The boeing 757-200 er airliner carries 200 passengers and has doors with a height of 72 inches. heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. if half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 inches.

a)0.0001

b)0.8577

c)0.

d)0.1432

b)0.8577

Step-by-step explanation:

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

z = (x - u)/s

Where

x = heights of men

u = mean height

s = standard deviation

From the information given,

u = 69 inches

s = 2.8 inches

We want to find the probability that the mean height of the 100 men is less than 72 inches.. It is expressed as

P(x < 72)

For x = 72

z = (72 - 69)/2.8 = 1.07

Looking at the normal distribution table, the probability corresponding to the z score is 0.8577

P(x < 72) = 0.8577

you can not solvr this without numbers