Chapter 9 Free Response Practice
1. Suppose that you and your lab partner flip a coin 20 times and you calculate the proportion of tails to be 0.8.
Your partner seems surprised at these results and suspects that the coin is not fair. Write a brief statement that
describes why you either agree or disagree with him. Your response should include statistical evidence that the
coin probably isn’t fair. Use the principles learned in Ch. 9.
AGREE: You would expect 50% heads and tails over the LONG RUN and even though 16 out of 20 heads is
unlikely this could occur due to sampling variability, so we really need to see what happens over a greater number
of flips before making a conclusion about the bias of the coin.
DISAGREE: If the coin were indeed fair, getting 16 out of 20 flips would have a probability of only .005,
[binompdf(20,.5,16)]. This probability of it occurring simply by chance is so low it would lead you to believe that
the coin is indeed biased.
2. A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax
from 5% to 6%, with the additional revenue going to education. Let p̂ denote the proportion in the sample
who say they support the increase. Suppose that 40% of all adults in Ohio support the increase.
(a) If p̂ is the proportion of the sample who support the increase, what is the mean of the sampling
distribution of p̂ ?
The mean for the statistic pɵ is the population parameter p, which is .40
(b) What is the standard deviation of the sampling distribution of p̂ ?
The standard deviation of a proportion uses the formula:
p(1 − p)
.4*.6
this gives the result: σ pɵ =
=.0126
n
1500
(c) Explain why you can use the formula for the standard deviation of p̂ in this setting.
There are more than ten times 1500 adults in Ohio.
(d) Check that you can use the Normal approximation for the distribution of p̂ .
np>10 and n(1-p)>10
1500(.4)=600 1500(.6)=900
(e) Find the probability that p̂ takes a value between 0.37 and 0.43.
normalcdf(.37, .43, .4, .0126)=0.9827
(f) How large a sample would be needed to guarantee that the standard deviation of p̂ is no more than 0.01?
Explain.
(.4)(.6)
≤.01 Solve for n. n>2400
n
3. ACT scores for the 1,171,460 members of the 2004 high school graduating class who took the test closely
followed the Normal distribution with mean 20.9 and standard deviation 4.8. Choose two students
independently and at random from this group.
(a) What is the expected sum of their scores?
20.9 + 20.9 = 41.8
(b) What is the expected difference of their scores?
20.9 - 20.9 = 0
(c) What is the standard deviation of the difference in their scores?
Assuming that their scores are independent, σ x − y = 4.82 + 4.82 = 6.788
(d) Find the probability that the sum of their scores is greater than 50. Show your method.
normalcdf(50, 100, 41.8, 6.788) = .1135
4. A certain beverage company is suspected of under filling its cans of soft drink. The company advertises that its
cans contain, on the average, 12 ounces of soda with standard deviation 0.4 ounce. For the questions that follow,
suppose that the company is telling the truth.
(a) Can you calculate the probability that a single randomly selected can contains 11.9 ounces or less? If so,
do it. If not, explain why you cannot.
They have not told us that the distribution of individual cans was normal, and we can’t assume it is so
therefore we can not calculate its probability.
(b) A quality control inspector measures the contents of an SRS of 50 cans of the company’s soda and
calculates the sample mean x . What are the mean and standard deviation of the sampling distribution of
x for samples of size n = 50?
σ
.4
µ x = µ = 12 σ x =
=
=.0566
n
50
(c) The inspector in part (b) obtains a sample mean of x = 11.9 ounces. Calculate the probability that a
random sample of 50 cans produces a sample mean fill of 11.9 ounces or less. Be sure to explain why you
can use a Normal calculation.
Because our sample size is large (n=50), the Central Limit Theorem indicates that the sample mean
distribution is approximately normal.
normalcdf(0, 11.9, 12, .0566) = .0386
(d) What would you conclude about whether the company is underfilling its cans of soda? Justify your answer.
If the company is telling the truth, a sample size of 50 cans should have an average weight of 11.9 oz or
lower less than 4% of the time. Either we got a very unlucky sample or the company is not telling the truth.
5. A box contains ten $1 bills, five $2 bills, three $5 bills, one $10 bill, and one $100 bill. A person is charged
$20 to select one bill.
(a) Identify the random variable. X = value of the bill selected
(b) Construct a probability distribution for these data (Table or Histogram).
Value
$1
$2
$5
$10 $100
Probability
.5
.25
.15
.05
.05
(c) Find the expected value.
1(.5)+2(.25)+5(.15)+10(.05)+100(.05) = $7.25
(d) Is the game fair? Explain briefly.
No, for every $20 you spend you would only expect to get back $7.25, a loss of $12.75 each time you play