Assume that womens heights are normally distributed with a m

Assume that women?s heights are normally distributed with a mean given by s 63.8 in, and a standard deviation given by a2.6in. Complete parts a and b. a. If 1 woman is randomly selected, find the probability that her height is between 63.1 in and 64.1 in. The probability is approximately 0.1542 (Round to four decimal places as needed.) b. If 5 women are randomly selected, find the probability that they have a mean height between 63.1 in and 64.1 in. The probability is approximately [ ]. (Round to four decimal places as needed.)

Solution

a) P( 63.1 < Z < 64.1)

= P( (63.1 - 63.8)/2.6 < Z < (64.1 - 63.8)/2.6)

= P( -0.27 < Z < 0.12)

= 0.8936 - 0.5478

=0.3458 Answer

b)P( 63.1 < Z < 64.1)

= P( (63.1 - 63.8)/(2.6/sqrt(5)) < Z < (64.1 - 63.8)/(2.6/sqrt(5)))

= P( -0.6 < Z < 0.27)

= 0.2743 -0.1064

=0.1679 Answer