Download Introduction to Statistics Sample Exam Chapters 5 – 7

Introduction to Statistics
Sample Exam Chapters 5 – 7
1. Which of the following are continuous variables, and which are discrete?
(a) Continuous
(b) Discrete
(c) Discrete
(d) Continuous
(e) Discrete
(f) Continuous
(g) Continuous
2. The Recharge Company makes cheap alternators for cars. About 15% of the alternators
they make are defective. Al’s Automotive Parts and Service has ordered several cases of
alternators. Al will randomly select one case of 12 alternators to test. If more than 3 of the
alternators are defective, Al will reject the shipment.
(a) What is the probability that Al rejects the shipment of alternators?
For n = 12 and p = 0.15,
P (r > 3) = .068 + .019 + .004 + .001 + .000 + .000 + .000 + .000 + .000 = .092 or 9.2%
(b) What is the expected number of defective alternators in a case of 12?
µ = np = 1.8 out of every 12.
(c) What is the standard deviation in the number of defective alternators in a case of 12?
√
σ = npq = 1.24 alternators.
3. Chauncey is a 90% free throw shooter (he makes 90% of the free throws he attempts).
(a) If Chauncey is fouled on a three point shot (and therefore gets three free throws) what
is the probability that he makes all three free throws?
n = 3, p = 0.90, P (r = 3) = 0.729.
(b) What is the expected number of free throws out of three?
µ = 2.7 free throws.
(c) If Chauncey takes 20 free throws in a game, what is the probability that he makes all
20? What is the probability that he makes more than 15?
n = 20, p = 0.90, P (r = 20) = 0.122 or 12.2%.
P (r > 15) = .090 + .190 + .285 + .270 + .122 = 0.957 or 95.7% of the time Chauncey will
make more than 15 out of 20.
4. Sam is a 40% free throw shooter (he makes only 40% of the free throws he attempts).
(a) If Sam is fouled and takes two free throws, what is the probability that he makes both
of them?
n = 2, p = 0.40. P (r = 2) = 0.160, or 16% of the time Sam will make both free throws.
(b) What is the probability that he makes neither of them?
n = 2, p = 0.40. P (r = 0) = 0.360, or 36% of the time Sam will miss both free throws.
(c) What is the expected number of made free throws (out of 2). µ = np = 0.8 free throws.
(d) If Sam takes 10 free throws, what is the probabiltiy that he makes at least one? What
is the probability that he makes at least two?
n = 10, p = 0.40. P (r ≥ 1) = 1 − P (r = 0) = 1 − .006 = 0.994 or 99.4% of the time Sam
will make at least one out of ten free throws.
n = 10, p = 0.40. P (r ≥ 2) = 1 − P (r = 0) − P (r = 1) = 1 − .006 − .040 = 0.954 or
95.4% of the time Sam will make at least two out of ten free throws.
5. Jane solicits contributions for charity by phone. She has a history of being successful at
getting a donor on 55% of her calls. If Jane has a quota to get at least four donors each day,
how many calls must she make to be 96.4% sure of meeting her quota?
p = 0.55. Guess and check to find n:
If n = 10, P (r ≥ 4) = 1 − P (r ≤ 3) = 1 − .075 − .023 − .004 − .000 = 0.898 or 89.8%. Not
96.4% or more so we need a larger n.
If n = 11, P (r ≥ 4) = 1 − P (r ≤ 3) = 1 − .046 − .013 − .002 − .000 = 0.939 or 93.9%. Not
96.4% or more so we need a larger n.
If n = 12, P (r ≥ 4) = 1 − P (r ≤ 3) = 1 − .028 − .007 − .001 − .000 = 0.964 or 96.4% and this
is exactly what we needed so n = 12 works.
6. Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that
is approximately normally distributed with mean µ = 27.2 kilograms and standard deviation
σ = 4.3 kilograms.
(a) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected fawn weighs between 30 and 35 kilograms. Find the probability.
P (30 ≤ x ≤ 35) = P (0.65 ≤ z ≤ 1.81) = 0.9649 − 0.7422 = 0.2227 or 22.27%.
(b) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected fawn weighs less than 25 kilograms. Find the probability.
P (x < 25) = P (z < −0.51) = 0.3050 or 30.5%.
(c) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected fawn weighs more than 22 kilograms. Find the probability.
P (22 < x) = P (−1.21 < z) = 1 − 0.1131 = 0.8869 or 88.69%.
7. A grocery store produce manager has decided to reject the smallest (by weight) 7.5% of
melons the store receives. If the melons weights are approximately normally distributed with
mean µ = 1.5 pounds and standard deviation σ = 0.3 pounds, what is the minimum weight
of a melon that the store will accept?
Smallest 7.5% corresponds to the z score with P (z < zc ) = 0.0750. The closest z score is
zc = −1.44 which gives probability 0.0749.
The raw data value corresponding to a z score of −1.44 is given by x = zσ + µ = 1.068
pounds.
8. Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that
is approximately normally distributed with mean µ = 27.2 kilograms and standard deviation
σ = 4.3 kilograms.
(a) For samples of size 5, what can we say about the sampling distribution (the distribution
of x)? Is it normal? What is the mean and standard deviation?
Since the underlying x distribution is approximately normal, the sampling distribution
4.3
is also normal with mean µx = 27.2 and σx = √
= 1.9.
5
(b) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected sample of 5 fawns has mean weight between 28 and 30 kilograms.
Find the probability.
P (28 ≤ x ≤ 30) = P (0.42 ≤ z ≤ 1.47) = 0.9292 − 0.6628 = 0.2664 or 26.64%.
(c) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected sample of 5 fawns has mean weight less than 27 kilograms. Find
the probability.
