Question

The heights of women aged 30 to 35 are approximately Normal with mean 63 inches and standard deviation 2.4 inches. Men the same age have mean height 70 inches with standard deviation 2.5 inches. If one woman and one man are randomly selected, what is the probability that heights of both are greater than 66 inches tall? Assume that heights of women and men are independent.

Answer 1

Answer

given,

mean of women  (μ)=63 inches

standard deviation of women (σ)= 2.4 inches

mean of men (μ)=70 inches

standard deviation of women (σ)= 2.5 inches

for women

$P(x>66)$

$P(x>66) = 1-P( \dfrac{x - \mu}{\sigma}$

$P(x>66) = 1-P( z$

$P(x>66) = 1-P( z$

using z-table

$P(x>66)= 1 - 0.8944$

$P(x>66)=0.1056[/tex]for men[tex]P(x>66)$

$P(x>66) = 1-P( \dfrac{x - \mu}{\sigma}$

$P(x>66) = 1-P( z$

$P(x>66) = 1-P( z$

using z-table

$P(x>66)= 1 - 0.0548$

$P(x>66)=0.9452$