25.5 girls is 35 kg. Find the mean weight of the class. 5\t 0.\t The mean of a set of 12 observations is 10 and \t\t The following are the steps involved in solving the another set of 8 observations is 12. The mean of above problem. Arrange them in sequential order. combined set is ______. \t\t(A) \u0007The total weight of 25 boys = 25 \u00d7 40 = \t\t (a) 11\t\t (b) 10.8 1000\u00a0kg \t\t(c) 11.2\t\t (d) 0.6 PRACTICE QUESTIONS Level 3 \t51.\t A class of 40 students is divided into four groups 5\t 2.\t Life (in hour) of 10 bulbs from each of four differ- named as A, B, C and D. Group-wise percentage ent companies A, B, C and D are given below in of marks scored by them are given below in the the table. table. ABC D ABCD \u2002120 \u2002700 \u2002950 530 20 42 10 21 30 51 25 69 1600 \u2002502 \u2002330 650 40 45 85 70 25 58 73 86 \u2002280 1430 1200 720 22 53 98 53 45 64 43 68 \u2002420 \u2002625 \u2002500 550 65 72 64 99 \u2002800 \u2002780 \u2002445 748 \u2002770 \u2002335 1260 570 \u2002270 \u2002224 \u2002375 635 \u2002455 1124 1130 804 \t\tBy using the coefficient of range find which of the \u2002150 \u2002473 \u2002185 500 group has shown good performance. \t\tBy using the coefficient of range find which com- pany has shown the best consistency in its quality? \t\t(a) A\t\t (b) B \t\t(a) A\t\t (b) B \t\t(c) C\t\t (4) D \t\t(c) C\t\t (d) D","Statistics 9.35 \t53.\t If the mode of the observations 5, 4, 4, 3, 5, x, 3, \t \t (a) 11.5\t\t (b) 12.5 4, 3, 5, 4, 3 and 5 is 3, then find the median of the \t\t(c) 13.5\t\t (d) 14.5 observations. 5\t 9.\t The mean of the following distribution is 4. Find \t\t(a) 3\t\t (b) 4 the value of q. \t\t(c) 5\t\t (d) 3.5 x 23457 f 4423q \t54.\t In a colony, the average age of the boys is 14 years and the average age of the girls is 17 years. If the \t\t(a) 2\t\t (b) 3 average age of the children in the colony is 15 years, \t\t(c) 0\t\t (d) 4 find the ratio of number of boys to that of girls. \t\t(a) 1 : 2\t\t (b) 2 : 1 6\t 0.\t If the ratio of mean and median of a certain data is 5 : 7, then find the ratio of its mode and mean. \t\t(c) 2 : 3\t\t (d) 3 : 2 \t55.\t Find the median of the following data. \t\t(a) 2 : 5\t\t (b) 11 : 5 Class Interval f \t\t(c) 6 : 5\t\t (d) 2 : 3 0\u20134 3 4\u20138 6 \t61.\t Find the mode of the following discrete series. 8\u201312 6 6 x 1234567 8 9 12\u201316 8 f 3 8 15 1 9 12 14 5 7 16\u201320 \t \t (a) 7\t\t (b) 5 \t \t (a) 13\t\t (b) 12 \t\t(c) 2\t\t (d) 3 \t\t(c) 11\t\t (d) 10 \t62.\t Find the median of the following data. \t56.\t In a class of 20 students, 10 boys brought 11 books x 12 15 18 21 24 each and 6 girls brought 13 books each. Remaining f 47234 students brought atleast one book each and no two students brought the same number of books. If the \t\t(a) 12\t\t (b) 16 average number of books brought in the class is a positive integer, then what could be the total num- \t\t(c) 18\t\t (d) 15 PRACTICE QUESTIONS ber of books brought by the remaining students? \t63.\t Find the mean deviation about the median for the \t\t(a) 12\t\t (b) 16 following data. \t\t(c) 14\t\t (d) 8 x 1234 56 f 3758 25 \t57.\t The mean of a set of 20 observations is 8 and another set of 30 observations is 10. The mean of \t\t(a) 1\t\t (b) 0.7 combined set is ______. \t\t(c) 3\t\t (d) 1.3 \t\t(a) 9.2\t\t (b) 10.8 \t64.\t Find the mode for the following data. \t\t(c) 11.2\t\t (d) 9.8 5\t 8.\t Find the approximate value of mean deviation Class Interval f about the mode of the following data. 0\u20139 2 4 Class Interval f 10\u201319 7 0\u201310 4 20\u201329 5 6 30\u201339 3 10\u201320 3 40\u201349 20\u201330 9 30\u201340 5 \t \t (a) 30\t\t (b) 25.5 40\u201350 \t\t(c) 32\t\t (d) 33","9.36 Chapter 9 \t65.\t Find the quartile deviation of the following dis- 6\t 8.