Triola Assignment C
Section 6-2 – Basic Skills and Concepts
1) When we refer to a “normal” distribution, does the word “normal” have the same meaning as
in ordinary language, or does it have a special meaning in statistics? What exactly is a normal
distribution?
The term normal does not have the same meaning as in ordinary language. In statistics,
normal refers to a distribution of data that is bell-shaped and symmetric about a mean
value. There are many scenarios in every day life where the data is normally
distributed; therefore, it is an important type of distribution to understand. Check: OK
3) What requirements are necessary for a normal probability distribution to be a standard normal
probability distribution?
In order for a normal probability distribution to be standard, the mean must by zero
and the standard deviation must be one. Check: OK
9) Assume that the readings on the thermometers are normally distributed with a mean of 0 and
a standard deviation of 1.00 C. A thermometer is randomly selected and tested. In each case,
draw a sketch, and find the probability of the reading: Less than –1.00.
The probability of reading less than –1.00 is: 0.1587 Check: OK
–1.00
0
11) Assume that the readings on the thermometers are normally distributed with a mean of 0
and a standard deviation of 1.00 C. A thermometer is randomly selected and tested. In each
case, draw a sketch, and find the probability of the reading: Less than 1.00.
The probability of reading less than 1.00 is: 0.8413 Check: OK
0
1.00
13) Assume that the readings on the thermometers are normally distributed with a mean of 0
and a standard deviation of 1.00 C. A thermometer is randomly selected and tested. In each
case, draw a sketch, and find the probability of the reading: Greater than 1.25.
The probability of reading greater than 1.25 is: 1 – 0.8944 = 0.1056 Check: OK
0
1.25
17) Assume that the readings on the thermometers are normally distributed with a mean of 0
and a standard deviation of 1.00 C. A thermometer is randomly selected and tested. In each
case, draw a sketch, and find the probability of the reading: Between 1.00 and 2.00.
The probability of reading between 1.00 and 2.00 is: 0.9772 – 0.8413 = 0.1359
Check: OK
0
1.00
2.00
Section 6-3 – Basic Skills and Concepts
1) What is the difference between a standard normal distribution and a non-standard normal
distribution?
For a standard normal distribution, the mean must be zero and the standard deviation
must be one. For a nonstandard normal distribution (which is more common in most
data sets), the mean may be non-zero, the standard deviation may be something other
than one, or there is a combination of both. Check: OK
5) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a
standard deviation of 15. Find the probability that a randomly selected adult has an IQ that is
less than 130.
130  100 30

 2.00
 Convert 130 to a z-score:
15
15
 Find the probability of a score less than 130: 0.9772. Check: OK
7) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a
standard deviation of 15. Find the probability that a randomly selected adult has an IQ between
90 and 110.
110  100 10
90  100  10

