Question

Assume that the heights of men are normally distributed with a mean of 70.9 inches and a standard deviation of 2.1 inches. If 36 men are randomly selected, find the probability that they have a mean height greater than 71.9 inches. 0.9979 0.0021 0.9005 0.0210

Answer 1

Solution :

Given that ,

mean = $\mu$ = 70.9

standard deviation = $\sigma$ = 2.1

n = 36

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 2.1 / $\sqrt$ 36 = 0.35

P($\bar x$ > 71.9) = 1 - P($\bar x$ < 71.9)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (71.9 - 70.9) / 0.35]

= 1 - P(z < 2.8571)

= 1 - 0.9979

= 0.0021

Probability = 0.0021