\t Find the mean of the following continuous crete series. distribution. x 8 10 13 16 19 22 Class Interval f f 4 \u20027 \u20028 \u20023 \u20025 \u20024 0\u201310 8 4 \t\t(a) 3.5\t\t (b) 6 10\u201320 6 20\u201330 3 \t\t(c) 5\t\t (d) 4.5 30\u201340 4 40\u201350 \t66.\t Find the arithmetic mean of the observations x + 5, x + 6, x + 10, x + 11, x + 14, x + 20 (where \t\t(a) 20.8\t\t (b) 21.4 x is any real number). \t\t(a) x + 11\t\t (b) x + 5 \t\t(c) 21.8\t\t (d) 22.2 \t\t(c) x + 13\t\t (d) x + 7 6\t 9.\t Which of the following is not changed for the observations 31, 48, 50, 60, 25, 8, 3x, 26, 32? \t67.\t Find the mode of the following discrete series. (where x lies between 10 and 15). x 1 2 \u20023 4 5 \u20026 \u20027 8 9 \t\t(a) Arithmetic mean\t f 3 8 15 1 9 12 17 5 7 \t\t(b) Range \t\t(a) 7\t\t (b) 5 \t\t(c) Median \t\t(c) 2\t\t (d) 3 \t\t(d) Quartile deviation PRACTICE QUESTIONS","Statistics 9.37 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 40.5 \t17.\t 1 24 \t2.\t 13 \t18.\t (x \u2212 3) \t3.\t 7 \t19.\t 10 \t4.\t 40 \t20.\t 7 \t5.\t 24 \t21.\t \u22122.006 \t6.\t size (length), frequency 2\t 2.\t 47 \t7.\t 3 2\t 3.\t 29 \t8.\t 10 \t24.\t 26 \t9.\t 2 \t25.\t 45 89 \t10.\t x = 4 \t26.\t 30 6 2 1\t 1.\t 3 \t27.\t 4 \t12.\t 5 2\t 8.\t 148 cm \t13.\t y\u2212x 2\t 9.\t 3 1 2 2 \t14.\t 2 \t30.\t 16 \t15.\t 5x \t16.\t a + c = 2b Shot Answer Type Questions 3\t 8.\t 122.8 cm ANSWER KEYS 3\t 9.\t 50 \t31.\t 30 kg \t40.\t 6 \t33.\t 20 \t41.\t 15, 2.9 \t34.\t 26 \t45.\t 6 3\t 5.\t `330 \t36.\t 30 3\t 7.\t 69 Essay Type Questions \t48.\t 54.5 \t49.\t 9.9 (approx) \t46.\t 13 \t47.\t 10.625","9.38 Chapter 9 CONCEPT APPLICATION Level 1 \t 1.\u2002 (c) \t 2.\u2002(c)\t 3.\u2002 (a) \t 4.\u2002(c)\t 5.\u2002 (b) \t\u2002 6.\u2002(d)\t 7.\u2002 (b) \t 8.\u2002(c)\t 9.\u2002 (d) \t 10.\u2002(b) \t11.\u2002 (d) \t 12.\u2002(c)\t 13.\u2002 (b) \t 14.\u2002(a)\t 15.\u2002 (c) \t 16.\u2002(b)\t 17.\u2002 (c) \t 18.\u2002(b)\t 19.\u2002 (b) \t 20.\u2002(c) \t21.\u2002 (b) \t 22.\u2002(b)\t 23.\u2002(b)\t 24.\u2002(a)\t 25.\u2002 (c) \t 26.\u2002(b)\t 27.\u2002 (d) \t 28.\u2002(a)\t 29.\u2002 (a) \t 30.\u2002(b) Level 2 33.\u2002 (b) \t 34.\u2002(b)\t 35.\u2002 (a) \t 36.\u2002(d)\t 37.\u2002 (b) \t 38.\u2002(a)\t 39.\u2002 (d) \t 40.\u2002(b) 43.\u2002 (c) \t 44.\u2002(d)\t 45.\u2002 (a) \t 46.\u2002(a)\t 47.\u2002 (b) \t 48.\u2002(c)\t 49.\u2002 (d) \t 50.\u2002(b) \t31.\u2002 (d) \t 32.\u2002(b)\t \t41.\u2002 (c) \t 42.\u2002(d)\t Level 3 53.\u2002 (b) \t 54.\u2002(b)\t 55.\u2002(c)\t 56.\u2002 (a) \t 57.\u2002(a)\t 58.\u2002 (b) \t 59.\u2002(b)\t 60.\u2002(b) 63.\u2002(d)\t 64.\u2002 (b) \t 65.\u2002(d)\t 66.\u2002 (a) \t 67.\u2002(a)\t 68.\u2002 (b) \t 69.\u2002(b) \t51.\u2002 (b) \t 52.\u2002(d)\t \t61.\u2002(d)\t 62.\u2002 (d) \t ANSWER KEYS","Statistics 9.39 CONCEPT APPLICATION Level 1 \t1.\t Sum of the first n natural numbers is n(n + 1) . \t22.\t Mean deviation = \u2211|xi \u2212 A| 2 n \t2.\t If each observations divided by x, then the new \t\tWhere A = Arithmetic mean. mean is \uf8eb m\uf8f6 . \t23.\t \t(i)\t \u2211 fx = 5. \uf8ec\uf8ed x \uf8f7\uf8f8 N \t3.\t Use empirical formula. \t\t\t(ii)\t\u0007Find the value of \u2211 fx and \u2211 f .\u03c4 \u03c4 \t4.\t Substitute x = 1. \u2211 fx \u2211 f \t5.\t Observe median after including new observations. \t\t(\t iii)\t\u0007 Mean = . \t6.\t Recall the formula for quartile deviation. \t24.\t Mean deviation of x and y is |x \u2212 y| . 2 \t7.\t For inclusive type class, given values are the limits. \t25.\t \t (i)\t Find the sum of observations. \t8.\t Semi-quartile range is the quartile deviation. Sum of observations Number of observations \t9.\t The \u2018x\u2019 value of the highest frequency class is the \t\t\t (ii)\t\u0007Use, x +2= . mode. 2\t 6.\t \t(i)\tConsider x as the least value. |x \u2212 y|. 1\t 0.\t Mean deviation of x and y is 2 \t\t\t (ii)\t\u0007Use, Range = Highest score \u2013 Lowest score 1\t 1.