Suppose x has a distribution with 24 and 22 a If a random

Suppose x has a distribution with = 24 and = 22.

(a) If a random sample of size n = 44 is drawn, find x, x and P(24 x 26). (Round x to two decimal places and the probability to four decimal places.) x = x = P(24 x 26) =

(b) If a random sample of size n = 59 is drawn, find x, x and P(24 x 26). (Round x to two decimal places and the probability to four decimal places.) x = x = P(24 x 26) =

Solution

Mean ( u ) =24
Standard Deviation ( sd )= 22/ Sqrt ( 44 ) = 3.3166
Number ( n ) = 44
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 24) = (24-24)/22/ Sqrt ( 44 )
= 0/3.3166
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 26) = (26-24)/22/ Sqrt ( 44 )
= 2/3.3166 = 0.603
= P ( Z <0.603) From Standard Normal Table
= 0.72675
P(24 < X < 26) = 0.72675-0.5 = 0.2268                  

b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 24) = (24-24)/22/ Sqrt ( 59 )
= 0/2.8642
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 26) = (26-24)/22/ Sqrt ( 59 )
= 2/2.8642 = 0.6983
= P ( Z <0.6983) From Standard Normal Table
= 0.7575
P(24 < X < 26) = 0.7575-0.5 = 0.2575