Q)If the mean height of college males was 70\" with a standard deviation of 3\",
ID: 3336385 • Letter: Q
Question
Q)If the mean height of college males was 70" with a standard deviation of 3", approximately what percentage of college males would be between 6'1" and 6'4" (Note: 12 inches = 1 foot)? A)68%. B)14%. C)48%. D)2%. E)No students in this range.Q)Using the data from question above, determine the z score for a height of 5'8" (show the work and select the answer) A)2/3. B)0.667. C) -2/3. D)-1.0 E)0. Q)If the mean height of college males was 70" with a standard deviation of 3", approximately what percentage of college males would be between 6'1" and 6'4" (Note: 12 inches = 1 foot)? A)68%. B)14%. C)48%. D)2%. E)No students in this range.
Q)Using the data from question above, determine the z score for a height of 5'8" (show the work and select the answer) A)2/3. B)0.667. C) -2/3. D)-1.0 E)0. Q)If the mean height of college males was 70" with a standard deviation of 3", approximately what percentage of college males would be between 6'1" and 6'4" (Note: 12 inches = 1 foot)? A)68%. B)14%. C)48%. D)2%. E)No students in this range.
Q)Using the data from question above, determine the z score for a height of 5'8" (show the work and select the answer) A)2/3. B)0.667. C) -2/3. D)-1.0 E)0.
Explanation / Answer
Solution:
1) option B) 14%
Explanation:
Let X be the height of college males.
Given that, mean(X) = 70, standard deviation(X) = 3.
X follows approximetely Normal distribution with mean=70, s.d.=3.
Z=(X-70)/3 follows standard normal distribution.
6'1" = 73", 6'4" = 76"
First we have to find P(73 < X < 76).
= P(73-70/3 < X- x/ < 76-70/3) = P(1< z< 2)
= P(0 < Z < 2)-P( 0 < Z < 1) = 0.4772-0.3413 = 0.1359
Percentage = 0.1359*100 = 13.59%
percentage of college males would be between 6'1" and 6'4" is 13.59% 14%
2) option E) 0
Explanation:
X follows approximetely Normal distribution with mean=70, s.d.=3.
Z=(X-70)/3 follows standard normal distribution.
5'8" = 69 inches
P(X < 69.6) = P[(X- x)/ < 70-70/3)
= P(Z < 0)
= 0.5
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