Download Test 3 - La Sierra University

Math 251, 19 November 2002, Exam III
Name: ANSWERS (The class average, etc. is listed on the last page)
Instructions. Do each of the following 6 problems. Please justify your answers by
showing all appropriate work. Thank you and good luck!
1. Suppose the distribution of heights of 11-year-old girls is normally distributed with a
mean of 62 inches and a standard deviation of 2.5 inches.
(a) (2 pts) What height has a percentile rank of 70?
Solution: z .52, and so x=62+.522.5  63.3, or approximately 63.3 inches.
(b) (2 pts) What proportion of 11 year-old-girls are between 60 and 65 inches tall?
Solution: z = (60-62) 2.5 = -.8 and z = (65-62)  2.5 = 1.2. Now
P(-.8 < z < 1.2) = .8849 - .2119 = .673
Thus the answer is approximately .673 or 67.3%.
(c) (3 pts) In a group of 800 randomly selected 11-year-old girls, how many would you
expect find that are (i) less than 60 inches tall; and (ii) more than 65 inches tall?
Solution: (i) P(z < -.8) = .2119, and .2119800 = 169.52; thus we would expect to find
about 169 or 170 girls that are less than 60 inches tall.
(ii) P(z > 1.2) = 1 - .8849 = .1151, and .1151 800  92; thus we would expect to find
about 92 girls that are more than 65 inches tall.
2. A farm in Idaho claims that the mean weight of a box of potatoes that they ship is 40
lbs. The quality control person on this farm collected a random sample of 60 boxes and
found a mean weight of 40.2 lbs with a standard deviation of .7 lbs.
(a) (3 pts) If the true mean is exactly 40 lbs, what is the probability of the quality control
person finding a random sample of 60 boxes with a mean weight of 40.2 lbs or more
assuming the population standard deviation is .7 lbs?
Solution: z 
40.2  40
 2.21
.7 60
Now, P(z > 2.21) = 1-.9864 = .0136
(b) (2 pts) If you are the quality control person, would you be satisfied that mean weight
of the boxes you ship is at least 40 lbs? Explain.
Solution: Yes, if the true mean weight was 40 lbs or less, there would only be a 1.36%
chance of finding the sample with a mean weight of 40.2 lbs as above. Therefore, we are
98.64% sure that the mean weight is, in fact, more than 40 lbs.
3. Alaska Airlines has found that 94% of people with tickets will show up for their Friday
afternoon flight from Seattle to Ontario. Suppose that there are 128 passengers holding
tickets for this flight, and the jet can carry 120 passengers, and that the decisions of
passengers to show up are independent of one another.
(a) (2 pts) Verify that the normal approximation of the binomial distribution can be used
for this problem.
Answer: We check that np = 128.94 = 120.32  5 and nq = 128.06 = 7.68  5.
(b) (4 pts) What is the probability that more than 120 passengers will show up for the
flight (i.e., not everyone with a ticket will get a seat on the flight)?
Solution: We approximate the binomial distribution with the normal random variable
with
 = 120.32
and
 = (128.94.06)1/2  2.687
using the continuity correction, we compute
P(x > 120.5 ) = P(z > (120.5 – 120.32)2.687) = P(z > .07) = 1 - .5279 = .4721
That is, there is a 47.21% chance that the airline will have to bump passengers.
4. A recent survey of 65 randomly selected gas stations in California found that the mean
price for unleaded gasoline in the sample was $1.47 per gallon with a sample standard
deviation of $.07.
(a) (2 pts) Why is the central limit theorem important in constructing confidence
intervals for means from large samples?
Answer: It allows us to assert that the sampling distribution for the mean is approximately
normal when the sample size is at least 30, and it gives us the formula for the standard
deviation of the sampling distribution. Using that we can compute the probability that the
population mean is within a given distance of the sample mean.
(b) (4 pts) Find a 96% confidence interval for the mean price of unleaded gasoline in
California.
Solution: For c = .96, zc = 2.05 (approximately), look at z value corresponding to an area
of .98 on table. Thus the confidence interval, using the large sample method (n is at least
30) yields endpoints:
1.47  2.05.07(651/2)
and, so the confidence interval is:
(1.4522,1.4878)
(c) (3 pts) Assuming the population standard deviation is .07, what sample size would be
needed to have a margin of error E=.01 with 96% confidence?
2
zc   
zc  
Solution: The formula to use is: E =
which implies n 
also, and so we
 E 
n
compute n = (2.05.07 .01)2 = 205.9225, thus we need a sample of size 206.
5. A recent (November 11-14, 2002) Gallup poll of 1001 adult Americans found that 14%
said fear of war as the most important issue facing America.
(a) (2 pts) What conditions are necessary in order to construct a confidence interval for a
population proportion?
Answer: nˆp  5 and nqˆ  5 .
(b) (4 pts) Find a 92% confidence interval for the proportion of adult Americans that
believe fear of war is the most important issue facing America.
Solution: zc=1.75 and so the confidence interval is .15 1.75(.14.86001)1/2, that is
(.120807, .159193)
(c) (2 pts) Based on the answer to (b), would it be appropriate for the Gallup organization
to state that less than 15% of adult Americans believe fear of war is the most important
issue facing America? Explain.
Answer: no, the confidence interval goes all the way up to 15.9%, so it would be more
appropriate to say that less than 16% of Americans feel that way.
6. (a) (3 pts) What size of sample is needed by the Gallup organization to estimate a
population proportion within .03 with 99% confidence? In your calculation assume the
worst case scenario that p = .5 and q = .5.
z 
pˆ qˆ
or n  pˆ  qˆ  c
E 
n
2
Thus n = .55(2.575.03) = 1841.84. Thus a sample s tc when
2
Solution: the relevant formula is E  zc 
(b) (2 pts) What value should be used for tc when constructing a 95% confidence interval
for the mean of an approximately normal population when using a sample size of 17?
Assume the population standard deviation is not known.
Solution: n=17 and so d.f.=16, the table thus shows with c = .95 that tc = 2.120
(c) (3 pts) What values of L2 and R2 should be used when constructing a 99%
confidence interval for variance from a normal population given a sample size of 14?
Solution: n = 14 and so d.f. = 13; in our case,  = .01, so we find L2 and R2 in the .995
and .005 columns respectively. Consequently,
L2 = 3.565 and R2 = 29.819.
Results on Test:
Mean = 61.2%, Standard Deviation = 22.7%, Low = 23.3%, High = 93.9%
First Quartile = 40.9%, Median = 63.4%, Third Quartile = 84.4%