\t Mean depends on the value of k. and find x. \t12.\t Range = Maximum value \u2013 Minimum value. 2\t 7.\t \t(i)\tW\u0007 rite the cumulative frequencies of each Hints and Explanation class. \t13.\t Use the properties of mean. \t14.\t Mean deviation = \u2211|xi \u2212 A|, where A = \t\t\t(ii)\t\u0007Trace the median class, i.e., the class corre- Arithmetic mean. n sponding the commutative frequency equal to \t15.\t The \u2018x\u2019 value of the highest frequency class is the \uf8eb N \uf8f6 . mode. \uf8ed\uf8ec 2 \uf8f7\uf8f8 \t16.\t Use the formula for coefficient of range. \t\t\t(iii)\tE\u0007valuate median, by using median \t17.\t Find the mode by using mean and median then \uf8eb N \u2212 F \uf8f6 find mean deviation. \uf8ec 2 \uf8f7 = L + \uf8ec \uf8f7 \u00d7 C. \uf8ed\uf8ec f \uf8f8\uf8f7 1\t 8.\t Mean = Sum of the observations . \t28.\t \t (i)\t\u0007Find the average (mean) pass percentage of all Total number of observations the schools. 1\t 9.\t AM = Sum of all observation . \t\t\t(ii)\t\u0007The school with the maximum pass percent- Total number of observations age is more consistent. \t20.\t Mode = 3 median \u2013 2 mean. 2\t 9.\t \t(i)\tFind N and proceed. 2 \t21.\t Inter-quartile range = Q3 \u2212 Q1 2 \t\t\t(ii)\t\u0007Find the cumulative frequency. Q3 = 3(n + 1) th observation \t\t(\tiii)\t\u0007Identify the value of x corresponding to the \t\t = 4 \uf8ebN \uf8ed\uf8ec 2 \uf8f6\uf8f8\uf8f7th observation. Q1 (n 1) + th observation. 4 \t\t\t(iv)\t\u0007Use the formula to find median.","9.40 Chapter 9 \t30.\t \t (i)\t Recall the properties of central tendencies. \t\t(\tiii)\t\u0007The maximum and minimum values are not changed in the given interval of x. \t\t\t(ii)\t\u0007For different values of x, AM, Median, and QD will be changed. Level 2 \t31.\t \t (i)\t\u0007Substitute, median 3 mean in empirical \t37.\t \t (i)\t Use deviation method. relation. =2 \t\t\t(ii)\t\u0007Find the mid values of the class intervals. \t\t\t (ii)\t\u0007Mode = 3 Median \u2013 2 Mean. \t\t(\t iii)\t\u0007Find \u2211 fx and \u2211 f .\u03c4 \u03c4 \t\t\t(iii)\t\u0007Express median in terms of mean according \t\t\t(iv)\t\u0007 Mean = \u2211 fx . to the given data. \u2211f \t\t\t(iv)\t\u0007Now write the ratio of the mode and mean N \u2212 m using the relation in (i). 2 f \u2211 fx = 13. \t38.\t \t(i)\t Median =l + \u00d7 C. N \t32.\t \t(i)\t \t\t\t(ii)\t\u0007Find the cumulative frequency. \t\t\t (ii)\t\u0007Find \u2211 fx and \u2211 f .\u03c4 \t\t(\t iii)\t\u0007Find N , m, f and c. \u03c4 2 \t\t\t(iii)\t\u0007 Mean = \u2211 fx , where \u2211f = N. \t\t\t(iv)\t\u0007Find the median by the formula, \u2211f N 3\t 3.\t \t(i)\tMode = 3 Median \u2013 2 Mean. 2 \u2212 m f \t\t\t(ii)\t\u0007Divide the above equation by median. \t\t\t\tMedian =l + \u00d7 C. Hints and Explanation \t\t(\t iii)\t\u0007Substitute, Mode = 6 and obtain the \t39.\t \t(i)\t QD = Q3 \u2212 Q1 . Median 5 2 required ratio. \t\t(ii)\t Find the cumulative frequency. \uf8ebN\uf8f6 3\t 4.\t \t(i)\t \u2211 fx = 8.2. \t\t(\t iii)\t\u0007Identify the x value corresponding to \uf8ec\uf8ed 2 \uf8f7\uf8f8 th N 3N \t\t\t (ii)\t\u0007Find \u2211 fx and \u2211 f .\u03c4 observation and \uf8eb 2 \uf8f6 th observation. \u03c4 \uf8ec\uf8ed \uf8f7\uf8f8 \u2211 fx \t40.\t \t (i)\t Inferences from the graph. \u2211f \t\t(\t iii)\t\u0007 Mean = . \t\t\t(ii)\t\u0007Use less than cumulative frequency curve. \t35.\t \t (i)\t Recall the properties of central tendencies. \t\t\t(iii)\tI\u0007dentify the cumulative frequency corre- sponding to 50. \t\t\t(ii)\t\u0007Arrange the given observations (except x) in ascending order. \t41.\t \t (i)\t\u0007Inferences from the graph. \t\t(\tiii)\t\u0007Find the median for different values of x. \t\t\t(ii)\t\u0007Use greater than cumulative frequency curve and identify the frequency corresponding to \t\t\t(iv)\t\u0007Write median interval for all values of x. 90. \t36.