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CHAPTER 9 REVIEW
1) The average adult has completed an average of 11.25 years of education with a standard deviation of
1.75 years. A random sample of 90 adults in obtained. What is the probability that the sample will have a
mean
(a) greater than 11.5 years?
(b) between 11 and 11.5 years?
ANS:
1.75
The sampling distribution of x-bar has  x  11.25,  x 
 0.184
90
Because the sample size is large (n=90), the central limit theorem tells us that large sample techniques are
appropriate. Accordingly,
11.5  11.25
 1.36)  0.0869 or normalcdf(11.5, 1E99, 11.25, 0.184)
(a) P( x  11.5)  P( z 
0.184
(b) normalcdf(11, 11.5, 11.25, 0.184) = 0.8258
2) Over the years, the scores on the final exam for AP Calculus have been normally distributed with a
mean of 82 and a standard deviation of 6. The instructor thought that this year’s class was quite dull and,
in fact, they only averaged 79 on their final. Assuming that this class is a random sample of 32 students
from AP Calculus, what is the probability that the average score on the final for this class is no more than
79? Do you think the instructor was right?
ANS:
79  82
)  2.83)  0.0023 or normalcdf(-1E99, 79, 82, 1.06)
P( x  79)  P( z 
6
32
If this group really were typical, there is less than a 1% chance of getting an average this low by random
chance alone. That seems unlikely, so we have some pretty good evidence that the instructor was correct.
3) Harold fails to study for his statistical final. The final has 100 multiple choice questions, each with 5
choices. Harold has no choice but to guess randomly at all 100 questions. What is the probability that
Harold will get at least 30% on the test?
ANS:
Because 100(0.2) and 100(0.8) are both greater than 10, we can use the normal approximation to the
sampling distribution of p-hat. Because p = 0.2, the sampling distribution of p-hat has μ = 0.2
0.2(0.8)
and  pˆ 
 0.04
100
0.3  0.2
 2.5)  0.0062 or normalcdf(0.3, 1E99, 0.2, 0.04)
Therefore, P( pˆ  0.3)  P( z 
0.04
Harold should have studied.
4) What is the probability that a sample of size 35 drawn from a population with mean 65 and standard
deviation 6 will have a mean less than 64?
ANS:
The sample size is large enough that we can use large-sample procedures. Hence,
64  65
)  0.99)  0.1611 or normalcdf(-1E99, 64, 65, 1.014)
6
35
5) Approximately 10% of the population of the United States is known to have blood type B. If this is
correct, what is the probability that between 11% and 15% of a random sample of 50 adults will have type
B blood?
ANS:
0.10(0.90)
If p = 0.10,  pˆ  0.10,  pˆ 
 0.042
50
0.11  0.10
0.15  0.10
z
)  P(0.238  z  1.19)  0.289 or
Then, P(0.11  pˆ  0.15)  P(
0.042
0.042
normalcdf(0.11, 0.15, 0.10, 0.042)
P( x  64)  P( z 
6) A tire manufacturer claims that his tired will last 40,000 miles with a standard deviation of 5000 miles.
(a) Assuming that the claim is true, describe the sampling distribution of the mean lifetime of a random
sample of 160 tires. “Describe” means discuss center, spread, and shape.
(b) What is the probability that the mean lifetime of the sample of 160 tires will be less than 39,000
miles? Interpret the probability in terms of the truth of the manufacturer’s claim.
ANS:
5000
 x  40000,  x 
 395.28 miles
160
(a) With n = 160, the sampling distribution of x-bar will be approximately normally distributed with mean
equal to 40,000 miles and standard deviation 395.28 miles.
39000  40000
 2.53)  0.006 or normalcdf(-1E99, 39000, 40000, 395.28)
(b) P( x  39000)  P( z 
395.28
If the manufacturer is correct, there is only about a 0.6% chance of getting an average this low or lower.
That makes is unlikely to be just a chance occurrence and we should have some doubts about the
manufacturer’s claim.
7) Crabs off the coast of Northern California have a mean weight of 2 lbs. with a standard deviation of 5
oz. A large trap captures 35 crabs.