 0.67

 0.67
 Convert 90 and 110 to z-scores:
15
15
15
15
 Find the probability of a score between 90 and 110: 0.7486 – 0.2514 = 0.4972.
Check: OK
9) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a
standard deviation of 15. Find P10, which is the IQ score separating the bottom 10% from the top
90%.
 Find the z-score corresponding to a probability of 0.10: –1.28.
 Convert –1.28 to a non-standard score:
x  100
 1.28 
 15(1.28)  x  100  19.2  x  100  x  80.8 Check: OK
15
11) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a
standard deviation of 15. Find the IQ score separating the top 35% from the others.
 Since 35% represents the portion of the scores above the cutoff, 65% of the scores fall
below this point.
 Find the z-score corresponding to a probability of 0.65: 0.39.
 Convert 0.39 to a non-standard score:
x  100
0.39 
 15(0.39)  x  100  5.85  x  100  x  105.85 Check: OK
15
17) Birth weights in the U.S. are normally distributed with a mean of 3420 grams and a standard
deviation of 495 grams. If a hospital plans to set up special observation conditions for the
lightest 2% of babies, what weight is used for the cutoff separating the lightest 2% from the
others.
 Find the z-score corresponding to a probability of 0.02: –2.05.
 Convert –2.05 to a non-standard score:
x  3420
 2.05 
 495(2.05)  x  3420  1014.75  x  3420  x  2405.25 grams
495
Check: OK
Section 6-4 – Basic Skills and Concepts
1) Use your own words to answer this question: “What is a sampling distribution?”
In a population, there are many different samples of a certain size that can be drawn.
Although the different samples would be the same size, the values (i.e. means, standard
deviations, etc.) would most likely vary from sample to sample. Thus, the sample
distribution is a way of depicting all of the possible samples and the values associated
with those samples. In general, sample distributions tend toward a normal distribution.
Check: OK
3) Which of the following statistics are unbiased estimators?
a) Sample mean used to estimate the population mean.
YES, this is an unbiased estimator.
b) Sample median used to estimate the population median.
NO, this is not an unbiased estimator.
c) Sample proportion used to estimate the population proportion.
YES, this is an unbiased estimator.
d) Sample variance used to estimate the population variance.
YES, this is an unbiased estimator.
e) Sample standard deviation used to estimate the population standard deviation.
Technically, standard deviation is not an unbiased estimator. However, if the
sample is large enough, the level of bias is relatively low. Therefore, under certain
conditions is can be treated as an unbiased estimator.
f) Sample range used to estimate population range.
NO, this is not an unbiased estimator. Check: OK
5) Based on a random sample of n = 400 voters, the NBC news division predicts that the
Democratic candidate for the presidency will get 49% of the vote, but actually gets 51%. Should
be conclude that the survey was done incorrectly? Why or why not?
The survey was most likely done correctly. First, a different sample of 400 voters would
probably produce a result that was similar but not exactly the same. Different samples
will produce varying results. Another sample might have produced percentages of
48%, 50%, 51%, 53%, etc. The key is that if every possible sample were averaged
together, the results would have been 51%. Second, the survey result was pretty close
to the actual result; therefore, an assessment of the results would indicate that the
survey was fairly accurate. Check: OK
9) The assets (in billions of dollars) of the five wealthiest people in the U.S. are 47, 43, 21, 20,
and 20. Assume that samples of size 2 are randomly selected with replacement from this
population of five values.
a) After listing the 25 different possible samples and finding the mean of each sample, use a
table to describe the sample distribution of the sample means.
SAMPLES
20, 20
20, 20
20, 21
20, 43
20, 47
20, 20
20, 20
20, 21
20, 43
20, 47
21, 20
21, 20
21, 21
21, 43
21, 47
43, 20
43, 20
43, 21
43, 43
43, 47
47, 20
47, 20
47, 21
47, 43
47, 47
20
20.5
31.5
33.5
20
20
20.5
31.5
33.5
20.5
20.5
21
32
34
31.5
31.5
32
43
45
33.5
33.5
34
45
47
MEANS
20
SAMPLING DISTRIBUTION OF SAMPLE MEANS
Means
Probability
20.0
4/25 = 0.16
20.5
4/25 = 0.16
21.0
1/25 = 0.04
31.5
4/25 = 0.16
32.0
2/25 = 0.08
33.5
4/25 = 0.16
34.0
2/25 = 0.08
43.0
1/25 = 0.04
45.0
2/25 = 0.08
47.0
1/25 = 0.04
b) Find the mean of the sampling distribution.
  x  Px  (20  0.16)  (20.5  0.16)  (21 0.04)  ...  (47  0.04)  30.2
c) Is the mean of the sampling distribution equal to the mean of the population of five listed
values? Are those means always equal?
20  20  21  43  37 151


