Question

Scores on an exam are normally distributed with a mean of 65 and a standard deviation of 9. Find the percent of the scores that satisfies the following:

(a) Less than 54 (b) At least 80 (c) Between 70 and 86

Answer 1

Solution :

Given that ,

mean = =65

standard deviation = = 9

P(X<54 ) = P[(X- ) / < (54-65) / 9]

= P(z <-1.22 )

Using z table

= 0.1112

answer=11.12%

(b)

P(x > 80) = 1 - P(x<80 )

= 1 - P[(x -) / < (80-65) /9 ]

= 1 - P(z <1.67 )

Using z table

= 1 - 0.9525

= 0.0475

answer= 4.75%

(b)

P(70< x <86 ) = P[(70-65) /9 < (x - ) / < (86-65) /9 )]

= P( 0.56< Z < 2.33)

= P(Z <2.33 ) - P(Z <0.56)

Using z table   

= 0.9901-0.7123

=0.2778

answer=27.78%