Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.1 in​, and a standard deviation given by sigma equals 2.7 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. ​(b) If 32 women are randomly​ selected, find the probability that they have a mean height less than 63 in.

Answer 1

Solution :

Given that ,

mean = = 62.1

standard deviation = = 2.7

(a)

P(x < 63) = P((x - ) / < (63 - 62.1) / 2.7)

= P(z < 0.33)

Using standard normal table,

P(x < 63) = 0.6293

Probability = 0.6293

(b)

n = 32

= 62.1 and

= / n = 2.7 / 32 = 0.4773

P( < 63) = P(( - ) / < (63 - 62.1) / 0.4773)

= P(z < 1.89)

Using standard normal table,

P( < 63) = 0.9706

Probability = 0.9706