Normal Distribution Practice/Review
Answers
1. Draw two normal curves that have the same mean, but different standard deviations. Describe the similarities and
differences.
They are centered around the same point (highest point in the middle), but they have a different spread (one
distribution will be wider than another).
2. Draw two normal curves that have different means but the same standard deviations. Describe the similarities and
differences.
They will have the same spread, but one will be centered at a larger number than the other.
3. Consumer Reports Magazine wrote an article stating that monthly charges for cell phone plans in the U.S.
are normally distributed with a mean of $62 and a standard deviation of $18.
a) Draw a picture of the normal curve with the cell
b) What percent of people in the U.S. have a cell
phone charges for 1, 2 and 3 standard deviations above phone bill between $62 and $80 per month? How do
and below the mean.
you know?
34%; Empirical rule: 68% between -1SD and 1SD, so
68% is between 44 and 80. 62-80 is half of that.
68/2=34
8
26
44
62
80
98
116
c) Between what two cell phone charges are the
middle 95% of people? How do you know?
$26 and $98; Empirical rule: 95% between -2SD and
2SD, which are $26 and $98
d) What percent of people in the U.S. have a monthly
cell phone bill between $26 and $44? How do you
know?
13.5%; 26 to 98 = 95%, 44 to 80 = 68%, 95-68 = 27%
so 27% is in the 26-44 & 80-98 range together and
half of 27% is 13.5%
4. A survey was conducted to measure the height of U.S. men. In the survey, respondents were grouped by age.
In the 20–29 age group, the heights were normally distributed, with a mean of 69.2 inches and a standard
deviation of 2.9 inches. A study participant is randomly selected. (Source: U.S. National Center for Health
Statistics)
a) Find the percentile for a height of 66 inches. What
does this tell us about how tall this person is compared
to other men?
z = (66-69.2)/2.9 = -1.1 -> 0.1357 (proportion from
table) -> 13th to 14th percentile
b) Find the percent of heights between 66 and 72
inches. How do you know?
About 71%
z = (72-69.2)/2.9 = 0.96 = 1.0 -> 0.8413
This man is taller than 13 to 14% of other men his age. 0.8413 – 0.1357 = 0.7056
(from part a)
c) Find the percent of heights that are greater than 72
d) Find the height of a man who is taller than or as tall
inches. How do you know?
as 20% of adult men. How do you know?
About 67 inches
About 16%; from part b, 0.8413 below 72 inches.
1 – 0.8413 = 0.1587
0.2119 -> z = -0.8 -> (-0.8)(2.9)+69.2 = 66.88
5. The lengths of Atlantic croaker fish are normally distributed, with a mean of 10 inches and a standard
deviation of 2 inches. An Atlantic croaker fish is randomly selected. (Adapted from National Marine Fisheries
Service, Fisheries Statistics and Economics Division)
a) Find the probability that the length of the fish is
less than 7 inches. How do you know?
7%
z = (7-10)/2 = -1.5 -> 0.0668
b) Find the probability that the length of the fish is
between 7 and 15 inches. How do you know?
93%
z = (15-10)/2 = 2.5 -> 0.9938
0.9938 – 0.0668 = 0.927
c) Find the probability that the length of the fish is
more than 15 inches. How do you know?
1%
1 – 0.9938 = 0.0062
d) If 200 Atlantic croakers are randomly selected,
about how many would you expect to be shorter than 8
inches? How do you know?
31
z = (8-10)/2 = -1 -> 0.1587 or 15.87%
15.87% of 200 = 31.74
6. In a recent year, the ACT scores for high school students with a 3.50 to 4.00 grade point average were
normally distributed, with a mean of 24.1 and a standard deviation of 4.3. (Source: ACT, Inc.)
a) Find the percentile of a student whose ACT score is
a 20. What does this tell you about how well the
student did compared to others?
16th percentile; scored better than 16% of peers
z = (20-24.1)/4.3 = -0.95, -1.0 -> 0.1587
b) Find the percent of scores between 22 and 27.
38%
z = (27-24.1)/4.3 = 0.7 -> 0.7580
z = (22-24.1)/4.3 = -0.5 -> 0.3821
0.7580 – 0.3821 = 0.3759
c) If a student is in the 14th percentile what score did
they receive on the ACT?
19
14th -> 0.1357 -> -1.1 = z
(-1.1)(4.3)+24.1 =19.37
d) If a student’s score is in the 83rd percentile what
score did they receive on the ACT?
28
83rd -> between 0.8159 and 0.8413 -> 0.95 = z
(0.95)(4.3)+24.1 = 28.185
7. The weights of adult male rhesus monkeys are normally distributed, with a mean of 15 pounds and a
standard deviation of 3 pounds. A rhesus monkey is randomly selected.
a) If a monkey weighs 13 pounds, how many standard
deviations above the mean was this?
-2/3
(13-15)/3 = -2/3
b) George the monkey weighs 19 pounds. Find the
standardized value for 19 pounds. What does this tell
us about George’s weight?
1.33; George’s weight is 1.33 standard deviations
above the mean
(19-15)/3 = 1.33
Additional Practice: p. 254 (15-16), p. 250 (3), p. 252 (6), p. 254 (12)
Answers to Additional Practice:
p. 254
p. 250
p. 252
p. 254