1. Lesson 6 - 1 Discrete Random Variables

2. Objectives Distinguish between discrete and continuous random variables Identify discrete probability distributions Construct probability histograms Compute and interpret the mean of a discrete random variable Interpret the mean of a discrete random variable as an expected value Compute the variance and standard deviation of a discrete random variable

3. Vocabulary Random variable – a numerical measure of the outcome of a probability experiment, so its value is determine by chance. Discrete random variable – has finite or countable number of values Continuous random variable – has infinitely many values Probability distribution of a discrete random variable – provides all possible values of the random variable and their corresponding probabilities (can be in the form of a table or graph – or a mathematical formula) Probability histogram – histogram with y values being probability and x axis being the random variable (similar to relative frequency histogram) Expected value –the mean of a random variable, E(x) Variance of a discrete random variable – weighted average of the squared deviations where the probabilities are the weights

4. Rules for a Discrete Probability Distribution Let P(x) denote the probability that the random variable X equals x, then The sum of all probabilities of all outcomes must equal 1 ∑ P(x) = 1 The probability of any value x, P(x), must between 0 and 1 0≤ P(x) ≤ 1

5. Discrete Random Variable - Mean The mean, or expected value [E(x)], of a discrete random variable is given by the formula μ x = ∑ [x ∙P(x)] where x is the value of the random variable and P(x) is the probability of observing x Mean of a Discrete Random Variable Interpretation: If we run an experiment over and over again, the law of large numbers helps us conclude that the difference between x and u x gets closer to 0 as n (number of repetitions) increases

6. Discrete Random Variable - Variance Variance and Standard Deviation of a Discrete Random Variable: The variance of a discrete random variable is given by: σ 2 x = ∑ [(x – μ x ) 2 ∙ P(x)] = ∑[x 2 ∙ P(x)] – μ 2 x and standard deviation is √σ 2 Note: round the mean, variance and standard deviation to one more decimal place than the values of the random variable

7. Uniform PDF An experiment is said to be a Uniform experiment provided: 1.The probability of each value of the random variable is equal (like in a six-sided die) 2.The trials are independent of each other (what happened last does not affect what happens next)

8. Uniform PDF If X is a value of the uniform random variable, then probability formula for X is 1 P(x) = ------- x = 0, 1, 2, 3, …, n n where n is the total number of discrete values of the random variable x Mean: μ x = ∑ [x ∙P(x)] = (1/n)∑ x Standard Deviation: σ 2 x = ∑ [(x – μ x ) 2 ∙ P(x)] = (1/n) ∑ [(x – μ x ) 2 = ∑[x 2 ∙ P(x)] – μ 2 x = (1/n) ∑ [x 2 ] – μ x 2

9. Example 1 You have a fair 10-sided die with the number 1 to 10 on each of the faces. Determine the mean and standard deviation. Mean: ∑ [x ∙P(x)] = (1/10) (∑ x) = (1/10)(55) = 5.5 Var: ∑[x 2 ∙ P(x)] – μ 2 x = (1/n) ∑ [x 2 ] – μ x 2 = (1/10) (385) - 30.25) = (38.5 – 30.25) = 8.25 St Dev = 2.8723

10. Example 2 Below is a distribution for number of visits to a dentist in one year. X = # of visits to a dentist x 0 1 2 3 4 P(x).1.3.4.15.05 Determine the expected value, variance and standard deviation. Mean: ∑ [x ∙P(x)] = (.1)(0) + (.3)(1) + (.4)(2) + (.15)(3) + (.05)(4) = 0 +.3 +.8 +.45 +.2 = 1.75 Var: ∑[x 2 ∙ P(x)] – μ 2 x = ∑ [x 2 ∙ P(x)] – μ x 2 = (0 +.3 +.4(4) +.15(9) +.05(16) ) – 3.0625) = 4.05 – 3.8626 = 0.9875 St Dev = 0.9937

11. Example 3 What is the average size of an American family? Here is the distribution of family size according to the 1990 Census: # in family 2 3 4 5 6 7 p(x).413.236.211.090.032.018 Mean: ∑ [x ∙P(x)] = (.413)(2) + (.236)(3) + (.211)(4) + (.09)(5) + (.032)(6) + (.018)(7) =.826 +.708 +.844 +.45 +.192 +.126 = 3.146

12. Example 4 You are trying to decide whether to take out a $250 deductible which will cost you $90 per year. Records show that for this community the average cost of repair is $900. Records also show that 10% of the drivers have an accident during the year. If you have sufficient assets so that you will not be financially handicapped if you had to pay out the $900 or more for repairs, should you buy the policy? P(accident) = 0.1 so P(no accident) = 0.9 ave repair cost = $900 Yearly cost = $90 Expected Yearly with Policy: ∑ [x ∙P(x)] = (.1)(250) + (.9)(0) + 90 = 25 + 90 = $115 Expected Yearly without: ∑ [x ∙P(x)] = (.1)(900) + (.9)(0) = $90

13. Summary and Homework Summary –Discrete variables have a finite number of values –Expected value is the mean ∑ [x ∙P(x)] –Variance is ∑[x 2 ∙ P(x)] – μ 2 x Homework –pg 323-327; 7, 10 – 15, 18, 19, 23, 31, 35