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Lesson 6 - 1

Lesson 6 - 1. Discrete Random Variables. 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

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Lesson 6 - 1

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  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 μx = ∑ [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 ux 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: σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2x and standard deviation is √σ2 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: • The probability of each value of the random variable is equal (like in a six-sided die) • 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: μx = ∑ [x ∙P(x)] = (1/n)∑ x Standard Deviation: σ2x = ∑ [(x – μx)2 ∙ P(x)] = (1/n) ∑ [(x – μx)2 = ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx2

  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: ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx2 = (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: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx2 = (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) = 0.9ave repair cost = $900 Yearly cost = $90Expected 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 ∑[x2 ∙ P(x)] – μ2x • Homework • pg 323-327; 7, 10 – 15, 18, 19, 23, 31, 35

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