Random_Variables_and_Probability_Distribution

������ ˏˋ°•*⁀➷Random Variables and Probability Distribution Random Variables (Fair and Biased Experiments) VARIABLES It is a characteristic or attribute that can assume different values. Examples: age, gender, height, weight. Capital letters are used to denote or represent variables that are associated with probabilities. RANDOM VARIABLES These are variables that are associated with probabilities. It is a function that associates a real number to each element in the sample space. It is a variable whose values are determined by chance. FAIR EXPERIMENTS (Equally Likely Outcomes) ˏˋ°•*⁀➷Random Variables and Probability Distribution 1

Example 1: Testing of Cellphones Suppose three cellphones are tested at random. We want to find out the number of defective cellphones that occur. Thus, for each outcome in the sample space, we shall assign a value. If there is no defective cellphone, we assign the number 0 and so on. The number of defective cellphones is a random variable. The possible values of this random variable are 0, 1, 2, and 3. Possible Outcomes: (N - Non-defective, D - Defective) NNN, NND, NDN, DNN, DDN, DND, NDD, DDD. Value of the Random Variable X (no. of defective cellphone): 0, 1, 1, 1, 2, 2, 2, 3. ˏˋ°•*⁀➷Random Variables and Probability Distribution 2

The random variable is the capital letter X. The value of the random variable (0, 1, 2, 3) is a smaller x. Example 2: Tossing of Three Coins Suppose three coins are tossed. Let Y be the random variable representing the number of tails occur. Find the values of the random variable Y. Prepare the table. The sample space of getting a tail or a head when three coins are tossed: S = {HHH, HTH, HHT, THH, TTH, THT, HTT, TTT} SAMPLE SPACE TABLE Value of the Random Variable Y (number of tails) Possible Outcomes (H - Head, T - 0 Tail) 1 HHH 1 HTH 1 HHT 2 THH 2 TTH 2 THT 3 HTT TTT Untitled ˏˋ°•*⁀➷Random Variables and Probability Distribution 3

The random variable is the capital letter Y. The value of the random variable (0, 1, 2, 3) is a smaller y. BIASED EXPERIMENTS (Non-equally Likely Outcomes) ⚙ nCr = n!/(r!(n − r)! Combination Formula Example 3: Drawing balls from an urn Three balls are drawn in succession without replacement from an urn containing 5 red balls and 6 blue balls. Let Z be the random variable representing the number of blue balls. Find the values of the random variable Z. Possible outcomes are general description of an occurring event. Recall that Combination is a grouping of terms in which order does not matter. The number of combinations of n things taken r at a time is denoted nCr. This ˏˋ°•*⁀➷Random Variables and Probability Distribution 4

can be solved using your calculator. 5C3 = 10 5C2 ⋅6 C1 = 60 5C1 ⋅6 C2 = 75 6C3 = 20 The total number of possible outcomes (N): 11C3 = 165 Z={0, 1, 2, 3} Example 4: Testing of Computers Suppose that 3 computers are tested from a computer laboratory containing 18 computers of which 6 are malfunctioning. Let M be the random variable representing the number of malfunctioning computers. Find the values of the random variable M. Recall that Combination is a grouping of terms in which order does not matter. The number of combinations of n things taken r at a time is denoted nCr. This can be solved using your calculator. ˏˋ°•*⁀➷Random Variables and Probability Distribution 5

12C3 = 220 12C2 ⋅6 C1 = 396 12C1 ⋅6 C2 = 180 6C3 = 20 Total number of possible outcomes (N): 18C3 = 816. M={0, 1, 2, 3} Probability Distribution of a Discrete Random Variable TYPES OF RANDOM VARIABLE 1. Discrete Random Variable This set of possible outcomes is countable. Mostly, discrete random variables represent count data. Examples: the number of defective chairs produced in a factory; number of students 2. Continuous Random Variable It takes on values on a continuous scale. Often, continuous random variables represent measured data. Example: height, weight, and temperature Probability Distribution of the Discrete Random Variable The set of of ordered pairs (x, f(x)) is a probability function, probability mass function, or probability distribution of the discrete random variable X if, for each possible outcome x, Notation Description P (x) = P (X = x) = f (x) is the probability associated with each value of x of the random f (x) variable X ˏˋ°•*⁀➷Random Variables and Probability Distribution 6

