Download Testbank 5

AP Statistics Testbank 5
Name ____________________________________ Date _____________________ Period __________
Multiple-Choice Questions
1) Suppose that an NBA basketball player has a field goal percentage of 44% (that is, on average, he makes
44% of his shots). In a given game he will take 20 shots. Then the mean ( μ ) and standard deviation ( σ ) of
the number of shots that he will make is
a) μ = 12.2, σ = 2.22
b) μ = 8.8, σ = 2.22
c) μ = 8.8, σ = 4.93
d) μ = 12.2, σ = 4.93
e) μ = 4.4, σ = 2.22
2) Suppose that a television game has three payoffs with the following probabilities:
Payoff($)
0
200
2,000
Probability
.7
.25
.05
What are the mean ( σ
a) μ = $150, σ
b) μ = $150, σ
c) μ = $150, σ
d) μ = $2,000, σ
e) μ = $150, σ
) and standard deviation ( σ ) of the payoff variable?
= $50
= $187,500
= $433.02
= $433.02
= $433.02
3) A company manufactures parts for public-transit busses. It is known that on average, only about .5% of
these parts are defective. These parts are shipped in large quantities, but before the shipment is made, 20
parts are selected at random and inspected. If at most one of the parts is seen the be defective, the shipment
is approved and the parts are sent. What is the probability that any given shipment will be approved?
a) 0.4%
b) 99.6%
c) 73.6%
d) 98.3%
e) 90.5%
4) In Problem 3, assume that the company discontinues the inspection of its parts, and sends them out in
shipments of 1,000. Compute the expected number and standard deviation of the number of defective parts
in any given shipment.
a) μ = 50, σ = 7.05
b) μ = 50, σ = 2.23
c) μ = 5, σ = 4.97
d) μ = 5, σ = 2
e) μ = 5, σ = 2.23
5) Billy often finds it difficult to arrive to class on time. In fact, the attendance records show that he is tardy for
his statistics class approximately 38% of the time. Let x be the number tardies that Billy accumulates during
a school year of 100 classes. Compute the mean ( μ ) and standard deviation ( σ ) of x.
a) μ = 38, σ = 3.82
b) μ = 3.8, σ = 3.82
c) μ = 62, σ = 4.85
d) μ = 38, σ = 4.85
e) μ = 3.8, σ = 4.85
6) We continue with Billy's record of tardiness. Compute
a) 64.6%
b) 63.5%
c) 69.4%
d) 58.7%
e) Impossible to calculate.
P(μ − σ ≤ x ≤ μ + σ ) .
7) In Problem 6 above, replace the binomial distribution with a normal distribution with mean μ and variance
σ 2 , compute
a) 65.6%
b) 68.3%
c) 95%
d) 99.7%
e) 34.4%
P(μ − σ ≤ x ≤ μ + σ ) .
8) Suppose that you are going to flip a fair coin 40,000 times. Use the normal approximation to this binomial
distribution to approximate P ( μ − σ ≤ x ≤ μ + σ ) .
a)
b)
c)
d)
e)
70.3
68.3
31.7
68.5
31.5
9) Suppose that an insurance company charges $1,000 annually for car insurance. The policy specifies that the
company will pay $1,100 for a minor accident and $5,000 for a major accident. Assume that during a given
year, your likelihood of having a minor accident is 20% and your likelihood of having a major accident is
5%. How much profit per year does the insurance company expect to make from each client?
a) $500
b) $510
c) $520
d) $530
e) $540
10) Suppose that it has been determined that if x measures the number of minutes that Billy is late for his
statistics class, then x is a uniformly distributed random variable with − 2 ≤ x ≤ 10 (thus, it's actually
possible for Billy to arrive early!). Based on this, on any given day, we expect Billy to arrive
a) on time.
b) 2 minutes late.
c) 4 minutes late.
d) 6 minutes late.
e) not at all.
11) Continuing on the theme of Problem 10, the probability that Billy will arrive late is about
a) 17%
b) 50%
c) 83%
d) 97.5%
e) 100%
12) Based on the above assumptions about Billy's tardiness, the probability that he arrives more than 10
minutes late for his statistics class is
a) 0%
b) 20%
c) 50%
d) 97.5%
e) 100%
13) Based on the above assumptions about Billy's tardiness, the probability that he arrives early for class is
about
a) 17%
b) 50%
c) 83%
d) 97.5%
e) 100%
14) Suppose that it has been determined that if x measures the number of minutes that Billy is late for his
statistics class, then x is a normally distributed random variable with a mean of 4 minutes and a standard
deviation 2 minutes. Based on this, the probability that Billy will arrive late is about
a) 17%
b) 50%
c) 83%
d) 97.7%
e) 100%
15) Based on the assumptions in Problem 14 about Billy's tardiness, the probability that he arrives late but no
more than 8 minutes late is approximately
a) 50%
b) 68%
c) 95%
d) 99.7%
e) 100%
16) Suppose that if y is the number of minutes that Kaz is late for his statistics class, then y is also a normally
distributed random variable with a mean of 4 minutes and a standard deviation 2 minutes. If we assume that
the random variables x and y are independent, then the probability that on a given day, both Billy and Kaz
arrive late for class is
a) 97.7%
b) 95.5%
c) 90.2%
d) 1
e) 4.5%
17) Suppose that we have data whose histogram is symmetrical about the mean and is roughly "bell shaped."