(a) Describe the sampling distribution for the average weight of a random sample of 35 crabs taken from
this population.
(b) What would the mean weight of a sample of 35 crabs have to be in order to be in the top 10% of all
such samples?
ANS:
5
 x  32, x 
 0.338
35
(a) With samples of size 35, the central limit theorem tells us that the sampling distribution of x-bar is
approximately normal with mean 32 oz. and standard deviation 0.338 oz.
(b) In order for x-bar to be in the top 10% of samples, it would have to be at the 90th percentile, which
x  32
tells us that its z-score is 1.28. Hence, z x  1.28 
0.338
Solving, we have x-bar = 32.43 oz. A crab would have to weigh at least 32.43 oz., or about 2 lb. 7 oz., to
be in the top 10% of samples of this size.
8) A certain type of light bulb is advertised to have an average life of 1200 hours. If, in fact, light bulbs
of this type only average 1185 hours with a standard deviation of 80 hours, what is the probability that a
sample of 100 bulbs will have an average life of at least 1200 hours?
80
ANS:  x  1185,  x 
8
100
1200  1185
P( x  1200)  P( z 
 1.875)  0.03 or normalcdf(1200, 1E99, 1185, 8)
8
9) Opinion polls in 2002 showed that about 70% of the population had a favorable opinion of President
Bush. That same year, a simple random sample of 600 adults living in the San Francisco Bay Area
showed found only 65% that had a favorable opinion of President Bush. What is the probability of getting
a rating of 65% or less in a random sample of this size if the true proportion in the population was 0.70?
ANS:
0.7(0.3)
If p = 0.70,  pˆ  0.70,  pˆ 
 0.019
600
0.65  0.70
 2.63)  0.004 or normalcdf(-1E99, 0.65, 0.70, 0.019)
Then, P( pˆ  0.65)  P( z 
0.019
10) A polling organization asks an SRS of 1500 employees if they took things from their place of work
for personal use. It is believed that on a national level, 35% of employees engage in white-collar crimes
and take things from their place of employment. What is the probability that a random sample of 1500
will give a result within 2 percentage points of the true population proportion of 35%?
ANS:
n = 1500 and p = 0.35. We are interested to find P(0.33  pˆ  0.37)
Can we use the normal distribution to approximate the sampling distribution of p-hat?
np = 1500(0.35) = 525 and n(1-p) = 1500(0.65) = 975. Both are much larger than 10.
Since the conditions are met, the normal approximation is an appropriate model.
0.35(0.65)
The sampling distribution of p-hat has  pˆ  0.35,  pˆ 
 0.0123
1500
0.33  0.35
0.37  0.35
P(0.33  pˆ  0.37)  P(
z
)  P(1.63  z  1.63)  0.8968 or normalcdf(0.33,
0.0123
0.0123
0.37, 0.35, 0.0123)
11) In a large high school of 2500 students, 21% of them are seniors. A simple random sample of 150
students is taken and the proportion of seniors calculated. What are the mean and standard deviation of
the sample proportion, p-hat?
ANS:
In this case, the population proportion is p = 0.21 and the sample size is n = 150. The mean of the sample
0.21(0.79)
proportion p-hat is  pˆ  p  0.21 and the standard deviation is  pˆ 
 0.033
36
12) In a large high school of 2500 students, 21% of them are seniors. A simple random sample of 150
students is taken and the proportion of seniors calculated. What is the probability that the sample will
contain less than 15% seniors?
ANS:
Since the binomial conditions np = 150(0.21) = 31.5 and n(1-p) = 150(0.79) = 118.5 are both greater than
10—the sampling distribution of p-hat can be approximated by a normal distribution with mean 0.21 and
standard deviation 0.033 (the results from (11)).
0.15  0.21
P( pˆ  0.15)  P( z 
 1.80)  0.0359 or normalcdf(-1E99, 0.15, 0.21, 0.033)
0.033
Thus, the chance of obtaining a sample with less than 15% seniors is about 3.6%--not very likely.
