Question

iq scores are normally distributed with a mean of 100 and a standard deviation of 15. a person who's score is higher than 84% has iq of?

Answer 1

Given that,

mean = =100

standard deviation = =15

Using standard normal table,

P(Z > z) = 84%

= 1 - P(Z < z) = 0.84

= P(Z < z ) = 1 - 0.84

= P(Z < z ) = 0.16

= P(Z < -0.99 ) = 0.16

z = -0.99 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= -0.99 *15+100

x= 85.15

x=85