Question

Normally distributed scores on Social Psych Final Exam had a mean of 235 and a standard deviation of 52. What is the minimum score needed to be in the highest 10%? Round to the nearest whole number.

Answer 1

Given that,

mean = $\mu$ = 235

standard deviation = $\sigma$ =52

Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10

= P(Z < z ) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z <1.28) = 0.90

z = 1.28 (using standard normal (Z) table )

Using z-score formula  

x = z * $\sigma$ + $\mu$

x= 1.28 *52+235

x= 301.56

x=302