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.

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

nope

step-by-step explanation:

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.