Graded Practice Set #13 Statistics Name:

1. The probability distribution below is for the random variable X = number of mice caught in trapsduring a single night in small apartment building.

X 0 1 2 3 4 5 P(X) 0.12 0.20 0.31 0.14 0.16 0.07

(a) Make a histogram of this probability distribution in the grid:

(b) Describe ( )2P X ≥ in words and find its value.

(c) Express the event “trapping at least one mouse” in terms of X and find its probability.

2. The total sales on a randomly-selected day at Joy’s Toy Shop can be represented by the continuousrandom variable S, which has a Normal distribution with a mean of $3600 and a standard deviationof $500. Find and interpret ( )$4000P S > .

3. Joe the barber charges $32 for a shave and haircut and $20 for just a haircut. Based on experience,he determines that the probability that a randomly selected customer comes in for a shave andhaircut is 0.85, the rest of his customers come in for just a haircut. Let J = what Joe charges arandomly-selected customer.

(a) Give the probability distribution for J.

(b) Find and interpret the mean of J, Jµ .

(c) Find and interpret the standard deviation of J, Jσ .

4. The probability distribution below is for the random variable X = number of medical testsperformed on a randomly selected outpatient at a certain hospital.

X 0 1 2 3 4 5 P(X) 0.33 0.20 0.18 0.14 0.12 0.03

(a) Make a histogram of this probability distribution in the grid:

(b) Describe ( )3P X ≤ in words and find its value.

(c) Express the event “performing at least two tests” in terms of X and find its probability.

5. The mean height of players in the National Basketball Association is about 79 inches and thestandard deviation is 3.5 inches. Assume the distribution of heights is approximately Normal. LetH = the height of a randomly-selected NBA player. Find and interpret ( )74P H > .

3. Man Hong is running the balloon darts game at the school fair. He has blown up hundreds of balloons with notes about prize tickets inside them. Twelve percent of the notes say “You win 5 tickets,” twenty percent say “You win 3 tickets,” and the rest say “Sorry, try again!” After each play, he replaces the popped balloon with another one bearing the same note. Let T = the number of tickets won by a randomly selected player of this game.

(a) Give the probability distribution for T.

(b) Find and interpret the mean of T, Tµ .

(c) Find and interpret the standard deviation of T, Tσ .

7. A four sided die shaped like an asymmetrical tetrahedron has the following roll probabilities.

Let X = the result of a single roll.

(a) Find ( )1 4P X< <

(b) Find ( )3P X ≠

(c) Describe ( )3| 2P x x= ≥ in words and find its value.

(d) Find the smallest value A for which ( ) 0.6P X A< >

(e) If T = the sum of two rolls, find ( )4P T =

Number on Die 1 2 3 4 Probability 0.4 0.3 0.2 0.1

8. (Continued) Below is a copy of the table from the first page, showing the probability distribution ofX = the number rolled on an asymmetrical four-sided die.

(f) Find and interpret the mean and standard deviation of X.

9. As of December 2008, various polls indicate that 35% of people who use the internet have profileson at least one social networking site. We will discover later in the text that if you take a sample of50 internet users, the proportion of the sample who have profiles on social networking sites can beconsidered a random variable. Moreover, assuming the 35% is accurate, this random variable willbe approximately Normally distributed with a mean of 0.35 and a standard deviation of 0.067.What is the probability that the proportion of a sample of size 50 who have profiles on socialnetworking sites is greater than 0.5?

X 1 2 3 4 P(X) 0.4 0.3 0.2 0.1

©BFW Publishers The Practice of Statistics for AP*, 5/e

Chapter 6 Solutions

Quiz 6.1A 1. (a) See histogram at right. (b) The probability that two or moremice are caught during a single night; 0.68. (c) ( )1 0.88P x ≥ = .

2. ( ) ( )4000 3600$4000 0.8 0.2119.500

P S P z P z−⎛ ⎞> = > = > =⎜ ⎟⎝ ⎠

This is the probability that the total sales on a randomly-selected day exceeds $4000. 3. (a)

(b) µJ = the mean amount of money Joe can expect to make percustomer in the long run = ( ) ( )20 0.15 32 0.85 $30.20.+ =

(c) σJ = the “typical” distance from the mean ($30.20) for eachindividual customer =

( ) ( )2 20.15 20 30.20 0.85 32 30.20 $4.28.− + − =

Quiz 6.1B 1. (a) See histogram at right. (b) The probability that an outpatientundergoes no more than 3 medical tests; 0.85. (c) ( )2 0.47P x ≥ = .

2. ( ) ( )74 7974 1.43 0.9236.3.5

P H P z P z−⎛ ⎞> = > = > − =⎜ ⎟⎝ ⎠

This

is the probability that a randomly-selected NBA player’s height is greater than 74 inches. 3.

(b) µT = the mean number of tickets won per contestant in the longrun = ( ) ( ) ( )0 0.68 3 0.2 5 0.12 1.2 tickets.+ + = (c) σT = the “typical”distance from the mean (1.2) for each individual contestant =

( ) ( ) ( )2 2 20.68 0 1.2 0.2 3 1.2 0.12 5 1.2 1.833 tickets.− + − + − =

0 1 2 4 3 5

0.1

0.2

0.3

0 1 2 4 3 5

0.1

0.2

0.3

J 20 32 P(J) 0.15 0.85

T 0 3 5 P(T) 0.68 0.20 0.12

©BFW Publishers The Practice of Statistics for AP*, 5/e

Quiz 6.1C 1. (a) ( ) ( )1 4 2 or 3 0.3 0.2 0.5P X P X X< < = = = = + = . (b) ( ) ( )3 1 3 1 0.2 0.8P X P X≠ = − = = − =

(c) The probability of rolling a 3, given that the roll is 2 or greater; 0.2 10.6 3

= .

(d) ( )3 0.4 0.3,P X < = + so A = 3. (e) The event T = 4 can happen three ways:{1, 3}, {3, 1}, and {2, 2}. Probabilities for these events are, respectively,( )( ) ( )( ) ( )( )0.4 0.2 0.08; 0.2 0.4 0.08; and 0.3 0.3 0.09.= = = Hence the total probability is 0.25.

(f) ( ) ( ) ( ) ( )1 0.4 2 0.3 3 0.2 4 0.1 2Xµ = + + + = ;

( ) ( ) ( ) ( )2 2 2 20.4 1 2 0.3 2 2 0.2 3 2 0.1 4 2 1Xσ = − + − + − + − = Xµ is the expected mean roll if the

die is rolled many times, or the expected long-run value of a single roll. Xσ is the typical distance

each roll is from the mean roll. 2. ( ) ( )0.5 0.350.5 2.24 0.0125.0.067

P X P z P z−⎛ ⎞> = > = > =⎜ ⎟⎝ ⎠