Question

Scores on a statistics final in a large class were normally distributed with a mean of 72 and a standard deviation of 4 . Use the Cumulative Normal Distribution Table to answer the following.

(a) Find the 37 th percentile of the scores.

(b) Find the 70 th percentile of the scores.

(c) The instructor wants to give an A to the students whose scores were in the top 12 % of the class. What is the minimum score needed to get an A?

(d) Between what two values is the middle 40 % of the scores? Round the answers to at least two decimal places.

Answer 1

solution

Given that,

mean = = 72

standard deviation = = 4

Using standard normal table,

P(Z < z) = 37%

= P(Z < z) = 0.37

z = -0.33 Using standard normal z table,

Using z-score formula  

x= z * +

x= -0.33*4+72

x= 70.68

b.

Using standard normal table,

P(Z < z) = 70%

= P(Z < z) = 0.70

z = 0.52 Using standard normal z table,

Using z-score formula  

x= z * +

x= 0.52*4+72

x= 74.08

c.

Using standard normal table,

P(Z > z) = 12%

= 1 - P(Z < z) = 0.12  

= P(Z < z) = 1 - 0.12

= P(Z < z ) = 0.88

= P(Z < 1.18 ) = 0.88

z =1.18

Using z-score formula  

x= z * +

x= 1.18*4+72

x= 76.72

d.

middle 40% of score is

P(-z < Z < z) = 0.40

P(Z < z) - P(Z < -z) = 0.40

2 P(Z < z) - 1 = 0.40

2 P(Z < z) = 1 + 0.40 = 1.40

P(Z < z) = 1.40/ 2 = 0.7

P(Z < 0.52) = 0.7

z  ±0.52 using z table

Using z-score formula  

x= z * +

x= ±0.52 *4+72

x= 69.92 and 74.08