13) In a large high school of 2500 students, the mean number of cars owned by students’ families is 2.35
with a standard deviation of 1.06. A simple random sample of 36 students is taken and the mean number
of cars owned is calculated.
(a) What are the mean and standard deviation of the sample mean, x-bar?
(b) What is the probability that the sample mean is greater than 2.5 cars?
(c) Below what value is the lowest 5% of all possible sample means?
ANS:
(a) The population mean is μ = 2.35 cars, the standard deviation is σ = 1.06 cars, and the sample size is n
1.06
= 36. The mean of the sample mean x-bar is  x  2.35,  x 
 0.177
36
2.5  2.35
 0.85)  0.1977 or normalcdf(2.5, 1E99, 2.35, 0.177)
(b) P( x  2.5)  P( z 
0.177
x  2.35
(c) The 5th percentile in Table A corresponds to a z-score of -1.65. Therefore, z  1.65 
0.177
Solving for x-bar, we get 2.958. The lowest 5% of all sample means will be below 2.958 cars.
14) The GPAs of graduating students at a large university are normally distributed, with a mean GPA of
2.8 and a standard deviation of 0.5. A random sample of 50 students is taken from all the graduating
students.
(a) Find the probability that the mean GPA of the sampled students is above 3.0.
(b) Find the probability that the mean GPA of sampled students is between 2.7 and 3.0.
ANS:
Sample size 50 is sufficiently large for us to assume approximate normality for the sampling distribution
0.5
of x-bar with  x  2.8,  x 
 0.071. In other words, X ~ N(2.8, 0.071) approximately.
50
3.0  2.8
 2.82)  0.0024 or normalcdf(3.0, 1E99, 2.8, 0.071)
(a) P( x  3.0)  P( z 
0.071
There is less than a 1% chance (0.24%) that the mean GPA of the 50 sampled students will exceed 3.0.
2.7  2.8
3.0  2.8
z
)  P(1.41  z  2.82)  0.9183 or normalcdf(2.7, 3.0,
(b) P(2.7  x  3.0)  P(
0.071
0.071
2.8, 0.071)
There is almost a 92% chance that the mean GPA of the 50 sampled students will be between 2.7 and 3.0.
15) The mathematics department at a state university notes that the SAT math scores of high school
seniors applying for admission into their program are normally distributed with a mean of 610 and
standard deviation of 50.
(a) What is the probability that a randomly chosen applicant to the department has an SAT math score
above 700?
(b) What is the shape, mean, and standard deviation of the sampling distribution of the mean of a sample
of 40 randomly selected applicants?
(c) What is the probability that the mean SAT math score in an SRS of 40 applicants is above 625?
(d) Would your answers to (a), (b), or (c) be affected if the original population of SAT math scores were
highly skewed instead of normal? Explain.
ANS:
700  610
 1.8)  0.0359 or normalcdf(700, 1E99, 610, 50)
(a) P( X  700)  P( z 
50
50
(b) It is roughly normal with N(610,
) =N(610, 7.91)
40
625  610
 1.90)  0.0287 or normalcdf(625, 1E99, 610, 7.91)
(c) P( x  625)  P( z 
7.91
(d) The answer to part (a) would be affected because it assumes a normal population. The other answers
would not be affected because for large enough n, the central limit theorem gives that the sampling
distribution will be roughly normal regardless of the distribution of the original population.
16) A jar of Jiffy Peanut Butter (labeled as a 32 oz. jar) is selected randomly off the end of the assembly
line. The weight of the jar is measured. If X measures this weight then X has the N(32.3, 0.40) distribution
(the filling machine is calibrated this way). That is, the mean fill is 32.3 oz., the standard deviation of the
fills is 0.40 oz.
(a) Identify the response variable being measured. Is it categorical or quantitative?
(b) Identify the population and parameters.
(c) What is the average fill of all Jiffy jars? Is this value a statistic?
(d) What’s the probability a randomly selected jar is filled with less than 32.0 fl oz? Such a can would be
“under volume.”
(e) What proportion of all Jiffy jars are under-volume?
(f) Consider a simple random sample of 6 jars. Let x-bar be the sample mean weight for these 6 jars. Is
x-bar a parameter or a statistic?