Assume that the mean of the data is m, that the sample deviation is s, and that roughly 49% of the data falls
between m − s and m + s . Then
a) The data very likely comes from a normal distribution.
b) The data might have come from a normal distribution, but is highly concentrated about the mean.
c) It is unlikely that the data comes from a normal distribution since those data in the range m ± s is much
less than what would be predicted by a normal distribution.
d) As soon as we know that the general shape of the data is bell shaped, we are guaranteed that it comes
from a normal distribution.
e) None of the above responses is correct.
18) Assume that a binomial distribution is given with mean μ and standard deviation σ . Assume further that
0 ≤ μ − 3σ < μ < μ + 3σ ≤ n , where n is the number trials for the binomial random variable. Then
a) The spread of the binomial data is very wide.
b) It is highly unlikely that a normal distribution could be used to model the binomial data.
c) A normal distribution with mean μ and variance σ 2 will approximate the given binomial distribution.
d) A uniform distribution might be a better continuous approximation to the given binomial distribution.
e) None of the above responses is correct.
19) Assume that we have a binomial distribution with n = 25 , and with a success probability of p = .75 . Then
the normal distribution with mean μ = 75 / 4 and variance σ 2 = 75 / 16 is not a good model because
a) μ − 3σ < 0
b) μ + 3σ > 0
c) μ + 3σ ≤ 25
d) μ + 3σ > 25
e) The binomial distribution is not symmetric.
20) Assume that x is a binomial random variable with mean μ and standard deviation σ . Assume that n is a
normal random variable also with mean μ and standard variation σ . If k is a positive integer and if we take
into account the "correction for continuity," then the normal approximation to the probability P ( x ≤ k ) is
given by
a) P ( x ≤ k ) ≈ P (n ≤ k )
b) P ( x ≤ k ) ≈ P ( n ≤ k − 0.5)
c) P ( x ≤ k ) ≈ P ( n ≤ k + 0.75)
d) P ( x ≤ k ) ≈ P ( n ≤ k + 0.5)
e) P ( x ≤ k ) ≈ P ( n ≤ k − 0.5)
21) Assume that x is a binomial random variable with mean μ and standard deviation σ . Assume that n is a
normal random variable also with mean μ and standard variation σ . If k and l are two positive integers and
if we take into account the "correction for continuity," then the normal approximation to the probability
P (l ≤ x ≤ k ) is given by
a) P (l ≤ x ≤ k ) ≈ P (l ≤ n ≤ k )
b) P (l ≤ x ≤ k ) ≈ P (l − 0.5 ≤ n ≤ k − 0.5)
c) P (l ≤ x ≤ k ) ≈ P (l − 0.5 ≤ n ≤ k + 0.5)
d) P (l ≤ x ≤ k ) ≈ P (l − 0.5 ≤ n ≤ k )
22) Consider the binomial random variable x, with n=1000 and p=.4. Using the normal approximation to this
binomial, one obtains P (x ≤ 390) ≈
a) .512
b) .456
c) .259
d) .730
e) .270
23) Consider the binomial random variable x, with n=10,000 and p=.7. Using the normal approximation to this
binomial, one obtains P (6,950 ≤ x ≤ 7,050) ≈
a) .683
b) .730
c) .317
d) .725
e) .500
24) Suppose that we have a large population of two types of objects (say white marbles and black marbles),
from which we sample relatively few. In computing the probability of having in our sample a number of
objects of a given type,
a) it makes a great difference whether the sampling is with replacement as compared with sampling
without replacement.
b) It doesn't make much difference whether the sampling is with replacement or without replacement.
c) It is easier to regard the sampling as being without replacement.
d) One should never proceed without drawing an appropriate tree diagram.
e) One needs to take into account Billy's attendance records.
25) Suppose that x is a random variable whose distribution is a continuous distribution with domain 0 ≤ x ≤ 4 .
Suppose now that we randomly (and independently) select a single sample from this population. Then the
probability that this sample falls somewhere between 1 and 2.5 is
a) .625
b) .125
c) 1
d) .375
e) .275
26) Suppose that x is a random variable whose distribution is a continuous distribution with domain 0 ≤ x ≤ 4 .
