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HON 180 Spring 2010 Homework 3 Brief Answers
p. 93 1, 2, 3, 6, 7, 9, 11
p. 104 1, 3, 4, 5
p. 105 4, 11, 13
Extra credit p. 105 3
45 total points plus up to 2 point for the extra credit
1. In a certain lake, the weights of stocked adult golden trout are normally distributed
with a mean of 3 pounds and a standard deviation of 1.3 lbs.
a. Approximately what percentage of adult golden trout in this lake weigh under 1 lb?
Using the table, look up z = (1 − 3)/1.3 = −1.54. The subtract from 100 and divide by 2.
Using a TI-calculator, type in normalcdf(−1E+99,1, 3, 1.3)  6.2%.
b. Between which two values do the middle 95% of adult golden trout weights lie?
Looking 95% in the area column in the table, we z  2. So the middle 95% will lie
between 3 – 2*1.3 = 0.4 and 3 + 2*1.3 = 5.6 lbs. Using a TI calculator, we find
InvNorm(0.025,3,1.3)  0.45 lbs and InvNorm(0.975,3,1.3)  5.55 lbs.
c. What is the interquartile range for the weights?
Looking up 50% in the area column we get z is between 0.65 and 0.7. In fact z  0.67.
So the answer is 3  0.67*1.3, 2.1 to 3.9 lbs. Using a TI calculator, you can calculate
InvNorm(0.25, 3, 1.3)  2.1 and InvNorm(0.75,3, 1.3)  3.9 lbs.
d. When fishing, golden trout weighing less than 2 pounds must be thrown back.
Approximately what percentage of adult golden trout need to be thrown back?
On the TI-calculator, the percentage is normalcdf(−1E+99, 2, 3, 1.3)  22%. Using the
table, you would look up z = (2 – 3)/1.3 = 0.77. You will get about 56%. Then subtract
from 100% and divide by 2 to get 22%.
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p. 93
1. a. (2 points)
Looking up z = 1.25 in the table or using normalcdf(-1.25,1.25, 0,1) on a TI calculator,
you get 78.87%. So there should be about 0.7887*25 or 20 scores in this range.
b. (1 point) 18
2. (2 points) Something is wrong with the computer. The entries should be around 1 in
size with some being perhaps as large as 3, but the numbers are much too large.
3. a) (2 points)
Looking up z  (700  543) /110  1.43 in the table and subtracting from 100% and
dividing by 2 or using normalcdf(700,1E+99, 543,110), you get 7.7%.
b) (2 points)
Looking up z  (700  499) /110  1.83 in the table or using normalcdf(700,1E+99,
499,110) on a TI calculator, you get 3.4%.
6. (2 points) No. For example, the normal curve says that about 16% of the scores should
be more than 1 SD above the average, and there are none!
7. a. (2 points) Looking up z  (400  550) /100  1.5 in the table or using
normalcdf(-1E+99, 400, 550,100) on a TI calculator, you get about the 7th percentile.
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b. (2 points) Looking up 0.5 in the area column of the table or using
invNorm(0.75,550,100), you get 617.
9. a. (1 point) False. For example, the list 1, 2, 99 has a median of 2 and an average of
34.
b. (1 point) False. For the list 1,2, 99, two-thirds of the entries are below the average.
c. (1 point) False. For example, income data (p. 88) have a long right-hand tail;
education levels have a long left-hand tail as well as bumps at 8, 12, and 16 years.
d. (1 point) False. If the histogram for a list follows the normal curve, the percentage of
entries within 1 SD of the average will be around 68%. Otherwise, the percentage could
be a lot different. The percentage within 1 SD of the average depends on the shape of the
histogram; the mean and SD do not determine the shape (i.e., it is possible for histograms
to have the same average and SD but have quite different shapes). For instance, take a
list of eight numbers: 30, 50, 50, 50, 50, 50, 50, 70. The average is 50, the SD is 10, and
75% are within 1 SD of the average.
A histogram that is more peaked than a normal curve such as the one below
About 78% of
the data is
within 1 SD of
the mean.
has more than 68% of the area between −1 and 1 SD’s from the mean. (For reference the
distribution above is  8 2 1 (41 x )4 .)
A histogram that is flatter than a normal curve such as the one depicted below
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Only about 58% of
the data is within 1
SD of the mean
has less than 68% of the histogram with one SD of the mean.
11. (2 points) Histogram (i) is correct. With (ii), the average of 1.1 would be right in the
middle, and a lot of people would be taking fewer than 1.1 – 1.5 = −0.4 courses. A
similar problem occurs with (iii).
p. 104
1. (1 point) False. Each measurement is thrown off by chance error, and this changes
from measurement to measurement.
3. a. (1 point) False. Chance errors are sometimes positive and negative. Bias pushes in
one direction.
b. (1 point) False.
c. (1 point) True.
4. (1 point) The expected chance error should be around the standard deviation of the
measurements. The standard deviation is
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3
(0.042  0.012  0.032 )  0.03 inches
5. a. (1 point) Person 2 and 10 probably copied from each other, but otherwise they seem
to have worked independently.
b. (1 point) Not only do different individuals get different measurements, but each
individual gets different measurements when he or she measures twice.
p. 105
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4. a. (2 points) Looking up z = 1.5 in the table or using normalcdf(400,700,550,100), you
get 86.6%.
b. (2 points) 68% of the scores should lie between 450 and 650. To find out how many
percent lie between 500 and 600 we use normalcdf(500,600,550, 100) ≈ 38.2%.
So there should be 1000*0.382/0.68 ≈ 560 students with scores between 500 and 600.
Some of you used the proportionality
1000
x

68% 38.2%
11. (2 points) Looking up 0.50 in table, you get 0.67.
This means 65.8 − 62.2 = 3.6 is 2*(0.67) = 1.34 SD’s. So the SD is about 2.7. The mean
is about (65.8 + 62.2)/2 = 64. The ninetieth percentile occurs at 1.28 standard deviations
from the mean (look up 0.2 in the area or use InvNorm(0.9,0,1) ). So the ninetieth
percentile occurs at around 64 + 1.28*2.7 = 67.5 inches.
13. a. (1 point) Heart disease rates go up with age: the drivers could be older, which could
account for the difference in rates. That explanation has been eliminated.
b. (1 point) You want the difference in rates to be due to exercise on the job, rather than
pre-existing factors. It might take some time for exercise to have its presumed beneficial
effect.
c. (1 point) Drivers and conductors are going to be similar with respect to age, education,
income, etc.
d. (1 point) This is an observational study, so there may be some confounding. Drivers
may be more at risk than conductors to start with. For example, drivers may be heavier to
begin with.
e. (1 point) You could look to see if the drivers and conductors had similar body sizes
when they were hired; this would suggest that the two groups were comparable at time of
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hire, and strengthen the argument that exercise matters. If drivers were heavier than
conductors at time of hire, the argument for exercise is weaker.
As it turned out, the drivers were heavier than the conductors at time of hire, explaining
the difference in rates of heart disease.
Extra credit p. 105 3. (2 points)
One way to do this is solve the system of equations
m  1.8SD  79

m  0.8SD  64
You will get a mean of 52 and a SD of 15. So the standard score of 52 is 0, the standard
score of 72 is 1.33, and 31 has a standard score of −1.4.
You can do this without solving systems of equations by noting that 79 and 64 are 1 SD
apart so SD =15. Then subtract 1.8*15 = 27 from 79 to get the mean of 52.
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