 30.2
5
5
The mean of the population is 30.2. The mean of the sampling distribution is also
30.2. As long as all of the possible samples of the same size are included, they will
always be equal. The mean of a sample distribution will target the population
mean. Check: OK
11) A genetics experiment involves a population of fruit flies consisting of 1 male named Mike
and 3 females named Anna, Barbara, and Chris. Assume that two fruit flies are randomly
selected with replacement.
a) After listing the 16 different possible samples, find the proportion of females in each
sample, then use a table to describe the sampling distribution of the proportions of females.
SAMPLES
M, M
M, A
M, B
M, C
A, M
A, A
A, B
A, C
B, M
B, A
B, B
B, C
C, M
C, A
C, B
C, C
0.5
0.5
0.5
0.5
1.0
1.0
1.0
0.5
1.0
1.0
1.0
0.5
1.0
1.0
1.0
MEANS
0
SAMPLING DISTRIBUTION OF SAMPLE MEANS
Proportion
Probability
0
1/16 = 0.0625
0.5
6/16 = 0.3750
1.0
9/16 = 0.5625
b) Find the mean of the sampling distribution.
  x  Px  (0  0.0625)  (0.5  0.375)  (1  0.5625)  0.75
c) Is the mean of the sampling distribution equal to the mean of the population of five listed
values? Are those means always equal?
The proportion of the population that was female was 3/4 = 0.75. The mean of the
sampling distribution is also 0.75. The sample proportion is always an unbiased
estimator of the population proportion. Check: OK
Section 6-5 – Basic Skills and Concepts
1) What is the standard error of the mean?
The standard error of the mean is the same as the standard deviation of the sample
means. It is found by dividing the standard deviation of the population by the square
root of the sample size. Check: OK
5) Assume that women’s heights are normally distributed with a mean given by  = 63.6 inches
and a standard deviation given by  = 2.5 inches.
a) If 1 woman is randomly selected, find the probability that her height is less than 64 inches.
 Since you are selecting from the population, you use the normal procedure for
calculating probability.
64  63.6 0.4

 0.16
 Convert 64 to a z-score:
2 .5
2.5
 Find the probability of a height of less than 64 inches: 0.5636. Check: OK
b) If 36 women are randomly selected, find the probability that they have a mean height less
than 64 inches.
 Since you are using a sample, you use the sample procedure for calculating
probability.
x   64  63.6
0.4


 0.96
 Convert 64 to a z-score: z 

2.5
0.416666
n
36
 Find the probability the sample mean will be less than 64 inches: 0.8315.
Check: OK
7) Assume that women’s heights are normally distributed with a mean given by  = 63.6 inches
and a standard deviation given by  = 2.5 inches.
a) If 1 woman is randomly selected, find the probability that her height is between 63.5 and
64.5 inches.
 Since you are selecting from the population, you use the normal procedure for
calculating probability.
64.5  63.6 0.9

 0.36
2.5
2.5
63.5  63.6  0.1

 0.04
2.5
2.5
 Find the probability of a height between 63.5 and 64.5 inches:
0.6406 – 0.4840 = 0.1566 Check: OK
 Convert 63. 5 and 64.5 to z-scores:
b) If 36 women are randomly selected, find the probability that they have a mean height less
than 64 inches.
 Since you are using a sample, you use the sample procedure for calculating
probability.
 Convert 63.5 and 64.5 to z-scores:
x   64.5  63.6
0.9
z


 1.08

2.5
0.833333
n
9
x   63.5  63.6
 0.1
z


 0.12

2.5
0.833333
n
9
 Find the probability the sample mean will be between 63.5 and 64.5 inches:
0.8599 – 0.4522 = 0.4077. Check: OK
c) Why can the central limit theorem be used in part (b), even though the sample size does
not exceed 30?
The population is normally distributed, therefore the sample distribution will be
normally distributed. Check: OK
13) An aircraft strobe light is designed so that the times between flashes are normally distributed
with a mean of 3.00 seconds and a standard deviation of 0.40 seconds.
a) Find the probability that an individual time is greater than 4.00 seconds.
 Since you are selecting from the population, you use the normal procedure for
calculating probability.
4.00  3.00 1.0

 2.50
 Convert 4.00 to a z-score:
0 .4
0.4
 Find the probability of a time that is greater than 4.00 seconds:
1 – 0.9938 = 0.0062 Check: OK
b) Find the probability that the mean for 60 randomly selected times is greater than 4.00
seconds.
 Since you are using a sample, you use the sample procedure for calculating
probability.
x   4.00  3.00
1.00