(g) What are the mean and standard deviation of the (sampling) distribution for the sample mean?
(h) Find the probability the sample mean fill of the 6 jars is less than 32.0 fl oz.
(i) Now take a simple random sample of 24 jars. What are the mean and standard deviation of the
distribution for x-bar?
(j) Find the probability the sample mean fill of the 24 jars is less than 32.0 fl oz.
ANS:
(a) Response variable = weight of the jar (quantitative.)
(b) Population = measurement of jars of Jiffy Peanut Butter.
Parameter = 32 oz.
(c) Average = 32.3 oz.
No, it’s not a statistic.
(d) P(X<32.0) = P( z 
32.0  32.3
 0.75)  0.2266 or normalcdf(-1E99, 32, 32.3, 0.4)
0.40
(e) 0.2266
(f) x-bar is a statistic.
(g)  x    32.3,  x 
0.4
 1.63
6
32.0  32.3
 0.18)  0.4286 or normalcdf(-1E99, 32, 32.3, 1.63)
(h) P( x  32.0)  P( z 
1.63
0.40
(i)  x  32.3,  x 
 0.08
24
32.0  32.3
 3.75)  essentially 0.
(j) P( x  32.0)  P( z 
0.08
17) The weights, in pounds, of Portuguese Water Dogs has the N(45, 3) distribution.
(a) If one dog is selected at random and X is its weight, find the probability X is within 1 lb. of the mean
weight of 45.
(b) Draw a simple random sample of 40 dogs. Find the probability the sample mean weight is within 1 lb.
of the mean weight of 45.
(c) Draw a simple random sample of 80 dogs. Find the probability the sample mean weight is within 1 lb.
of the mean weight of 45.
ANS:
(a) P(44≤X≤46) = normalcdf(44, 46, 45, 3) = 0.2611
3
(b)  x  45, x 
 0.4743
40
P(44  x  46)  normalcdf(44, 46, 45, 0.4743) = 0.9650
3
(c)  x  45,  x 
 0.33541
80
P(44  x  46) = normalcdf(44, 46, 45, 0.33541) = 0.9971
18) A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are
supposed to contain 300 millilters (ml). In fact, the contents vary according to a normal distribution with
mean  = 303 ml and standard deviation  = 3 ml.
a. What is the probability that an individual bottle contains less than 300 ml?
b. Now take a random sample of 10 bottles. What are the mean and standard deviation of the sample
mean contents x-bar of these 10 bottles?
c. What is the probability that the sample mean contents of the 10 bottles is less than 300 ml?
ANS:
(a) P(X<300) = P( z 
300  303
 1)  0.1587 or normalcdf(-1E99, 300, 303, 3)
3
(b)  x  303,  x 
3
10
(c) P( x  300)  P( z 
 0.94868
300  303
 3.16)  0.0008 or normalcdf(-1E99, 300, 303, 0.94868)
0.94868
19) For 1998 as a whole, the mean return of all common stocks listed on the New York Stock Exchange
(NYSE) was  = 16% and standard deviation  = 26%. Assume that the distribution of returns is roughly
normal.
a. What % of stocks lost money?
b. Suppose we create a portfolio of 8 stocks by randomly selecting stocks from the NYSE and
investing equal amounts of money in each stock. What are the mean and standard deviation of the
sample mean returns x-bar for these 8 stocks?
c. What is the probability the portfolio loses money? Explain the difference between this result and
that of part (a).
d. The probability is 0.05 that a portfolio constructed this way has a return of more than ________ ?
(This would be the 95th percentile of portfolio returns; however, remember that these portfolios
form a hypothetical population -- no one actually owns such a portfolio.)
ANS:
0  0.16
 0.62)  0.2676 or normalcdf(-1E99, 0, 0.16, 0.26)
0.26
0.26
(b)  x  0.16,  x 
 0.09192
8
0  0.16
 1.74)  0.0409 or normalcdf(-1E99, 0, 0.16, 0.09192)
(c) P( x  0)  P( z 
0.09192
This is the probability that the average of 8 randomly selected stocks loses money; the 0.2676 in part (a) is
the probability a single stock loses money.
x  0.16
 x  0.3112 or invNorm(0.95, 0.16, 0.09192)
(d) 0.9500  1.645  z 
0.09192
That is, 5% of these portfolios will make more than 31.12%.