Notation Description 0 ≤ f(x) ≤ 1 the value of the probabilityf (x) is between 0 and 1 inclusive Σf(x) = 1 the sum of all the probabilities of x of the random variable X is equal to 1 DISCRETE PROBABILITY DISTRIBUTION (Fair Experiment) A discrete random variable assumes each of its values with a certain probability. Example 1: Consider the experiment of tossing a coin three times. Suppose the variable X represents the number of heads that occur. The sample of space of getting a tail or head when three coins are tossed: S={HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} 1. Prepare the table to find the values of the random variable. Sample Space x HHH 3 HHT 2 HTH 2 THH 2 TTH 1 THT 1 HTT 1 TTT 0 2. Use the values on the first table to compete the probability distribution. X frequency Probability P(x) Probability P(x) fraction decimal ˏˋ°•*⁀➷Random Variables and Probability Distribution 7

X frequency Probability P(x) Probability P(x) 01 1/8 0.125 13 3/8 0.375 23 3/8 0.375 31 1/8 0.125 Total 8 1 1 Suppose the variable x represents the number of heads that occur, we have: For x = 0, the probability is 1⁄8 → P(0) = 1⁄8 For x = 1, the probability is 3⁄8 → P(1) = 3⁄8 For x = 2, the probability is 3⁄8 → P(2) = 3⁄8 For x = 3, the probability is 1⁄8 → P(3) = 1⁄8 Recall that the probability of an event considers the number of occurrences of an event relative to the number of possible outcomes. ⚙ P(X) = no. of favorable outcomes (frequency) / total no. of possible outcomes (total) ⚙ Σ f(x)=1 → P(0) + P(1) + P(2) + P(3) = 1/8 + 3/8 + 3/8 + 1/8 = 8/2 or 1; f(x) is the same with P(x). DISCRETE PROBABILITY DISTRIBUTION (Biased Experiment) 8 Example 2: Drawing balls from an urn Three balls are drawn in succession without replacement from an urn containing 5 red balls and 6 blue balls. Let Z be the random variable representing the number of blue balls. Find the values of the random variable Z. 1. Prepare the table to find the values of the random variable. ˏˋ°•*⁀➷Random Variables and Probability Distribution

Possible Outcomes (5 red balls, 6 blue Value of the Random Variable Z (number of blue balls) balls) All are red 0 Two red, one blue 1 Two blue, one red 2 All are blue 3 5C3 = 10 5C2 ⋅6 C1 = 60 5C1 ⋅6 C2 = 75 6C3 = 20 The total number of possible outcomes (N): 11C3 = 165 2. Use the values on the first table to complete the probability distribution. Random Variable Values Z Frequency (freq) Probability P(z) 0 10 10/165 1 60 60/165 2 75 75/165 3 20 20/165 Total 165 1 P(z) = frequency / total Mean, Variance, and Standard Deviation 9 FINDING THE MEAN AND THE VARIANCE Just like the frequency distribution, the probability distribution can be described by computing its mean and variance. The mean or the expected value of a probability distribution tells the value of a random variable that is expected to be obtained if the experiment is done ˏˋ°•*⁀➷Random Variables and Probability Distribution

repeatedly. In other words, if the experiment or process is repeated long enough, there is likelihood or chance that the average of the outcomes will begin to approach the expected value of the mean value. ⚙ Mean of a Discrete Probability Distribution - To find the mean µ or expected value E(X) of a discrete probability distribution, we use the following formula: µ = E(x) = Σx ⋅ P (x) Example 1: Find the mean of the discrete probability distribution X with the following probability distribution. STEP 1. Multiply the value of X by its corresponding probability value P(x). Write the product on the third column. x P(x) x * P(x) 3 1/8 3/8 2 3/8 6/8 1 3/8 3/8 0 1/8 0 TOTAL 1 12/8 ˏˋ°•*⁀➷Random Variables and Probability Distribution 10

