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

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

answer=92.7%

(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

answer=99.38%

(C)P(x  9)

= P[(x - ) /   (9-11) / 2]

= P(z -1)

Using z table,

=0.1587

answer=15.87%