Question

A normally distributed set of scores has mean 120 and standard deviation 10.

What score cuts off the top 15%?

What is the probability of a randomly chosen score being between 105 and 125?

What percentage of scores are less than 100?

Answer 1

Given,

= 120 , = 10

We convert this to standard normal as

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

a)

We have to calculate x such that

P (X > x) = 0.15

That is

P (X < x) = 0.85

P( Z < x - / ) = 0.85

From Z table, z-score for the probability of 0.85 is 1.0364

x - / = 1.0364

x - 120 / 10 = 1.0364

x = 130.364

b)

P (105 < X < 125) = P( X < 125) - P (X < 105)

= P (Z < 125 - 120 / 10) = P( Z < 0.5) - P( Z < -1.5)

= 0.6915 - ( 1 - 0.9332)

= 0.6247

c)

P( X < 100) = P( Z < 100 - 120 / 10)

= P( Z < -2)

= 1 - P( Z < 2)

= 1 - 0.9772

= 0.0228

= 2.28%