STATISTICS

+1 MATHS IMPROVEMENT CLASS STATISTICS

Measures of Dispersion Range, Quartile deviation, mean deviation, variance, standard deviation are measures of dispersion. Range Range = Maximum Value – Minimum Value Eg, The runs scored by a batsman in their last ten matches as follows. Find range of the data Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 Maximum Value = 117 Minimum Value = 0 Range = 117-0 = 117

Mean Deviation Mean deviation for ungrouped data Let x1 , x2 , x3 , ...., xn be n observations , x̅ and M are mean and median of the data. Mean deviation about the mean is given by MD ( ¯x )= ∑ |xi−¯x| n Mean deviation about the median is given by MD ( M )= ∑ |xi− M| n

Median for ungrouped data Let x1 , x2 , x3 , ...., xn be n observations.Write the observations in a ascending order Case 1 If n is odd n + 1 th 2 Median=( ) observation Case 2 If n is even n th n th 2 2 Median= ( ) observation +( + 1 ) observation 2

Mean deviation for grouped data 1. Discrete frequency distribution Let the given data consist of n distinct values x1 , x2 , ...,.... xn occurring with frequencies f1 , f2 , ....... fn . This data can be represented in the tabular form as given below, and is called discrete frequency distribution (i) Mean deviation about mean MD ( ¯x )= ∑ f i|x i−¯x| , where N ∑¯x=f i xi N N is the sum of the frequencies and

(ii) Mean deviation about median First to find the median of the given discrete frequency distribution For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify the observation whose cumulative frequency is equal to or just greater than N/2, where N is the sum of frequencies. Let M be the median of the data The mean deviation about the median is given by MD ( M )= ∑ f i|xi− M| N

2. Continuous frequency distribution Let x1 , x2 , ...,.... xn be the midpoints of classes of the data with corresponding frequencies f1 , f2 , ....... fn . ie, (i) Mean deviation about mean MD ( ¯x )= ∑ f i|x i−¯x| , where N ∑¯x=f i xi N N is the sum of the frequencies and

(ii) Mean deviation about median First to find the median of the given continuous frequency distribution. For this write the cumulative frequencies of the classes. Then, we identify the class(median class) whose cumulative frequency is equal to or just greater than N/2, where N is the sum of frequencies. If M is the median of the data and l, f, and h are, respectively the lower limit , the frequency, the width of the median class and C the cumulative frequency of the class just preceding the median class. The median is given by The mean deviation about the median is given by MD ( M )= ∑ f i|xi− M| N

Q1:Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12 Ans: Given 6, 7, 10, 12, 13, 4, 8, 12 x 6 7 10 12 13 4 8 12 |xi -x̅ | 3 2 1 3 4 5 1 3 22 MD ( ¯x )= ∑ |xi− ¯x| n=8 22 n 8 MD (¯x )= MD ( ¯x )=2.75

Q2:Find the mean deviation about the median for the following data 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Ans; Arranging the data into ascending order is 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 Median=( 11+ 1 th observation =6th observation =9 2 ) 18 19 21 9 10 12 58 xi 3 3 4 5 7 9 10 12 |xi - 6 6 5 4 2 0 1 3 M| MD ( M )= ∑ |xi− M| n=11 MD ( M )= 58 n ∑ |x i − M|= 58 11 =5.27

Q4: Find mean deviation about the mean for the following data : Ans: N = 40 ∑¯x= f i xi N 300 ¯x = 40 ¯x =7.5 The mean deviation about the mean is given by MD ( ¯x )= ∑ f i|x i− ¯x| = 92 =2.3 40 N

Q2:Find the mean deviation about median for the following data Ans: Now, N=30 which is even

. N =30 ∑ f i|xi−M|=149 MD ( M )= ∑ f i|xi− M| N MD ( M )= 149 30 MD (M )=4.97

Q5:Find the mean deviation about the mean for the following data Ans: N =40 ∑¯x= f i xi N ¯x = 1800 40 =45 MD ( ¯x )= ∑ f i|xi−¯x| = 400 =10 40 N

Q6:Find the mean deviation about median for the following data Ans: l = 20, C = 13, f = 15, h = 10 N = 50 Median M =l+ N −C × h 2 f Class =20 + 25−13 × 10 15 MD ( M )= ∑ f i|xi− M| = 508 =10.16 =20 + 120 50 15 N M =28