P (x < 27) = P (z < −.11) = 0.4562 or 45.62%.
(d) Sketch the normal distribution and shade the area corresponding to the probability that
a randomly selected sample of 5 fawns has mean weight more than 28 kilograms. Find
the probability.
P (x > 28) = P (z > 0.42) = 1 − 0.6628 = 0.3372 or 33.72%.
9. The amount of asprin in the tablets produced by a certain company has average µ = 325mg
and standard deviation σ = 5mg.
(a) If we take a random sample of 20 tablets, can we assume the sampling distribution is
normal? Why or why not?
No. The underlying distribution is not said to be normal, and the sample size is under
30.
(b) If we take a random sample of 75 tablets, can we assume the sampling distribution is
normal? Why or why not?
Yes. The sample size is larger than 30 so the sampling distribution is approximately
normal with mean µx = 325 and σx = √575 = 0.577.
(c) Find the probability that a random sample of 75 tablets averages more than 330mg.
P (x > 330) = P (z > 8.67) = 1 − 1.0000 = 0. There is essentially 0% chance that the
average of 75 tablets will be this large.
10. In a blind taste test with unlabeled identical glasses, 49 out of 93 people prefered Brand A to
Brand B. In another test where Brand A was served in a white glass and Brand B was served
in a black glass, it was found that 54 out of 89 people prefered Brand A.
(a) What is the point estimate for the percentage of those that prefer Brand A in the
sample with identical glasses? Find a 90% confidence interval for the percentage that
prefer Brand A in this test.
p̂1 = 49
93 = 0.5269 is the point estimate. The margin of error is
r
r
p̂q̂
0.5269 · 0.4731
= 1.645
= 0.0852
E = zc
n
93
We can be 90% confident that the interval 0.4417 to 0.6121 contains the true population
proportion for the test with identical glasses.
(b) What is the point estimate for the percentage of those that prefer Brand A in the sample
with colored glasses? Find a 75% confidence interval for the percentage that prefer Brand
A in this test.
54
p̂2 = 89
= 0.6067 is the point estimate. The margin of error is
r
r
p̂q̂
0.6067 · 0.3933
E = zc
= 1.15
= 0.0595
n
89
We can be 75% confident that the interval 0.5472 to 0.0.6662 contains the true population
proportion for the test with colored classes.
(c) What is the point estimate for the difference between the percentages? Find a 95%
confidence interval for the difference in percentages between the two taste tests.
(d) Does this indicate that the color of the glasses changed people’s preferences?
p̂1 − p̂2 = −0.0798 is the point estimate for the difference of proportions. The margin of
error is
r
r
p̂1 q̂1 p̂2 q̂2
0.5269 · 0.4731 0.6067 · 0.3933
E = zc
+
= 1.96
+
= 0.1435
n1
n2
93
89
We can be 95% certain that the interval −0.0798 ± 0.1435 contains the true population
difference in percentages. This interval goes from a 22.33% decrease to a 6.37% increase.
(That’s −0.2233 to 0.0637.) Since this interval contains 0 we cannot conclude from these
samples that the true population percentages are different.
11. (a) A random sample of 132 adult male wildebeest were (very carefullly!) randomly sampled
and found to have a mean weight of 411 pounds with a sample standard deviation of
s = 11 pounds. Find a 99% confidence interval for the population mean weight of adult
male wildebeest.
Point estimate: x = 411. Margin of error: E = tc √sn = 2.626 √11
= 2.5 pounds
132
We use Student’s t distribution because we know thet sample standard deviation s and
not the population standard deviation σ. We use d.f. = 100 because this is the next
smaller value on the table from n − 1 = 131. We use confidence level c = 0.990 for 99%.
We can be 99% certain that the true population mean weight of wildebeest is between
408.5 pounds and 413.5 pounds.
(b) A random sample of 81 adult male wildebeest from a different herd were found to have
a mean weight of 419 pounds and a sample standard deviation of s = 13 pounds. Find
a 95% confidence interval for the difference in the population means based on these two
samples. Does this provide evidence that one herd was heavier than the other?
Point estimate: x1 − x2 = 411 − 419 = −8. Margin of error:
s
r
s21
s22
112
132
E = tc
+
= 1.990
+
= 3.4
n1 n2
132
81
We can be 95% certain that the true difference in population means is −8.0±3.4 pounds,
or between −11.4 and −4.6 pounds. In other words, the second sample comes from a
population that averages between 4.6 and 11.4 pounds heavier than the first sample.
12. Suppose that we know from previous studies that the population standard deviation in the
heights of adult emperor penguins is σ = 5 cm. We wish to collect a random sample of
emperor penguins and estimate the population mean height using a 99% confidence interval.
How many individuals must we include in our sample if we wish to know the population mean
height to within 2 cm?
2
2
n = zEc σ = 2.58·5
= 41.6, so we need to sample at least 42 penguins.
2
13. We wish to estimate the number of voters that favor Ballot Measure 13b using a 95% confidence interval. How many voters must we include in our sample if we wish to know the
percentage to within plus or minus 2 percentage points?
2
2
Since we don’t have an estimate of p, we use the formula n = 14 zEc = 14 1.96
= 2401
0.02
14. Data from ten years ago indicated that about 13% of voters favored a ballot measure similar
to Ballot Measure 13b. Using this as a preliminary estimate, how many voters must we
include in our sample if we wish to know the percentage now in favor of Ballot Measure 13b
to within plus or minus 2 percentage points in a 95% confidence interval?
2
We have a preliminary estimate for p so we may use the formula n = p̂q̂ zEc
= 0.13 ·
1.96 2
0.87 .02 = 1086.2, so we need at least n = 1087 voters in our sample.