(a) P( X  0)  P( z 
20) The length of human pregnancies from conception to birth varies according to a distribution that is
approximately normal with mean 264 days and standard deviation 16 days. Consider 15 pregnant women
from a rural area. Assume they are equivalent to a random sample from all women.
a. What are the mean and standard deviation of the sample mean length of pregnancy x-bar of these
15 pregnancies?
b. If we want to predict, with 90% accuracy, the sample mean length of pregnancy for 15 randomly
selected women, what values do we use? (That is, find value L AND U such that there's a 90%
probability the sample mean x-bar lies between L and U.)
c. What's the probability the sample mean length of pregnancy lasts less than 250 days? (Contrast
this with the probability a single pregnant women is pregnant for less than 250 days, which is
0.1908.)
d. Toxic waste is believed to have effected the health of residents of this area. Suppose the sample
mean length of pregnancy is indeed 250 days; use the result of part (c) to argue that the waste has
an effect of length of pregnancy.
ANS:
(a)  x  264,  x 
16
15
 4.13119
x  264
 x  258 or invNorm (0.05, 264, 4.13119)
4.13119
x  264
 x  271 or invNorm (0.95, 264, 4.13119)
0.9500  1.645 = z 
4.13119
So, it will be between 258 and 271 days.
250  264
 3.39)  0.0003 or normalcdf(-1E99, 250, 264, 4.13119)
(c) P( x  250)  P( z 
4.13119
(d) Assume the toxin has no effect on length of pregnancy -- the average length of pregnancy for all
people (including people exposed to the toxin) is 264. The chance of an average length of pregnancy at
least as low as the observed 250 is very remote -- it should occur in 1 in 3333 trials on average. This leads
one to believe that perhaps the result isn't due to chance alone and, instead, that our assumption of 264
days on average is in question. (This result is "beyond a reasonable doubt.")
(b) 0.0500-1.645 = z 
21) An airplane is only allowed a gross passenger weight of 8000 kg. If the weights of passengers
traveling by air between Toronto and Vancouver have a mean of 78 kg and a standard deviation of 7 kg,
the approximate probability that the combined weight of 100 passengers will exceed 8,000 kg is:
(a) 0.4978
(b) 0.3987
(c) 0.1103
(d) 0.0044
(e) .0022
Solution: e
7
P (x > 8000/100) = normalcdf (80, 1E99, 78,
) = 0.002137
100
7 * 100
OR
P (x > 8000) = normalcdf (8000, 1E99, 78*100,
) = 0.002137
100
22) The time required to assemble an electronic component is normally distributed with a mean of 12
minutes and a standard deviation of 1.5 min. Find the probability that the time required to assemble all
nine components (i.e. the total assembly time) is greater than 117 minutes.
(a) 2514
(b) .2486
(c) .4772
(d) .0228
(e) .0013
Solution: d
1 .5
) = 0.02275
9
1.5 * 9
OR
P (x > 117) = normalcdf (117, 1E99, 12 * 9,
) = 0.02275
9
23) The sample mean is an unbiased estimator for the population mean. This means:
(a) The sample mean always equals the population mean.
(b) The average sample mean, over all possible samples, equals the population mean.
(c) The sample mean is always very close to the population mean.
(d) The sample mean will only vary a little from the population mean.
(e) The sample mean has a normal distribution.
Solution: b
P (x > 117/9) = normalcdf (13, 1E99, 12,
24) The average monthly mortgage payment for recent home buyers in Winnipeg is μ = $732, with
standard deviation of σ = $421 A random sample of 125 recent home buyers is selected. The approximate
probability that their average monthly mortgage payment will be more than $782 is:
(a) 0.9082
(b) 0.4522
(c) 0.4082
(d) 0.0478
(e) 0.0918
Solution: e