Variance and Standard Deviation Standard deviation for ungrouped data Mean of the squares of the deviations from mean is called the variance and is denoted by σ 2 (read as sigma square). The variance of n observations x1 , x2 ,..., xn is given by ∑σ 2= ( xi−¯x )2 n The positive square-root of the variance is called standard deviation and it is denoted by σ The standard deviation of n observations x1 , x2 ,..., xn is given by √ ∑σ = ( xi−¯x)2 n

Standard deviation for ungrouped data 1. Standard deviation of a discrete frequency distribution Let the given discrete frequency distribution be The standard deviation is given by √ ∑σ = f i( xi−¯x)2 N ∑¯x= f i xi where N is the sum of the frequencies and N The variance is given by ∑σ 2= f i(xi−¯x)2 N

2. Standard deviation of a continuous frequency distribution Let x1 , x2 , ...,.... xn be the midpoints of classes of the data with corresponding frequencies f1 , f2 , ....... fn . ie, The standard deviation is given by √ ∑σ = f i( xi−¯x)2 N ∑¯x = f i xi where N is the sum of the frequencies and N The variance is given by ∑σ 2= f i( xi−¯x)2 N

Q7:Find the variance and standard deviation of the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 Ans: n=10 Mean= 150 10 ¯x =15 ( x i−¯x )2 n ∑Variance , σ2= σ 2= 330 10 σ 2=33 Standard deviation , σ =√33 =5.74

Q8:Find the variance and standard deviation for the following data Ans: N =30 ∑¯x= f i xi N ¯x = 420 30 ¯x =14 ∑Variance , σ2= f i (x i−¯x )2 = 1374 =45.8 SD , σ =√45.8 =6.77 N 3

Q9:Calculate the mean, variance and standard deviation for the following distribution Ans: N =50 ∑¯x= f i xi N ¯x = 3100 50 ¯x =62 ∑Variance , σ2= f i( xi−¯x )2 10050 =201 SD , σ=√201 =14.18 N = 50

Analysis of Frequency Distributions The measure of variability which is independent of units is called coefficient of variation and it is denoted as C.V. The coefficient of variation is defined as where σ and x̅ are the standard deviation and mean of the data. For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Note 1. Let x1̅ and σ1 be the mean and standard deviation of the first distribution, and x̅2 and σ2 be the mean and standard deviation of the second distribution. 2. For two series with equal means, the series with greater standard deviation (or variance) is called more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other.

Q10: Two plants A and B of a factory show following results about the number of workers and the wages paid to them. In which plant, A or B is there greater variability in individual wages? Ans: The variance of the distribution of wages in plant A ( σ 1 2 ) = 81 . ‘ . standard deviation of the distribution of wages in plant A ( σ 1 ) = 9 Also, the variance of the distribution of wages in plant B ( σ 2 2 ) = 100 . ‘ . standard deviation of the distribution of wages in plant B ( σ 2 ) = 10 Since the average monthly wages in both the plants is same(Rs.2500), then the plant with greater standard deviation will have more variability. Thus, the plant B has greater variability in the individual wages.

Q11; The following values are calculated in respect of heights and weights of the students of a section of Class XI Can we say that the weights show greater variation than the heights? Ans: Given Variance of height = 127.69 cm 2 Therefore Standard deviation of height = √127.69cm = 11.3 cm Also Variance of weight = 23.1361 kg Therefore Standard deviation of weight = √23 . 1361 kg = 4.81 kg C.V. in heights C.V. in weights Clearly C.V in weights is greater than the C.V. in heights Therefore , we can say that weights show more variability than heights .

Q12:Coefficient of variation of two distributions are 60 and 70, and their standard deviations are 21 and 16, respectively. What are their arithmetic means. Ans: Given C.V.of 1st distribution = 60, σ 1 = 21 C.V. of 2nd distribution = 70, σ 2 = 16 LCet.xV̅ 1.aonf d1sxt ̅d2ibsteritbhuetimone=anσx¯s11o×f 1st and 2nd distribution, σ2 × 100 100 C .V . of 2nd distribution= x¯2 60= 21 × 100 70 = 16 × 100 x¯1 x¯2 x¯1= 21 × 100 x¯2= 16 × 100 60 70 x¯1=35 x¯2=22 . 85

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