\t \t(i)\t Mode = L + \u22061C . 4\t 2.\t \t (i)\t\u0007Use mean concept. \u22061 + \u22062 \t\t\t(ii)\t\u0007Let the two students received x number of \t\t\t (ii)\t\u0007Identify f1, f2 and L. books each. \t\t(\t iii)\t\u0007 Mode = L + D1 \u00d7C (Where D1 = f \u2212 \t\t\t(iii)\t\u0007According to the data 2x + 13p = 180. D1 + D2 \t\t\t(iv)\t\u0007From the options choose x such that p must f1 and D2 = f \u2013 f2). be an integer.","Statistics 9.41 \t43.\t \t(i)\t\u0007QD = Q3 \u2212 Q1 . \u2234 q = 36 2 p + q + 2p = 31 = 93 \t\t\t(ii)\t\u0007Write cumulative frequency. \t\t \t\t\u21d2 3 3p \t\t\t(iii)\t\u0007The corresponding values of x of N th and + 36 \u21d2 p = 19. 4 \u2234 Mean of 19 and 36 = 19 + 36 = 27.5. 2 3N th observation is Q1 and Q3 respectively. 4 4\t 9.\t As x, y and 2x are in ascending order, median is y. 4\t 4.\t \t (i)\t\u0007Find median and then mean deviation. \t\t\t(ii)\t\u0007Calculate the median of the given discrete y = 27. data. Mean = x + y + 2x = 33 3 \t\t\t(iii)\tE\u0007valuate mean deviation by using \t\t 3x + 27 \u2211|xi \u2212 A| f \u21d2 3 = 33 \u21d2 x = 24. \u2211f . \t45.\t \t (i)\t\u0007Find mean and then mean deviation. \u2234 Mean of x and y = x +y = 24 + 27 = 25.5. 2 2 \t\t\t(ii)\t\u0007Find the mid-values of the series. \t50.\t The mean of set of 12 observations is 10. \t\tS\t um of the observations = 12 \u00d7 10 = 120 \t\t\t(iii)\t\u0007Find the mean. \t\tT\t he mean of another set of 8 observations is 12 \t\tS\t um of the observations = 12 \u00d7 8 = 96 \t\t\t(iv)\t\u0007Find the mean deviation using the formula, \t\tT\t otal number of observations = 12 + 8 = 20 \t\tT\t otal sum of observations = 216 MD = \u2211 (xi \u2212 A) f \t\tC\t ombined me=an 2=2106 10.8. \u2211 f \t46.\t BCA is the required sequential order. Hints and Explanation \t47.\t ACB is the required sequential order. \t48.\t As p, q and 2p are in descending order, their median is 36. Level 3 \t51.\t \t (i)\t\u0007Find the coefficient of range of A, B, C and D \t\t(\tiii)\t\u0007Find the coefficient of range of A, B, C and D. and compare. \t\t\t(iv)\t\u0007The least coefficient of range implies good \t\t\t(ii)\t Coefficient of range = performance. 5\t 3.\t Given, Mode = 3 \t\t\t\t\u0007Maximum value \u2212 Minimum Value . \t \t \u2234x = 3 Maximum value + Minimum Value \t\tNow, arrange the given data in ascending order, \t\t3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5 \t\t(\tiii)\t\u0007Find the coefficient of range of A, B, C and D. \t \t \u2234 Median = 4 \t\t\t(iv)\t\u0007The least coefficient of range implies good \t\tHence, option (b) is correct. \t54.\t Given, performance. \t\tAvg. age of boys = 14 years \t52.\t \t (i)\t\u0007Find the coefficient of range of A, B, C and D \t\tAvg. age of girls = 17 years and compare. \t\t\t(ii)\t Coefficient of range = \t\t\t\t\u0007Maximum value \u2212 Minimum Value . Maximum value + Minimum Value","9.42 Chapter 9 \t\tAvg. age of children in society = 15 years \t57.\t The mean of set of 20 observations is 8. \t\tSum of the observations = 20 \u00d7 8 = 160 \t\tLet no. of boys be x, and \t\tThe mean of another set of 30 observations is 10 \t\tSum of the observations = 30 \u00d7 10 = 300 \t\tNo. of girls be y \t\tTotal number of observations = 20 + 30 = 50 \t\tTotal sum of observations = 460 \t \t \u21d2 Sum of boys = 14 \u00d7 no. of boys \t\tCombined mean== 4=5600 9.2. \t\tSum of girls = 17 \u00d7 no. of girls \t58. \t\tSum of children = 15 \u00d7 (no. of boys + no. of girls) \t \t \u21d2 14 \u00d7 no. of boys + 17 \u00d7 no. of girls = 15 \u00d7 (x + y) \t \t \u21d2 14x + 17y = 15x + 15y \t \t \u21d2 x - 2y = 0 Class Interval \t \t \u21d2 x = 2y f x D = |x - x| fD 0\u201310 \t \t\u2234x:y=2:1 10\u201320 45 31 124 20\u201330 \t\tHence, option (b) is correct. 