Question

1. Z is a standard normal variable. Find the value of Z in the area to the left of Z is 0.9279. The area to the right of Z is 0.1539.

2. Given that x is a Normal random variable with a mean of 10 and standard deviation of 4, find the following probability:

     P(8.8<x<12.5)

Answer 1

solution

(A)Using standard normal table,

P(Z < z) = 0.9279

= P(Z < z) = 0.9279

= P(Z <1.4603 ) = 0.9279

z=1.4603 ( rounded)

z = 1.46 (see the probability 0.9279 in standard normal (Z) table corresponding z value is 1.46 )

(B)

sing standard normal table,

P(Z > z) = 0.1539

= 1 - P(Z < z) = 0.1539

= P(Z < z ) = 1 - 0.1539=0.8461

= P(Z < z ) = 0.8461

= P(Z < 1.0198 ) = 0.8461  

z = 1.0198 ( rounded)

z=1.02 (see the probability 0.8461 in standard normal (Z) table corresponding value is 1.02 )

(C)

Solution :

Given that ,

mean = = 10

standard deviation = = 4

P(8.8< x <12.5 ) = P[(8.8 - 10) /4 < (x - ) / < (12.5 - 10) /4 )]

= P( -0.3< Z < 0.625)

= P(Z <0.625 ) - P(Z <-0.3 )

Using z table ( see the z value 0.625 in standard normal (z) table corresponding value is 0.7340) and ( see the z value -0.3 in standard normal (z) table corresponding value is 0.3821 )

= 0.7340 -0.3821   

=0.3519

probability=0.3519