X is a normally distributed random variable with mean 72 and

X is a normally distributed random variable with mean 72 and standard deviation 22. Find the probability indicated.

P(78<X<127)

P(60<X<90)

P(49<X<71)

Solution

a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 78) = (78-72)/22
= 6/22 = 0.2727
= P ( Z <0.2727) From Standard Normal Table
= 0.60747
P(X < 127) = (127-72)/22
= 55/22 = 2.5
= P ( Z <2.5) From Standard Normal Table
= 0.99379
P(78 < X < 127) = 0.99379-0.60747 = 0.3863                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-72)/22
= -12/22 = -0.5455
= P ( Z <-0.5455) From Standard Normal Table
= 0.29272
P(X < 90) = (90-72)/22
= 18/22 = 0.8182
= P ( Z <0.8182) From Standard Normal Table
= 0.79337
P(60 < X < 90) = 0.79337-0.29272 = 0.5007                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 49) = (49-72)/22
= -23/22 = -1.0455
= P ( Z <-1.0455) From Standard Normal Table
= 0.14791
P(X < 71) = (71-72)/22
= -1/22 = -0.0455
= P ( Z <-0.0455) From Standard Normal Table
= 0.48187
P(49 < X < 71) = 0.48187-0.14791 = 0.334