Question

A variable is normally distributed with mean 11 and standard deviation 2.

a. Find the percentage of all possible values of the variable that lie between 8 and 16.

b. Find the percentage of all possible values of the variable that are at least 6.

c. Find the percentage of all possible values of the variable that are at most 9.

Answer 1

Solution :

Given that ,

mean = = 11

standard deviation = = 2

P(8< x <16 ) = P[(8 - 11) / 2< (x - ) / < (16 - 11) /2 )]

= P( -1.5< Z <2.5 )

= P(Z <2.5 ) - P(Z < -1.5)

Using z table   

= 0.9938 - 0.0668

= 0.9270

=92.70%

b.

P(x > 6) = 1 - P(x< 6)

= 1 - P[(x -) / < (6 -11) /2 ]

= 1 - P(z <-2.5 )

Using z table

= 1 - 0.0062

=0.9938

=99.38%

c.

P(x< 9)

= P[(x -) / < (9 -11) /2 ]

= P(z <-1)

Using z table

=0.8413

=84.13%