Question

Imagine that self report measure of creativity is normally distributed with a mean of μ = 40 and a standard deviation of σ = 5

1.What is the score that cuts off the highest 10% of creative people if our sample consists of n = 100 individuals?

2.What is the score that cuts off the lowest 2% of creative people if our sample consists of 36 individuals?

Answer 1

Solution:-

Given that,

mean = = 40

standard deviation = = 5

1)n = 100

= = 40

= / n = 5/ 100 = 0.5

P(Z > z ) = 0.10

1- P(z < z) =0.10

P(z < z) = 1-0.10 = 0.90

z = 1.28

Using z-score formula,

= z * +   = 1.28 *0.5+40 = 40.64

2)

n = 36

= = 40

= / n = 5/ 36 = 0.8333

P(Z < z) = 0.02

z = -2.054

Using z-score formula,

= z * +   = -2.054 *0.8333+40 = 38.29