30\u201340 6 15 21 126 \t 40\u201350 3 25 11 33 55. Cumulative 9 35 1 9 Class Interval f Frequency 5 45 9 45 0\u20134 3 3 \t\t\tMode = L + \u22061 \u00d7 c. \u22061 + \u22062 4\u20138 5 8 m \t\tH\t ere, maximum frequency is 9. 8\u201312 6 14 \t\t\u2234\t Model class is 30\u201340 Hints and Explanation 12\u201316 3 17 \t\t\u2234\t L = 10 16\u201320 8 25 \u22061 = f \u2212 f1 and \u22062 = f \u2212 f2 \t\t\t\u22061 = 9 \u2212 3, \u22062 = 9 \u2212 5 = 4 and C = 10 \t\tHere, N = 25 \t\tN=2 25 12.5 . =2 6 6 \t\t\u2234 The median class = 8 \u2013 12 Mode = 30 + 6 + 4 \u00d7 10 = 30 + 10 \u00d7 10 \t\t\t \t\t\u2234 l = 8, f = 6, m = 8, c = 4. = 30 + 6 = 36. Median = l + N \u2212m \u00d7c = 8+ 12.5 \u2212 8 \u00d7 4 \u2211 fD = 337, N = 27 \t\t 2 f 6 4.5 \u2234 MD = \u2211 f \u22c5d = 337 \u223c 12.5. 3 N 27 = 8 + \u00d7 2 = 8 + 1.5 \u00d7 2 = 8 + 3 = 11. \t \t56.\t 10 boys brought 11 books each. Therefore, total 59. x f fx number of books brought by 10 boys = 110. Six 248 girls brought 13 books each. Therefore, total num- ber of books brought by 6 girls = 78. 3 4 12 \t\tRemaining number of students are 4 and the total 428 number of books brought by 4 students = x. 5 3 15 x + 78 + 110 7 q 7q 20 \t\tGiven that is a positive integer. \t\tHere \u2211 fx = 43 + 7q \t\t\u2211 f = 13 + q \t\tIf x = 12 then the above value is an integer.","Statistics 9.43 \t\tWe know that, AM = \u2211 fx \t\tHere N = 30 \u2211f \t\tN2 = 15 , which falls in the class with x = 3. \t\t\u2234 Median = 3, \u03a3f |D|= 40 4 = 43 + 7q \t\t 13 + q \u03a3f |D | 40 52 + 4q = 43 + 7q \u21d2 3q = 9 \u21d2 q = 3. \t\tMean deviation = N = 30 = 1.33. 6\t 0.\t Given that, Mean = 5 \t64.\t First we convert the given class intervals into Median 7 continuous class intervals by subtracting 0.5 and adding 0.5 to lower and upper limits of each class \t\tWe know that, Mode = 3 Median \u2013 2 Mean. respectively. Mode = Mean \uf8eb 3 Median \u2212 2 \uf8f6 \uf8ec\uf8ed Mean \uf8f7\uf8f8 Class Interval f \t\tMode = \uf8eb 3 \u22c5 7 \u2212 2 \uf8f6\uf8f7\uf8f8 = 21 \u2212 10 = 151 . 0\u20139 2 Mean \uf8ec\uf8ed 5 5 10\u201319 4f1 6\t 1.\t Mode = 3 20\u201329 7f 30\u201339 \t\t\u2234 Since the frequency of 3 is maximum. 40\u201349 5f2 3 \t 62. Cumulative \t \t Here, f = 7, f1 = 4, f2 = 5 and L = 19.5 (lower f Frequency boundary of the highest frequency class). x 12 44 Mode = L + \u22061 \u00d7 c \u22061 + \u22062 15 7 11 \u22061 = f \u2212 f1 = 3 Hints and Explanation 18 2 13 \t\t\u22062 = f \u2212 f2 = 2 21 3 16 3 24 4 20 x = 19.5 + 5 \u00d7 10 \t\tN = 20 19.5 + 6 = 25.5. \t\tThe value of x corresponding to N th value class \t65.\t x f Less than Cumulative 2 \u20028 4 Frequency in the cumulative frequency is the median. Here 10 7 \u20024 13 8 11 N = 10 and the corresponding x value is median, 16 3 19 2 19 5 22 22 4 27 i.e., 15. 31 \t 63. Cumulative xi fi Frequency D = xi \u2013 A fi|D| 13 \u20023 \u20132 \u20026 27 10 \u20131 \u20027 \t\tN = 31 35 15 \u20020 \u20020 \t\tQ1 = size of (n + 1) th item = 8th item = 10. 4 48 23 \u20021 \u20028 3(n + 1) 52 25 \u20022 \u20024 \t\tQ3 = size of 4 th item = 24th item = 19. 65 30 \u20023 15 Q3 \u2212 Q1 19 \u2212 10 2 2 40 \t\tQD = = = 4.5.","9.44 Chapter 9 \t66.\t AM = Sum of all observations \t\tHere \u03a3fx = 535 total number of observations \t\t\u03a3f = 25 \t\t= x +5+x +6+ x + 10 + x + 11 + x + 14 + x + 20 \t\tMean \u03a3fx 535 21.4. 6 \u03a3f 25 = = = \t\t= 6x + 66 = 6(x + 11) = x + 11. 6\t 9.\t If we take x value between 10 and 15. There is no 6 6 change in maximum value and the minimum value. 6\t 7.\t Mode = 7 \t\t\u2234 Range will not change. \t\t\u2234 Since the frequency of 7 is maximum. 6\t 8.