Note You can earn partial credit on this problem Assume that

Note: You can earn partial credit on this problem.

Assume that women\'s weights are normally distributed with a mean given by 174 lb Note: You can earn partial credit on this problem. 108 lb and 70 women are randomly selected, find the probability that they have a mean weight between If 174 lb (c) 108 lb and 3 women are randomly selected, find the probability that they have a mean weight between If 174 lb (b) 108 lb and If 1 woman is randomly selected, find the probability that her weight is between sigma=29 lb. (a) mu=143 lb and a standard deviation given by

Solution

a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 108) = (108-143)/29/ Sqrt ( 24 )
= -35/5.9196
= -5.9126
= P ( Z <-5.9126) From Standard Normal Table
= 0
P(X < 174) = (174-143)/29/ Sqrt ( 24 )
= 31/5.9196 = 5.2368
= P ( Z <5.2368) From Standard Normal Table
= 1
P(108 < X < 174) = 1-0 = 1                  
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 108) = (108-143)/29/ Sqrt ( 3 )
= -35/16.7432
= -2.0904
= P ( Z <-2.0904) From Standard Normal Table
= 0.01829
P(X < 174) = (174-143)/29/ Sqrt ( 3 )
= 31/16.7432 = 1.8515
= P ( Z <1.8515) From Standard Normal Table
= 0.96795
P(108 < X < 174) = 0.96795-0.01829 = 0.9497                  
c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 108) = (108-143)/29/ Sqrt ( 70 )
= -35/3.4662
= -10.0976
= P ( Z <-10.0976) From Standard Normal Table
= 0
P(X < 174) = (174-143)/29/ Sqrt ( 70 )
= 31/3.4662 = 8.9436
= P ( Z <8.9436) From Standard Normal Table
= 1
P(108 < X < 174) = 1-0 = 1