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Statistics Chapter 16: Review B
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1. The owner of a used car lot estimates that 50% of the cars on his lot are worth about $2500,
25% are worth about $5000, 10% are worth about $10,000, and the rest are worth about
$15,000. He is running a sweepstakes in which the winner blindly draws a key to one of these
cars from a fishbowl and gets to keep the car that matches that key.
a. Create a probability model for the random variable of the value of the prize car.
b. What is the expected value of the prize car?
c. What is the standard deviation?
2. Snickers candy bars are running a promotion in which 15% of all candy bars have a coupon for
a free candy bar printed inside the label.
a. What is the probability that you purchase 10 candy bars and find exactly 2 coupons.
b. What is the probability that 5 candy bars in a case of 24 have coupons in them?
c. What is the probability that at least one candy bar in a pack of 6 has a coupon in it?
d. What is the probability that at most 2 in a pack of 6 have coupons in them?
e. What is the probability that you don’t find a coupon until your 10th try?
f. What is the mean and standard deviation of the number of coupons in a pallet of 1200
candy bars?
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
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3. According to a Pew Research study in 2006, 75% of adults prefer to watch movies at home
over going to a theater. A group of theater owners selects a random sample of 200 adults in
their area to see how many prefer to watch movies at home.
a. Verify that a Normal model is a useful approximation for the Binomial in this situation.
b. Assuming the national percentage applies to this area, what is the probability that at least
80% (160 adults) of this sample will report preferring to watch movies at home?
c. Is the percentage of adults in this area who prefer to watch movies at home higher than it is
nationally? What response in this sample would it take for you to be convinced of that?
Explain.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
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Statistics Chapter 16: Review B – KEY
Show your work
1. The owner of a used car lot estimates that 50% of the cars on his lot are worth about $2500,
25% are worth about $5000, 10% are worth about $10,000, and the rest are worth about
$15,000. He is running a sweepstakes in which the winner blindly draws a key to one of these
cars from a fishbowl and gets to keep the car that matches that key.
a. Create a probability model for the random variable of the value of the prize car.
Outcome
$2500
$5000
$10,000
$15,000
Time X
$2500
$5000
$10,000
$15,000
Probability
P(X)
0.5
0.25
0.1
0.15
b. What is the expected value of the prize car? calculated using 1-Var Stats L1,L2
E(X) = 0.5($2500) + 0.25($5000) + 0.1($10000) + 0.15($15000) = $5750
c. What is the standard deviation?
Var ( X ) = (2500 − 5750) (0.50) + (5000 − 5750) (0.25)
2
2
+ (10000 − 5750) (0.1) + (15000 − 5750) (0.15) = $20, 062,500
2
SD(X)
2
= 20062500 = $4479.12 calculated using 1-Var Stats L1,L2
2. Snickers candy bars are running a promotion in which 15% of all candy bars have a coupon for
a free candy bar printed inside the label.
a. What is the probability that you purchase 10 candy bars and find exactly 2 coupons.
P ( X = 2 ) = 10 C2 ( 0.15 ) ( 0.85 ) = 0.2759
2
8
calculated using binompdf(10, 0.15, 2)
b. What is the probability that 5 candy bars in a case of 24 have coupons in them?
P ( X = 5) =
C5 ( 0.15 ) ( 0.85) = 0.1472
5
24
19
calculated using binompdf(24, 0.15, 5)
c. What is the probability that at least one candy bar in a pack of 6 has a coupon in it?
P ( X ≥ 1) = 1 − P ( X = 0 ) = 1 − 6 C0 ( 0.15 ) ( 0.85 ) = 0.6229
0
6
calculated using 1-binompdf(6, 0.15, 0)
d. What is the probability that at most 2 in a pack of 6 have coupons in them?
P ( X ≤ 2 ) = 0.9527
calculated using binomcdf(6, 0.15, 2)
e. What is the probability that you don’t find a coupon until your 10th try?
( 0.85 ) ( 0.15) = 0.0347
9
f. What is the mean and standard deviation of the number of coupons in a pallet of 1200
candy bars?
µ = (1200 )( 0.15 ) = 180 coupons
σ=
(1200 )( 0.15)( 0.85) = 12.3693 coupons
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
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3. According to a Pew Research study in 2006, 75% of adults prefer to watch movies at home
over going to a theater. A group of theater owners selects a random sample of 200 adults in
their area to see how many prefer to watch movies at home.
a. Verify that a Normal model is a useful approximation for the Binomial in this situation.
(200)(0.75) = 150 and (200)(0.25) = 50. Both are larger than 10, so proceed with a Normal model
approximation.
b. Assuming the national percentage applies to this area, what is the probability that at least
80% (160 adults) of this sample will report preferring to watch movies at home?
µ = 150 adults
σ =
(200)(0.75)(0.25) = 6.1237 adults
160 − 150
= 1.633
6.1237
P ( X ≥ 160) ≈ 0.0512 calculated using normalcdf(1.633,4)
z=
c. Is the percentage of adults in this area who prefer to watch movies at home higher than it is
nationally? What response in this sample would it take for you to be convinced of that?
Explain.
The percentage is higher than the national rate by about 1.6 standard deviations. For me to be
convinced that it is significantly higher, I would need to see a value 2 standard deviations or
higher than the national average, or more than 150 + 6.1237(2) = 162.25 adults.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
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