Question

If men's heights are normally distributed with mean 68 and standard deviation 3.1 inches, and women's heights are normally distributed with mean 65 and standard deviation 2.8 inches, if a random man and random woman are chosen, find the probability that the woman is taller than the man.

Answer 1

Answer:

the probability that the woman is taller than the man is  0.1423

Step-by-step explanation:

Given that :

the men's heights are normally distributed with mean $\mu$  68

standard deviation $\sigma$   = 3.1

And

the women's heights are normally distributed with mean $\mu$  65

standard deviation $\sigma$   = 2.8

We are to find the probability that the woman is taller than the man.

For woman now:

mean  $\mu$ = 65

standard deviation $\sigma$ = 2.8

$P(x > 68 ) = 1 - p( x< 68)$

$\\ 1 -p \ P[(x - \mu ) / \sigma < (68-25)/ 2.8]$

= 1-P (z , 1.07)

Using z table,

= 1 - 0.8577

= 0.1423

Thus, the probability that the woman is taller than the man is  0.1423