Question

The scores for the final exam in a particular class are approximately normally distributed with a meann of 78.4 points and standard deviation of 5.7.

A. What score would a student nneed to score inn the top 20% of sudent scores? Round two decimal places.

B. What is the probablity that a randomly selected group of 36 students will have a eman score of more than 80 points? Innclude a probability statement. Round four decimal places.

Answer 1

(A)

= 78.4

= 5.7

To 20% corresponds to area = 0.80 - 0.50 = 0.30 from mid value to Z on RHS.

Table of Area Under Standard Normal Curve gives Z = 0.84

So,

Z = 0.84 = (X - 78.4)/5.7

So,

X = 78.4 + (0.84 X 5.7)

= 83.188

So,

Answer is:

83.188

(B)

n = 36

SE = /

= 5.7/

0.95

To find P(>80):
Z = (80 - 78.4)/0.95

= 0.6316

Table of Area Under Standard Normal Curve gives area = 0.2357

So,

P(>80) = 0.5 - 0.2357

= 0.2643

So,

Answer is:

0.2643