Question

Exam grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of 11. Use the normal distribution to answer the following questions.

(a) What percent of students scored above an 88 ?Round your answer to one decimal place.

(b) What percent of students scored below a 59 ?Round your answer to one decimal place.

(c) If the lowest 7%of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?Round your answer to one decimal place.

(d) If the highest 9%of students will be given a grade of A, what is the cutoff to get an A? Round your answer to one decimal place.

Answer 1

Solution :

Given that ,

mean = = 72

standard deviation = = 11

(a)P(x >88 ) = 1 - P(x < 88)

= 1 - P[(x - ) / < (88 - 72) / 11]

= 1 - P(z <1.46 )

Using z table,

= 1 -0.9279

=0.0721

answer =7.2%

(b)P(x <59 ) = P[(x - ) / < (59 -72 ) /11 ]

= P(z <-1.18 )

Using z table,

=0.119

answer = 11.9%

(c)Using standard normal table,

P(Z < z) = 7%

= P(Z < z) = 0.07

= P(Z <-1.48 ) = 0.07

z = -1.48

Using z-score formula  

x= z * +

x= -1.48 *11+72

x= 55.7

(d)Using standard normal table,

P(Z > z) = 9%

= 1 - P(Z < z) = 0.09  

= P(Z < z) = 1 - 0.09

= P(Z < z ) = 0.91

= P(Z < 1.34) = 0.91

z =1.34

Using z-score formula,

x = z * +

x = 1.34*11+72

x = 86.7