Homework SourseUS adult male heights are on average 70 inches510 with a sta
U.S adult male heights are on average 70 inches(5\'10) with a standard deviation of 4 inches. Adult women are on average a bit shorter and less variable in height with a mean height of 65 inches (5\'5) and a standard deviation of 3.5 inches.
Calculate the percentile for the following
a) Lebron James who is 80 inches tall
b) Tom Cruise who is 67 inches tall
c) Nicku Minaj who is 62 inches tall
d) Aisha Tyler who is 72 inches tall
Solution
A)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 80
u = mean = 70
s = standard deviation = 4
Thus,
z = (x - u) / s = 2.5
Thus, using a table/technology, the left tailed area of this is
P(z > 2.5 ) = 0.993790335 OR 99.38th percentile [answer]
[Please round this down to the nearest ineteger if that\'s what you do in class.]
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B)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 76
u = mean = 64
s = standard deviation = 26
Thus,
z = (x - u) / s = 0.461538462
Thus, using a table/technology, the right tailed area of this is
P(z > 0.461538462 ) = 0.322206167 or 32.22th percentile [ANSWER]
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C)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 62
u = mean = 65
s = standard deviation = 3.5
Thus,
z = (x - u) / s = -0.857142857
Thus, using a table/technology, the left tailed area of this is
P(z > -0.857142857 ) = 0.195682969 = 19.57th percentile [ANSWER]
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D)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 72
u = mean = 65
s = standard deviation = 3.5
Thus,
z = (x - u) / s = 2
Thus, using a table/technology, the left tailed area of this is
P(z > 2 ) = 0.977249868 or 97.72th percentile [ANSWER]