\t Class Interval D x fx 0\u201310 8 \u20025 \u200240 10\u201320 4 15 \u200260 20\u201330 6 25 150 30\u201340 3 35 105 40\u201350 4 45 180 Hints and Explanation","1102CChhaapptteerr pKrionbemabatiilictsy REmEmBER Before beginning this chapter, you should be able to: \u2022 Understand the basic concepts of probability \u2022 Compute basic probability in daily life KEY IDEaS After completing this chapter, you should be able to: \u2022 Learn the concepts related to probability \u2022 Find probability of happening of any event \u2022 Apply the concept of probability in daily life and solve problems Figure 1.1","10.2 Chapter 10 INTRODUCTION In our daily life we frequently use some statements like the followings: 1.\t I will probably get a scholarship for my performance. 2.\t I doubt that I will catch the train. 3.\t The chances of the prices of all commodities going up are very high. 4.\t There is a fifty-fifty chance of Sania winning the match. 5.\t I will most probably return from my official tour in three days. 6.\t The chances of Sachin making a century in his next match are very few. In the above statements the words probably, chances, doubt and most probably involve an uncertainty. In statement (1), I will probably get a scholarship means he may get a scholarship or he may not get a scholarship. Here we are predicting a scholarship based on his performance. Similar predictions are also made in statements (2) to (6). In this chapter, we are going to learn how to measure the uncertainty (i.e., probability) numerically by using probability in many cases. Actually probability started with gambling but it has been used extensively in all fields like biological sciences, medical sciences, weather forecasting, commerce, etc. E\u0007 xample: Take any `1 coin and toss it 5 times, 10 times, 15 times, 20 times, etc., and record your observations in three columns as shown below: Number of Times Number of Times Number of Times Coin Tossed Head Shows up Tail Shows up 5 2 10 3 4 15 6 9 20 6 12 : 8 : : : : : Now find the values in each of the following case: 1.\t Number of times head shows up Total number of times the coin is tossed 2.\t Number of times a tail shows up Total number of times the coin is tossed Observations 1.\t A\u0007 s the number of times a head\/a tail shows up is less than the total number of times the coin is tossed, the values of the above two fractions are proper fractions. 2.\t T\u0007 he values of the above two fractions are positive but less than one. 3.\t W\u0007 e observe that, when the number of times the coin tossed becomes larger and larger, then 1 the value of the above two fractions approaches 2 or 0.5.","Probability 10.3 \u0007Example: A six-faced (well-balanced) cube with its six faces marked with numbers 1 to 6 is called a dice. Roll a dice 30 times and record the faces showing up with numbers 1 to 6 in each case as shown below: Number of Times Number of Times Each Face with the a Dice is Rolled Following Values Shows up 30 123456 854652 Find the values of the following: 1.\t Number of times 1 turned up Total number of times the dice is rolled Number of times 2 turned up 2.\t Total number of times the dice is rolled 3.\t Number of times 3 turned up Total number of times the dice is rolled 4.\t Number of times 4 turned up Total number of times the dice is rolled 5.\t Number of times 5 turned up Total number of times the dice is rolled Number of times 6 turned up 6.\t Total number of times the dice is rolled By repeating the above 20, 40 or 50 times, we observe the following: 1.\t The value of each fraction is a proper fraction and the value of it lies between 0 and 1. 2.\t The value of each fraction approaches 1 . 