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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