Assume that womens heights are normally distributed with a m

Assume that women?s heights are normally distributed with a mean given by mu = 62.9 in, and a standard deviation given by sigma =2.2 in. Complete parts a and b. a. If 1 woman is randomly selected, find the probability that her height is between 62.1 in and 63.1 in. The probability is approximately []. (Round to four decimal places as needed.)

Solution

P(62.1<X<63.1) = P((62.1-62.9)/2.2 <(X-mean)/s <(63.1-62.9)/2.2)

=P(-0.36<Z<0.09)

=0.1764 (from standard normal table)