Question

Assuming IQ scores are normally distributed in the population with a mean of 100 and a standard deviation of 16, what percentages of IQ scores are less than 120, between 110 and 130, and what IQ will place you in the top 8% of the population?

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = = 16

P(x < 120) = P((x - ) / < (120 - 100) / 16)

= P(z < 1.25)

= 0.8944 Using standard normal table,

Probability = 0.8944

P(110 < x < 130) = P((110 - 100)/ 16) < (x - ) /  < (130 - 100) / 16) )

= P(0.63 < z < 1.88)

= P(z < 1.88) - P(z < 0.63)

= 0.9699 - 0.7357

Probability = 0.2342

P( Z > z ) = 8 %

1 - P( Z < z ) = 0.08

P( Z <  ) = 1 - 0.08

P( Z < z ) = 0.92

P( Z < 1.41 ) = 0.92

z = 1.41

Using z - score formula,

X = z * +

= 1.41 * 16 + 100  

= 122.56

The 8% value is 122.56