Assume that women\'s heights are normally distributed with a mean given by mu =

Question

Assume that women's heights are normally distributed with a mean given by mu = 62.9 in, and a standard deviation given by sigma = 1.9 in. If 1 woman is randomly selected, find the probability that her height is between 62.6 in and 63.6 in. The probability is approximately. (Round to four decimal places as needed.) If 13 women are randomly selected, find the probability that they have a mean height between 62.6 in and 63.6 in. The probability is approximately. (Round to four decimal places as needed.) Why can the central limit theorem be used in part (b), even though the sample size does not exceed 30? The sample is normally distributed. The population is normally distributed. The population size is greater than 30. The sample size needs to be less than 30, not greater than 30.

Explanation / Answer

Here mean is 62.9 and sd is 1.9

a. Now we need to compute P(62.6<x<63.6)=>P(62.6-62.9/1.9<z<63.6-62.9/1.9)

P(-0.16<z<0.37)=P(z<0.37)-P(z<=-0.16)=0.2079

b. Now we need to find n is 13

So P(62.6<x<63.6)

=P((62.6-62.9)/(1.9/sqrt(13))<z<(63.6-62.9)/(1.9/sqrt(13))

=P(-0.57<z<1.33)

=P(z<1.33)-P(z<-0.57)

=0.6239

c. As per central limit theorem

Hence answer is b. The population is normally distributed.

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