Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.2 in​, and a standard deviation given by sigma equals 1.9 in.

Complete parts a and b.

a. If 1 woman is randomly​ selected, find the probability that her height is between 61.9 in and 62.9 in. The probability is approximately nothing. ​(Round to four decimal places as​ needed.)

b. If 14 women are randomly​ selected, find the probability that they have a mean height between 61.9 in and 62.9 in. The probability is approximately nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 62.2

standard deviation = = 1.9

a) P(61.9 < x < 62.9) = P[(61.9 - 62.2)/ 1.9) < (x - ) /  < (62.9 - 62.2) / 1.9) ]

= P(-0.16 < z < 0.37)

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

Using z table,

= 0.6443 - 0.4364

= 0.2079

b) n = 14

= = 62.2

= / n = 1.9/ 14 = 0.508

P(61.9 < < 62.9)  

= P[(61.9 - 62.2) /0.508 < ( - ) / < (62.9 - 62.2) / 0.508)]

= P(-0.59 < Z < 1.38)

= P(Z < 1.38) - P(Z < -0.59)

Using z table,  

= 0.9162 - 0.2776

= 0.6386