Heights of adult males are approximately normally distributed with a mean of 66.9 inches and a standard deviation of 1.7 inches. If a sample of 50 adult males are​ chosen, what is the probability the their mean height will be between than 67 inches and 67.4 ​inches? Report your answer to four decimal places.

Respuesta :

Answer: P(67 ≤ x ≤ 67.4) = 0.644

Step-by-step explanation:

Since the heights of adult males are approximately normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = the heights of adult males.

µ = mean height

σ = standard deviation

n = number of samples

From the information given,

µ = 66.9 inches

σ = 1.7 inches

n = 50

The probability that their mean height will be between than 67 inches and 67.4 ​inches is expressed as

P(67 ≤ x ≤ 67.4)

For x = 67,

z = (67 - 66.9)/(1.7/√50) = - 0.42

Looking at the normal distribution table, the probability corresponding to the z score is 0.3372

For x = 67.4,

z = (67.4 - 66.9)/(1.7/√50) = 2.08

Looking at the normal distribution table, the probability corresponding to the z score is 0.9812

Therefore,

P(67 ≤ x ≤ 67.4) = 0.9812 - 0.3372 = 0.644

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