Question #24288

suppose that heights of a 10 year old boys and girls closely follow a normal distribution with mean 51 inches and standard deviation 2 inches<br><br>a. what percentage of 10 year olds are shorter than 50 inches?<br>b. What percentage of 10 year olds fall between 48 and 54 inches?

Expert's answer

normal distribution with mean 51 inches andstandard deviation 2 inches:

f (x, mean, sigma)

where:

mean - 51 inches,

sigma - standard deviation - 2 inches

a)

percentage of 10 year olds is shorter than 50 inches:

x<50 => x (-Infinity, 50)

Therefore:

Integrate [ f(x, mean, sigma), {x, -Infinity, 50}] = 0.308538= 31%

b)

percentage of 10 year olds falls between 48 and 54inches:

48<x<54 => x (48, 54)

Therefore:

Integrate [ f(x, mean, sigma), {x, 48, 54}] = 0.866386 = 87%

f (x, mean, sigma)

where:

mean - 51 inches,

sigma - standard deviation - 2 inches

a)

percentage of 10 year olds is shorter than 50 inches:

x<50 => x (-Infinity, 50)

Therefore:

Integrate [ f(x, mean, sigma), {x, -Infinity, 50}] = 0.308538= 31%

b)

percentage of 10 year olds falls between 48 and 54inches:

48<x<54 => x (48, 54)

Therefore:

Integrate [ f(x, mean, sigma), {x, 48, 54}] = 0.866386 = 87%

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