Given a normal distribution with a mean of 100 and standard

Given a normal distribution with a mean of 100, and standard deviation of 10, if the meanselects a sample of n=25, what is the probabilit that Xbar is:

a. Less than 95?

b. Between 95 & 97.5<

c. Above 102.2?

d. There is a 65% chance that Xbar is above what value?

Solution

Normal Distribution
Mean ( u ) =100
Standard Deviation ( sd )=10
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 95) = (95-100)/10
= -5/10= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.3085                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 95) = (95-100)/10
= -5/10 = -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.30854
P(X < 97.5) = (97.5-100)/10
= -2.5/10 = -0.25
= P ( Z <-0.25) From Standard Normal Table
= 0.40129
P(95 < X < 97.5) = 0.40129-0.30854 = 0.0928                  
c)
P(X > 102) = (102-100)/10
= 2/10 = 0.2
= P ( Z >0.2) From Standard Normal Table
= 0.4207                  
d)
P ( Z < x ) = 0.97
Value of z to the cumulative probability of 0.97 from normal table is 1.881
P( x-u/s.d < x - 100/10 ) = 0.97
That is, ( x - 100/10 ) = 1.88
--> x = 1.88 * 10 + 100 = 118.81