Question

23. Suppose that a large Introduction to Psychology class has taken a midterm exam, and the scores are normally distributed with a mean of 75 and a standard deviation of 9. a. What proportion of the class got scores above 90? b. What percent of the class got scores below 78? c. What percent of the class got scores above 68? d. What proportion of the class got scores below 94

Answer 1

Solution :

Given that ,

mean = = 75

standard deviation = = 9

a) P(x > 90 ) = 1 - p( x< 90 )

=1- p [(x - ) / < (90 - 75) / 9]

=1- P(z < 1.67 )

= 1 - 0.9525 = 0.0475

proportion = 0.0475

b)

P(x < 78 ) = P[(x - ) / < (78 -75) /9 ]

= P(z < 0.33)

= 0.6293

probability = 0.6293 = 62.93%

c)

P(x > 68 ) = 1 - p( x< 68 )

=1- p [(x - ) / < (68- 75) / 9]

=1- P(z < -0.78 )

= 1 - 0.2177= 0.7823

proportion = 0.7823 = 78.23%

d)

P(x < 94 ) = P[(x - ) / < (94 -75) /9 ]

= P(z < 2.11)

= 0.9826

proportion = 0.9826