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Have you started the homework. Yes No With a little help from a friend. Upcoming In Class. Sunday Homework 6 Quiz 3 – Sept. 25th Exam 1 – October 9th. Chapter 14. Random Variables. Everything. More About Means and Variances. Adding or subtracting a constant from data
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Have you started the homework • Yes • No • With a little help from a friend
Upcoming In Class • Sunday Homework 6 • Quiz 3 – Sept. 25th • Exam 1 – October 9th
Chapter 14 Random Variables
More About Means and Variances • Adding or subtracting a constant from data E(X ± c) = E(X) ± c Var(X ± c) = Var(X) • Multiplying each value of a random variable by a constant multiplies the mean by that constant and the variance by the square of the constant: E(aX) = aE(X) Var(aX) = a2Var(X)
More About Means and Variances (cont.) • In general, • The mean of the sum of two random variables is the sum of the means. • The mean of the difference of two random variables is the difference of the means. E(X ± Y) = E(X) ± E(Y) • If the random variables are independent, the variance of their sum or difference is always the sum of the variances. Var(X ± Y) = Var(X) + Var(Y)
Insurance Policies • An insurance company estimates that it should make an annual profit of $130 on each homeowner’s policy, with a standard deviation of $4,000.
What is the mean and SD if the company write 3 policies? • Mean=3*130SD=sqrt(3*4,000) • Mean=130+130+130SD=sqrt(3*4,0002) • Mean=3*130SD=3*4,000 • Mean=130+130+130SD=3*4,0002
What is the mean and SD if the company write 10,000 policies? • Mean=10,000*130SD=sqrt(10,000*4,0002) • Mean=130+130+130SD=sqrt(10,000*4,0002) • Mean=10,000*130SD=10,000*4,000 • Mean=130+130+130SD=10,000*4,000
What assumption did you make? • The annual profit on a policy is a continuous random variable • Losses are dependent • Losses are independent of each other • The annual profit on a policy is a discrete random variable.
Will the company be profitable? • No. The variance is larger than the mean. • Yes, $0 is 3.25 standard deviations below the mean for 10,000 policies • Yes, the expected value is greater than zero. • No. Catastrophes are far too unpredictable to expect a profit.
Applying the normal model • Remember the z-score involves a mean and standard deviation, • So applying our expected value and standard deviation of our random variable…
Delivery Problem • A delivery company’s trucks occasionally get parking tickets, and based on past experience the company plans that the trucks will average 1.2 tickets a month, with a standard deviation of 0.7 tickets. • Suppose they have 17 trucks…
What is the expected number of tickets for 17 trucks? • 1.2 • 11.9 • 17 • 20.4
What is the ST. Dev. Of parking tickets for 17 trucks? • 0.7 • 2.89 • 11.9 • 17
What would be an unusually bad month for the company? Assume that a variation of 2 St. Dev. Is unusual. • 22 • 25 • 27 • 30
More About Means and Variances • In general, • The mean of the sum of two random variables is the sum of the means. • The mean of the difference of two random variables is the difference of the means. E(X ± Y) = E(X) ± E(Y) • If the random variables are independent, the variance of their sum or difference is always the sum of the variances. Var(X ± Y) = Var(X) + Var(Y)
Cereal Bowls • The amount of cereal that can be poured into a bowl varies… • Large Bowl • E(XL)=2.7 oz • SD(XL)=0.2 oz • Small Bowl • E(XS)=1.7 oz • SD(XS)=0.2 oz • You open a new box of cereal and pour one large bowl and one small bowl.
How much more cereal do you expect to eat by using the large bowl? • On average 2.7oz • On average 1.7oz • On average 2.7+1.7oz • On average 2.7-1.7oz
What is the SD of the difference between the large and small bowls? • SD=sqrt(0.2-0.2) • SD=sqrt(0.2+0.2) • SD=sqrt(0.22-0.22) • SD=sqrt(0.22+0.22)
Assuming normality, what is the prob. that the small bowl contain more cereal? • -3.57 • -3.25 • 0.0002 • 0.9998
What is the mean of the total amount in the two bowls? • Mean=2.7 • Mean=1.7 • Mean=2.7+1.7 • Mean=2.7+1.7
What is the SD of the total amount in the two bowls? • SD=sqrt(0.2-0.2) • SD=sqrt(0.2+0.2) • SD=sqrt(0.22-0.22) • SD=sqrt(0.22+0.22)
Assuming normality, what’s the prob. you poured more than 4.7 oz of cereal total? • 1.0714 • -1.0714 • .8577 • .1423
More Cereal, More Random Variables • The amount of cereal the manufacturer puts in the boxes is a random variable with • Mean=16.4 • SD=0.2
What is the expected Value? • 16.4-2.7 • 16.4-1.7 • 16.4+2.7+1.7 • 16.4-2.7-1.7
What is the Standard Deviation? • SD=sqrt(0.2-0.2-0.2) • SD=sqrt(0.2+0.2+0.2) • SD=sqrt(0.22-0.22-0.22) • SD=sqrt(0.22+0.22+0.22)
Assuming normality, what’s the prob. The box has more than 11 oz of cereal total? • 1.0714 • -1.0714 • .8577 • .1423
Applying the normal model • Remember the z-score involves a mean and standard deviation, • So applying our expected value and standard deviation of our random variable…
Coffee Problem • At a certain coffee shop, all the customers buy a cup of coffee and some also buy a doughnut. • The shop owner believes that the number of cups she sells each day is ~N(340, 23) • And the number of doughnuts is independent of the number of cups • And the number of doughnuts is ~N(160, 11) • The shop is open every day but Sunday.
What’s the probability it will sell over 2000 cups of coffee in a week? • -0.71 • 0.71 • 0.7611 • 0.2389
More coffee… • Suppose she makes a profit of $.50 on each cup of coffee and $.40 on each doughnut
Can she expect to make a profit of over $300 in one day? • Yes, her expected profit is over $300 • No, her expected profit is under $300 • No, $300 is more than 5 standard deviations above the mean. • Yes, $300 is more than 5 standard deviations below the mean.
What the probability that in one day she will sell a doughnut to more than half his coffee customers? • 0.735 • 0.628 • 0.5 • 0.265
Upcoming In Class • Sunday Homework 5 • Quiz 3 – Sept. 25th • Exam 1 – October 9th