6 In the above examples each toss of a coin or each roll of a dice is called a trial. When a coin is tossed, the possible outcome is a head or a tail. Getting a head or tail in a particular throw is called an event with outcome head or tail. When a dice is rolled, the possible outcome is 1, 2, 3, 4, 5 or 6. Getting a particular numbered face is an event. Getting an odd number is also an event and this event consists of three outcomes 1, 3 and 5. Getting a prime number is also an event and consists of three outcomes 2, 3 and 5. Getting a multiple of 3 is an event which consists of two outcomes 3 and 6. An event can be defined as the collection of some outcomes of a trial, or we can say an event is a subset of a set of all the outcomes of a trial. The probability of happening of any event (E\u2009) of a trial is denoted by P(E\u2009) and is defined as P(E) = Number of favourable cases for the event to happen . Total number of trials","10.4 Chapter 10 Example 10.1 A coin is tossed 1200 times whereby head occurred in 745 cases and in the remaining tail occurred. Find the probability of each event. Solution Given that the coin is tossed 1200 times. Total number of trials = 1200 Let the event of getting a head be H, and getting a tail be T. The number of times the event H occurred = 745 Probability of event H, i.e., P(H ) = NumTboertaol fntuimmebserheoafdtroiaclcsu=red 17=24050 149 240 Number of times tail, i.e., event T occurred = 1200 - 745 = 455. \u2234 Probability of event T, i.e., P(T ) = number of times tail occured total number of trials = 455 = 91 1200 240 \u2002Notes\u2002 1.\t The number of possible outcomes in each of the above trial is two (H and T\u2009). 2.\t P(H) + P(T ) = 149 + 91 = 240 = 1. 240 240 240 Example 10.2 A dice is rolled 100 times with the frequencies of the outcomes 1, 2, 3, 4, 5 and 6, which are given in the following table. Find the probability of getting each outcome. Outcome 1 2 3 4 5 6 Frequency 10 18 15 25 20 12 Solution Given that the total number of trials = 100 Let each event be denoted by Ei, where i = 1, 2, 3, 4, 5 and 6. Then, probability of getti=ng 1 P=(E1 ) Frequency of 1 = 10 . Total numbers of trials 100 Simila=rly, P(E2 ) 11=080 ,P(E3 ) 11=050 , P(E4 ) 1=2050 , P(E5 ) 20 and P(E6 ) = 12 . 100 100 \u2002Notes\u2002 1.\t The possible outcomes in each trial are 1, 2, 3, 4, 5 and 6. 2.\t Sum of the probabilities, P(E1 ) + P(E2 ) + P(E3 ) + P(E4 ) + P(E5 ) + P(E6 ) = 1.","Probability 10.5 Example 10.3 A bulb-manufacturing company kept a record of the number of hours it can glow before a bulb needed to be replaced. The following table shows the results of 500 bulbs: Life of the Bulb (in Less than 1500 1500 to 2500 2500 to 5000 More than 5000 hours) 10 150 250 90 Frequency If you buy a bulb of this company, then find the probability of the following events: (a)\t The bulb needs to be replaced before it glows for less than 1500 hours. (b)\t The bulb needs to be replaced when it glows between 1500 hours and 5000 hours. (c)\t The bulb lasts for more than 2500 hours. Solution Total number of bulbs tested = 500. (a)\t Number of bulbs replaced for less than 1500 hours = 10 \t P(bulb to be replaced before it glows for < 1500 hou=rs) 5=1000 0.02. (b)\t Number of bulbs replaced between 1500 hours and 5000 hours = 150 + 250 = 400 \t P(bulb to glow between 1500 hours and 5000 hours) = 400 = 0.8. 500 (c)\t Number of bulbs that last for more than 2500 hours = 250 + 90 = 340 \t P(bulb lasts for more than 2500 hours) = 340 = 0.68. 500 Example 10.4 The percentage of monthly targets achieved in producing a certain type of bolts in a company for different months is given in the following table: Month Jan Feb Mar April May June Target Achieved (in per cent) 55 87 73 82 74 80 Find the probability that the company achieved 80% of the monthly target in producing the bolts. Solution The total number of months for which target is fixed = 6. The number of months the company achieved 80% of the target fixed = 3. \\\\ P(achieving 80% of the target in producing bolts)= 36= 1 = 0.5. 