 19.364  19.36
 Convert 4.00 to a z-score: z 

0.40
0.0516397
n
60
 Find the probability the sample mean will be greater than 4.00 seconds:
1 – 0.9999 = 0.0001. Check: OK
c) Given that the strobe light is intended to help other pilots see an aircraft, which results is
more relevant for assessing the safety of the strobe light? Why?
The result from part (a) is the most relevant. The chance that a sample of 60 lights
will have a mean greater than 4.00 is extremely low. Therefore, it does not provide
much applicable information about the safety for an individual situation. Assessing
the safety for individual planes is the most beneficial. Check: OK
Unit 6 – Review Exercises
1) A scale is designed so that when items are weighed, the errors in the indicated weights are
normally distributed with a mean of 0 grams and standard deviation of 1 gram.
a) If an item is randomly selected and weighed, what is the probability that it has an error
between –0.5 grams and 0.5 grams?
The probability of reading between –0.5 and 0.5 is:
0.6915 – 0.3085 = 0.3830. Check: OK
b) If 16 items are randomly selected and weighed, what is the probability that the mean of the
errors is between –0.5 grams and 0.5 grams?
 Since you are using a sample, you use the sample procedure for calculating
probability.
 Convert –0.5 and 0.5 to z-scores:
x   0.5  0 0.5
x    0.5  0  0.5
z


 2.00
z


 2.00

1

1
0.25
0.25
n
16
n
16
 Find the probability the sample mean will be between –0.5 and 0.5:
0.9772 – 0.0228 = 0.9544. Check: OK
c) What is the 90th percentile for the errors?
 Find the z-score corresponding to a probability of 0.90: 1.28.
 Convert 1.28 to a non-standard score:
x0
1.28 
 1(1.28)  x  0  1.28  x  0  x  1.28 Check: OK
1
3) The serum cholesterol levels in men aged 18-24 are normally distributed with a mean of 178.1
and a standard deviation 40.7. Units are in mg/100 mL, and the data are based on the National
Health Survey.
a) If one man aged 18-24 is randomly selected, find the probability that his serum cholesterol
level is greater than 260, a value considered to be “moderately high.”
260  178.1 81.9

 2.0122  2.01
 Convert 260 to a z-score:
40.7
40.7
 Find the probability of a level greater than 260:
1 – 0.9778 = 0.0222. Check: OK
b) If one man aged 18-24 is randomly selected, find the probability that his serum cholesterol
level is between 170 and 200.
 Convert 170 and 200 to z-scores:
200  178.1 21.9

 0.53808  0.5381
40.7
40.7
170  178.1  8.1

 0.199017  0.1990
40.7
40.7
 Find the probability of a level between 170 and 200:
0.7054 – 0.4207 = 0.2847. Check: OK
c) If 9 men aged 18-24 are randomly selected, fin the probability that their mean serum
cholesterol level is between 170 and 200.
 Since you are using a sample, you use the sample procedure for calculating
probability.
 Convert 170 and 200 to z-scores:
x   200  178.1
21.9
z


