Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.6 inμ=62.6 in​, and a standard deviation given by sigma equals 2.2 inσ=2.2 in.

​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 6363 in.​

(b) If 3232 women are randomly​ selected, find the probability that they have a mean height less than 6363 in.

Answer 1

Solution :

Given that ,

mean = = 62.6

standard deviation = = 2.2

(a)

P(x < 63) = P((x - ) / < (63 - 62.6) / 2.2)

= P(z < 0.18)

Using standard normal table,

P(x < 63) = 0.5714

Probability = 0.5714

(b)

n = 32

= 62.6 and

= / n = 2.2 / 32 = 0.3889

P( < 63) = P(( - ) / < (63 - 62.6) / 0.3889)

= P(z < 1.03)

Using standard normal table,

P( < 63) = 0.8485

Probability = 0.8485