Question

21- Scores on an examination are assumed to have a normal distribution with mean 78 and standard deviation 6.

a) What is the proportion of scores that are higher than 75?

b) What is the proportion of scores that fall between 75 and 87?

c) Suppose that students scoring in the top 10% of this distribution are to receive an A grade. What is the minimum score a student must achieve to earn an A grade?

PLEASE EXPLAIN YOUR ANSWER

Answer 1

Given,

= 78 , = 6

We convert this to standard normal as

P(X < x) = P(Z < (x - ) / )

a)

P(X > 75) = P(Z > ( 75 - 78) / 6)

= P(Z > -0.5)

= P(Z < 0.5)

= 0.6915 (From Z table)

b)

P(75 < X < 87) = P(X < 87) - P(X < 75)

= P(X < 87) - P(X < 75)

= P(Z < ( 87 - 78) / 6) - P(Z < ( 75 - 78) / 6)

= P(Z < 1.5) - P(Z < -0.5)

= 0.9332 - 0.3085

= 0.6247

c)

We have to calculate X such that P(X > x) = 0.10

That is

P(X < x) = 0.90

P(Z < ( x - ) / ) ) = 0.90

From Z table, z-score for the probability of 0.90 is 1.2816

( x - ) / ) = 1.2816

Solve for x

( x - 78) / 6 = 1.2816

X = 85.69