 1.6142  1.61

40.7
13.566666
n
9
x   170  178.1
 8.1
z


 0.59705  0.60

40.7
13.566666
n
9
 Find the probability the sample mean will be between 170 and 200:
0.9463 – 0.2743 = 0.6720. Check: OK
d) The Providence Health Maintenance Organization wants to establish a criterion for
recommending dietary changes if cholesterol levels are in the top 3%. What is the cutoff for
men aged 18-24.
 Find the z-score corresponding to a probability of 0.97: 1.88.
 Convert 1.88 to a non-standard score:
x  178.1
1.88 
 1.88(40.7)  x  178.1  76.516  x  178.1  x  254.616
40.7
Check: I choose the wrong z-score nearest to a probability of 97%. I used 1.89
which made my result a little high. Otherwise the procedures were fine. OK
5) When several women are not hired at the Telektronics Company, they do some research and
find that among the many people who applied, 30% were women. However, the 20 people who
were hired consist of only 3 women and 17 men. Find the probability of randomly selecting 20
people from a large pool of applicants (30% of whom are women) and getting 3 or fewer women.
Based on the result, does it appear that the company is discriminating based on gender?
 Use the procedure for using a normal distribution to approximate a binomial
distribution.
 The requirements are: np  5 and nq  5. If n = 20, p = 0.3, and q = 0.7 then:
20(0.30) = 6  5 and 20(0.70) = 14  5.
 Calculate  and :
  n  p  20  0.30  6
  n  p  q  20  0.30  0.70  4.2  2.04939  2.05
 Since you want the value of 3 included in the probability and the data is a discrete
set, use 3.5 instead of three as the upper limit. Convert 3.5 to a z-score:
x   3.5  6  2.5
z


 1.2195  1.22

2.05
2.05
 Find the probability the sample proportion will be 3 or fewer: 0.1112.
 Because this probability not extremely low, it cannot be assumed that women are
being discriminated against. Check: OK I needed to use the answer in order to
extrapolate backwards. Even though I had the right idea for using the procedures for
using a normal distribution to approximate the binomial situation, I needed to use the
answer to figure our which procedure to use.
Unit 6 – Cumulative Review Exercises
1) Some standard food items are randomly selected and the carbohydrate contents (in grams) are
measured with the results listed below. (The items are regular coffee, one cup of whole milk,
one egg, one banana, one doughnut, one tablespoon of peanut butter, one carrot, and 10 potato
chips):
0, 12, 1, 27, 24, 3, 7, 10.
a) Mean:
b) Median:
c) SD:
0  12  1  27  24  3  7  10 84

 10.5 Check: OK
8
8
0 1 3 7 10 12 24 27
7  10 17

 8.5 Check: OK
The median is:
2
2
 x   0  1  3  7  10  12  24  27  1608
 x   (0  1  3  7  10  12  24  27)  84  7056
n x    x 
8(1608)  7056
12864  7056
s



2
2
2
2
2
2
2
2
2
2
nn  1
d) Variance:
2
2
2
2
88  1
87 
103.714285714  10.184  10.18
Check: OK
s 2  10.1840212  103.7142857  103.71 Check: OK
e) Convert the value of 3 grams to a z score:
3  10.5  7.5

 0.7367  0.74 Check: OK
10.18
10.18
5808

56
f) Find the actual percentage of these sample values that exceeds 3 grams:
5
 0.625  62.5% Check: OK
8
g) Assuming a normal distribution, find the percentage of population amounts that exceeds 3
grams. Use the sample values of x and s as estimates of  and .
 Since you do not want the value of 3 included in the probability and
the data is a discrete set, use 3.5 instead of three as the upper limit.
Convert 3.5 to a z-score:
x   3  10.5  7.5
z


 0.73673  0.74

10.18
10.18
 Find the probability the sample proportion will exceed 3 grams:
1 – 0.2296 = 0.7704.  77.04% Check: I used 3.5 instead of
three original because of the word “exceeds”. To me, that means
on lower portion of values, 3 should be part of that group. Once I
changed the value back to 3, I got the right answer. OK
h) What level of measurement describes this data set?
This data set is: ratio Check: OK
i) The listed measurements appear to be rounded to the nearest gram, but are the exact
unrounded amount discrete or continuous?
The unrounded data would be continuous since you could theoretically get any
decimal value. Check: OK
j) Does it make sense to use the sample mean as an estimate of the carbohydrate content of
the food items consumed by the average adult American? Why or why not?
Because the sample is very small, it would probably not be wise to use the sample
mean as an estimate. There are many different types of food with various amounts
of carbohydrate values. In addition, the eating patterns of Americans vary greatly.
Therefore, the food choices in the same may not be representative of the population
and the mean from this sample may not reflect the larger population. Check: OK