Suppose now that we randomly (and independently) select two samples from this population. Then the
probability that both of these samples fall somewhere between 1 and 2.5 is
a) .391
b) .141
c) .375
d) 1
e) 0
27) Suppose that x is a random variable whose distribution is a continuous distribution with domain 0 ≤ x ≤ 4 .
Suppose now that we randomly (and independently) select three samples from this population. Then the
probability that exactly one of these samples falls somewhere between 1 and 2.5 is
a) .439
b) .264
c) .244
d) .561
e) .736
Free-Response Questions
28) Suppose that we have a box with 12 marbles in it: 4 white, 5 red, and 3 blue. From this box, select three
marbles (without replacement).
a) In terms of binomial coefficients, express the probability that 2 white marbles would be selected. Then
compute this answer as a rational number.
(Binomial Coefficients): ____________________________
(Rational Number): ____________________________
b) Now model the above experiment in terms of a tree diagram, filling in the missing components. Express
the probability of selecting 2 white marbles as an appropriate sum of products, and then write the final
answer as a rational number.
nonwhite
white
(Sum of Products): ____________________________
(Rational Number): ____________________________
29) For the above experiment, let x be the random variable which counts the number of white marbles taken
from the box. Give the distribution for x below.
X
p(x)
30) Continuing with the above experiment, compute the mean μ x and standard deviation σ x of x.
μ x = ____________________________
σ x = ____________________________
31) Suppose that in the above experiment we introduce the new random variable y, as follows. If you draw 0
white marbles, y = −10 , if you draw one white marble, y = 0 ; if you draw two white marbles, y = 15 ;
and if you draw three white marbles, y = 20 . Compute the mean μ y and standard deviation σ y of y.
μ y = ___________________________
σ y = ____________________________
32) Finally, suppose that you have the same box of marbles, but this time you draw three marbles with
replacement. Compute the expected number and standard deviation of the number of white marbles drawn.
mean ____________________________
standard deviation____________________________
33) Suppose you roll two fair dice and let x be the sum of the number of dots showing. Compute the distribution
of x below.
X
p(x)
34) Compute E (x ) , where x is as in the problem above.
E ( x ) = ____________________________
35) Consider the following payoffs for a single roll of two dice. If you roll a total of 2, 3, or 12, you lose $50. If
you roll a 7 or an 11 you win $10. Otherwise you win (or lose) nothing. Compute the expected payoff for
this game.
expected payoff ____________________________
36) Assume that you have a box of marbles, 30% of which are blue, and that you randomly draw four marbles
without replacement. If x is the random number representing the number of blue marbles drawn, you wish to
compute P ( x ≤ 1) .
a) What else do you need to know to obtain an exact answer?
Ans: ____________________________
b) If you're not told this information, what is your approximate answer?
Ans: ____________________________
37) It is known that most snakes in Australia are poisonous. If, in fact, roughly 65% of all snakes in Australia
are poisonous, and if you encounter 15 snakes during a nature hike, what is the probability that at most 2 of
these snakes are poisonous?
Ans: ____________________________
38) Suppose that one encounters 100 snakes on a several-day nature hike in Australia. Denote by x be the
number of poisonous snakes encountered. Compute the mean and standard deviation of x.
mean ____________________________
standard deviation____________________________
σ be the mean and standard deviations that you computed in the above problem.
a) Compute P(σ − μ ≤ x ≤ σ + μ ) .
39) Let μ and
P(σ − μ ≤ x ≤ σ + μ ) = ____________________________
b) What would your answer to Part (a) be if instead of a binomial distribution, we were talking about a normal
distribution (with the same mean and variance)?
Ans: ____________________________
40) Suppose that an NBA basketball player has a field goal percentage of 44% (that is, on average, he makes
44% of his shots). In a given game he will take 20 shots. Let μ and σ be the mean and standard deviations
of the number of shots that he makes in a typical game.
a) Compute μ and σ .
μ = ____________________________
σ = ____________________________
b) Compute P(σ − μ ≤ x ≤ σ + μ ) .
P(σ − μ ≤ x ≤ σ + μ ) = ____________________________
41) Suppose that x is a random variable whose distribution is a uniform distribution with domain 0 ≤ x ≤ 10 .
Suppose now that we randomly (and independently) select two samples from this distribution. Compute the
probability that both of these samples are between 0 and 3.
Ans: ____________________________
42) Assume that x is a binomial random variable with number of trials 30 and success probability .8. Show that
μ + 3σ > 30 , where μ =E(x) and where σ is the standard variation of the above distribution. What does
this say about the appropriateness of approximating this binomial distribution with the normal distribution
also having mean μ and standard deviation σ ?