STEP 2. Find the mean or the expected value of the probability distribution by getting the sum of the values. µ = E(x) = Σx ⋅ P (x) µ = 12/8 = 1.50 ������ ANSWER: The sum of the values in the third column is 12/8 or 1.50. Hence, it is the mean of the probability distribution. Example 2: Find the mean or the expected value of the probability distribution Y shown below. STEP 1: Multiply the value of y by its corresponding probability value P(y). y P(y) y * P(y) 2 0.042 0.084 3 0.010 0.03 4 0.021 0.084 5 0.375 1.875 6 0.188 1.128 7 0.344 2.408 8 0.021 0.168 STEP 2: Find the mean or the expected value of the probability distribution by getting the sum of the values. ������ ANSWER: The sum of the values in the third column is 5.778 or 5.78. Hence, it is the mean of the probability distribution. µ = 5.78 ˏˋ°•*⁀➷Random Variables and Probability Distribution 11

VARIANCE AND STANDARD DEVIATION OF DISCRETE PROBABILITY DISTRIBUTION To find the variance σ2 standard deviation σ of a discrete probability distribution, we use the following formulas: ⚙ VARIANCE: σ2 = [Σx2 ⋅ P (x)] − µ2 STANDARD DEVIATION: σ= [Σx2 ⋅ P (x)] − µ2 Note: The standard deviation is just the square root of the variance. Example 3: Find the variance and the standard deviation of the discrete probability distribution X with the following probability distribution. STEP 1. Find the value of the mean by following the steps on the previous examples. x P (x) x ⋅ P (x) x2 x2 ⋅ P (x) 3 1/8 3/8 9 9/8 2 3/8 6/8 4 12/8 1 3/8 3/8 1 3/8 0 1/8 0 00 TOTAL 1 12/8 24/8 µ = 1.50 STEP 2. Find the values of x2. Write them on the next column. STEP 3. Multiply the value of x2 by its corresponding probability valueP (x). STEP 4. Find the sum of the product of x2 and P (x). STEP 5. Substitute the solved values from the formula. ˏˋ°•*⁀➷Random Variables and Probability Distribution 12

������ Mean: (µ) = 1.50 ; x2 ⋅ P (x) = 24 8 VARIANCE σ2 = [Σx2 ⋅ P (x)] − µ2 σ2 24 (1.50)2 σ2 = 8 − = 3 − 2.25 σ2 = 0.75 STANDARD DEVIATION σ = σ2 σ = 0.75 σ = 0.8660 The standard deviation is just the square root of the variance. Example 4: Probability distribution of the number of computers sold daily in a computer shop during the past several months: Number of Computers (x) Probability P(x) x ⋅ P (x) x2 x2 ⋅ P (x) 0 0.10 0 00 1 0.20 0.20 1 0.20 2 0.45 0.90 4 1.80 3 0.15 0.45 9 1.35 4 0.05 0.20 16 0.80 5 0.05 0.25 25 1.25 Total 1 2.00 5.40 ˏˋ°•*⁀➷Random Variables and Probability Distribution 13

������ µ = 2.00 (Mean) 14 σ = [Σx2 ⋅ P (x)] − µ2 σ = 5.40 − (2.00)2 σ = 1.40 σ = 1.1832 (Standard Deviation) What is the probability that on a given day— A. fewer than three computers will be sold? ������ P(X < 3) = 0.75 B. at most, two computers will be sold? ������ P(X ≤ 2) = 0.75 C. at least, four computers will be sold? ������ P(X ≥ 4) = 0.10 D. exactly three computers will be sold? ˏˋ°•*⁀➷Random Variables and Probability Distribution

������ P(X = 3) = 0.15 What is the mean of the probability distribution? ������ µ = 2.00 What is the standard deviation of the probability distribution? ������ σ = 1.1832 ˏˋ°•*⁀➷Random Variables and Probability Distribution 15