5 -Minute Check on section 7 -1 b

5 -Minute Check on section 7 -1 b 1. What are the two AP continuous probability distributions? Normal and uniform 2. If x is a continuous probability distribution, what is P(x=2)? P(x = 2) = 0 (height times a point width) If x is a normally distributed random variable with a mean of 6 and standard deviation of 2, find the following: 3. P(x < 4) = normalcdf(-e 99, 4, 6, 2) = 4. P(x > 5) = normalcdf(5, e 99, 6, 2) = If x is a continuous uniform variable from [2, 6], find 5. P(x < 5) = (1/(6 – 2)) (5 – 2) = (0. 25) (3) = 0. 75 6. P(x > 4) = (1/(6 – 2)) (4 – 2) = (0. 25) (2) = 0. 50 Click the mouse button or press the Space Bar to display the answers.

Lesson 7 – 2 a Means and Variances of Random Variables

Knowledge Objectives • Define what is meant by the mean of a random variable • Explain what is meant by a probability distribution • Explain what is meant by a uniform distribution • Discuss the shape of a linear combination of independent Normal random variables

Construction Objectives • Calculate the mean of a discrete random variable. • Calculate the variance and standard deviation of a discrete random variable. • Explain, and illustrate with an example, what is meant by the law of large numbers. • Explain what is meant by the law of small numbers. • Given µx and µy, calculate µa+bx, and µx+y. • Given x and y, calculate 2 a+bx, and 2 x+y (where x and y are independent). • Explain how standard deviations are calculated when combining random variables.

Vocabulary • Mean – balance point of the probability histogram or density curve. Symbol: μx • Standard Deviation – square root of the variance. Symbol: x • Variance – is the average squared deviation of the values of the variable from their mean. Symbol: σ²x

Means • Mean of a random variable is also known as its expected value

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

Variance

Discrete Random Variable - Variance and Standard Deviation of a Discrete Random Variable: The variance of a discrete random variable is given by: σ2 x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x 2 ∙ P(x)] – μ 2 x 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

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)

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: σ2 x = ∑ [(x – μx)2 ∙ P(x)] = (1/n) ∑ [(x – μx)2 = ∑[x 2 ∙ P(x)] – μ 2 x = (1/n) ∑ [x 2 ] – μx 2

Using your TI-83 calculator We can use 1 -Var-Stats to calculate the mean and standard deviation of a discrete random variable given it’s outcomes and probability • Type in outcomes (x values) in L 1 • Type in corresponding probabilities in L 2 • Use 1 -Var-Stats L 1, L 2 to get statistics We can graph the probability histograms by changing the frequency to L 2

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 ] – μx 2 = (1/10) (385) - 30. 25) = (38. 5 – 30. 25) = 8. 25 St Dev = 2. 8723

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)] – μx 2 = (0 +. 3 +. 4(4) +. 15(9) +. 05(16) ) – 3. 0625) = 4. 05 – 3. 8626 = 0. 9875 St Dev = 0. 9937

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

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

Summary and Homework • Summary – Expected value is the mean ∑ [x ∙ P(x)] – Variance is ∑[x 2 ∙ P(x)] – μ 2 x – Standard Deviation is variance – Use your calculator! • Homework – pg 486 -7; 7. 24 – 7. 30