2","10.6 Chapter 10 Example 10.5 A pack has 90 cards. Each card was marked with a different number among 110 to 199. A card was selected at random. Find the probability that the number on it is not a perfect square. (a) 37 \t\t (b) 13 \t\t (c) 41 \t\t (d) 43 45 15 45 45 Solution Among 110 to 199, the perfect squares are 121, 144, 169 and 196. Let the number on the card be x. Probability (x being a perfect squar=e) 9=40 2 . 45 \\\\ Required probability = 1\u2212 2 = 43 . 45 45 Example 10.6 A number was chosen at random from the first 300 three digits natural numbers. Find the probability of it ending with a zero. (a) 1 \t\t (b) 1 \t\t (c) 1 \t\t 1 15 25 10 (d) 20 Solution Let the number be x, 100 \u2264 x \u2264 399. A number ending with a 0 is divisible by 10. Least value of x, divisible by 10 = 100 = 10(10) Greatest value of x, divisible by 10 = 390 = 10(39) Number of values of x divisible by 10 = number of numbers from 10 to 39 = 30 \\\\ Required probabil=ity 3=3000 1 . 10","Probability 10.7 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t When two coins are tossed simultaneously, list out \t7.\t Let n be the number of trials that an event E the possible outcomes. occurred and m be the total number of trials, then find the probability of the event E. \t2.\t When a dice is rolled, what are all the possible outcomes? \t8.\t A dice is rolled. Find the number of outcomes of getting a composite number. \t3.\t Two coins are tossed. Find the number of out- comes of getting one head. \t9.\t When a dice is rolled the probability that the number on the face showing up is greater than 6 \t4.\t What is the sum of all the probabilities of trials of is _____. an experiment? 1\t 0.\t A coin is tossed 20 times and head occurred 12 \t5.\t When a dice is rolled, then what is the number of times. How many times did tail occur? possible outcomes of obtaining an even number? \t6.\t Probability of occurring of an event always lies between _____. Short Answer Type Questions \t11.\t A coin is tossed 500 times. Head occurs 343 times \t\tIf a student is selected at random, what is the prob- PRACTICE QUESTIONS and tail occurs 157 times. Find the probability of ability that he gets more than 18 marks? each event. 1\t 4.\t In a cricket match, Dhoni hits a six 4 times from 1\t 2.\t A day is selected in a week, find the probability 24 balls he plays. Find the probability of hitting a that the day is Monday. six. 1\t 3.\t In a monthly test, 10 students were awarded marks 1\t 5.\t Three coins are tossed 100 times, and three heads in a Mathematics examination as follows: occurred 25 times, two heads occurred 38 times, one head occurred 14 times and head did not occur 23, 25, 18, 15, 20, 17, 10, 24, 15, 19 23 times. Find the probability of getting more than one head. Essay Type Questions \t16.\t A dice is rolled 250 times, and the outcomes 1, 2, Male Child 210 3, 4, 5 and 6 occurred as given in the following Number of Families 27 38 10 table: \t\tIf a family is selected at random, what is the Outcome 123456 probability that the family has only one male child? Frequency 56 78 46 27 19 24 \t19.\t The percentage of marks obtained by a student in a monthly test is as follows: \t\tFind the probability of getting an odd number. Test I II III IV \t17.\t A Class IX English book contains 200 pages. A Marks 78% 63% 82% 65% page is selected at random. What is the prob- ability that the number on the page is divisible \t\tWhat is the probability that the student gets more by 10? than 75% marks in a test? \t18.\t In a colony there are 75 families and each family \t20.\t A box contains 50 tickets. Each ticket is numbered has two children. The number of male children of from 1 to 50. One ticket is selected at random, the families is as follows: find the probability that the number on the ticket is not a perfect square."]