Assume that women\'s heights are normally distributed with a mean given by u=63.
ID: 3356971 • Letter: A
Question
Assume that women's heights are normally distributed with a mean given by u=63.7 in, and a standard deviation given by o=2.2 in. Complete parts a and b.
a. If 1 woman is randomly selected, find the probability that her height is between 63.6 in and 64.6 in.
The probability is approx. ______ (Round to four decimal places as needed.)
b. If 15 women are randomly selected, find the probability that they have a mean height between 63.6 in and 64.6 in.
The probability is approx. _____ (Round to four decimal places as needed.)
Explanation / Answer
Mean is 63.7 and s is 2.2
a) P(63.6<x<64.6)=P((63.6-63.7)/2.2<z<(64.6-63.7)/2.2)=P(-0.05<z<0.41), or P(z<0.41)-(1-P(z<0.05))
from normal distribution table we get 0.6591-(1-0.5199)=0.179
b) P(53.6<xbar<64.6), standard error SE is s/sqrt(N)=2.2/sqrt(15)=0.568
thus P((63.6-63.7)/0.568<z<(64.6-63.7)/0.568)= P(-0.18<z<1.58) =P(z<1.58)-(1-P(z<0.18))
from normal table we get 0.9429-(1-0.5714)=0.5143