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QUIZ 6.3B
1. a) This is neither binomial nor geometric, because the probability of success (selecting a female
student) is probably different in each trial, unless every classroom has exactly the same proportion of
females, which is unlikely.
b) This is a geometric setting: BINARY outcomes (more than two occupants or not), INDEPENDENT
trials (the number of occupants of one car does not influence the number of occupants in the next
randomly-selected car), we are counting the number of TRIALS to the first car with more than two
occupants, and the probability of SUCCESS—finding a car with more than two occupants—is always
the same.
 20 
 (0.02)1 (0.98)19 = 0.2725
1
2. a) X is binomial with n = 20 and p = 0.02. P(X = 1) = 
b) P(X  2 )
c) µx
= binomcdf (20, 0.02, 2) = 0.9929
= np = (20)(0.02) = 0.4
3. a) (0.8)2(0.2) = 0.128
σx = np(1  p) = (20)(0.02)(0.98) = 0.63
(b) P(X  4) = 1-P(X  3) = 1-geomcdf (0.2, 3) = 0.512
QUIZ 7.1C
1. a) The parameter is the proportion of people in the entire community who would answer “Yes” to the
question. It’s equal to 0.40. The statistic is the proportion of people in the sample of 100 who would
answer “Yes” to the question.
b) The sampling distribution describes the distribution of the proportion of people who would answer
“Yes” to this question in all possible samples of size 100 from this population.
c) If a statistic is an unbiased estimator of the parameter, then the mean of the stastic’s sampling
distribution is equal to the parameter.
d) No. As long as the sample is less than 10% of the population, the size of the population from which
the sample is taken does not influence the sampling distribution.
e) Yes. The standard deviation of the sampling distribution would be larger if the sample size were
smaller.
3. a) The population is all the guppies in the pet store. We’ve been given the population mean,   5
cm, and the population standard deviation,   0.5 cm.
b) The sample mean is x = 4.8 cm and the sample size is n = 10.
c) No, it’s merely an approximation of a sampling distribution generated by simulating 200 sample
means. The actual sampling distribution includes the means from all possible samples of size 10 from
the population—many more than the 200 values shown!
d) 21 out of 200, or 10.5%, of the sample means in our simulation are as far or farther below 5.0 as our
sample was. Our sample is not sufficiently unusual to arouse suspicions about the store’s claim.
QUIZ 7.2B
1. a)  p-hat = p = 0.22 ; σ p-hat =
p(1  p)
=
n
(.22)(.78)
= 0.034
150
b) Since np = (150)(.22) = 33  10 AND n(1-p) = (150)(.78) = 117  10, the
distribution is approximately normal.
c) If the sample size were 36 instead of 150, the mean, µ p-hat would not change. The
standard deviation, σ p-hat would be larger (
(.22)(.78)
= 0.069), and the distribution
36
would be non-Normal, since np = (36)(0.22) = 7.92, which is less than 10.
d) The largest sample we can take is 60 (n 
1
N , where N = 600), otherwise the sample
10
would be more than 10% of the population, and sampling without replacement would
require a finite population correction to calculate standard deviation. (That is not in the
scope of this course!)
2.a) µ p-hat = p =0.22;
σ p-hat =
(.22)(.78)
= 0.0589
50
b) State: What is the probability that more than 30% of the 50 randomly-selected songs
on George’s mp3 player are Beatles songs: Plan: P (p-hat > 0.30) Check the Normal
condition: np = (50)(.22) = 11, and n(1-p) = (50)(.78) = 39. Both are greater than 10, so
the Normal approximation will be quite accurate. So. . .Do: Standardize! z 
0.30  0.22
0.0589
= 1.36 . DRAW and LABEL the Normal curve! Shade to the right. Using Table A, P (z
> 1.36) = 1- .9131 = .0869. Conclude: Therefore, there is approximately an 8.69%
chance that more than 30% of George’s randomly selected songs are Beatles songs
played on his mp3 player.
QUIZ 7.3A
1. a) µ x = µ = 12 minutes
b) Yes. It seems reasonable to assume that the sample of 10 is less than 10% of the entire
population calls. σ x =
5
10
= 1.58
c) No. The population distribution is skewed right, and n = 10, which is not large enough
for the CLT to apply.
2. a) P( x>400) = P  z 
400  380 
 = P( z > 0.71) = 1 - .7611 = 0.2389
28


There is a 23.89% chance that a single randomly selected apple will weigh more than 400 gm.
400  380 
b) P( x > 400) = P  z 
 = P( z > 1.24) = 1 - .8925 = .1075
16.165 

There is a 10.75% chance that the mean weight of the SRS of 3 apples will weigh more than 400 gm.
c) The mean weight of a random sample of three apples is less variable than the weight of
a single randomly-selected apple, so we are less likely to get a mean weight that is 20 gm
above the mean when we take a sample of three apples.