Question

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.

Part B

Find the probability that the person has an IQ greater than 110.

What is the probability? (Round your answer to four decimal places.)

Part C

The middle 60% of IQs fall between what two values?

State the two values. (Round your answers to the nearest whole number.)

What is the probability? (Round your answer to four decimal places.

x1	=
x2	=

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = = 15

B) P(x > 110) = 1 - p( x< 110)

=1- p P[(x - ) / < (110 - 100) / 15]

=1- P(z < 0.67)

Using z table,

= 1 - 0.7486

= 0.2514

C) Using standard normal table,

P( -z < Z < z) = 60%

= P(Z < z) - P(Z <-z ) = 0.60

= 2P(Z < z) - 1 = 0.60

= 2P(Z < z) = 1 + 0.60

= P(Z < z) = 1.60 / 2

= P(Z < z) = 0.80

= P(Z < 0.84) = 0.80

= z  ± 0.84

Using z-score formula,

x = z * +

x = -0.84 * 15 + 100

x = 87.4

x₁ = 87

Using z-score formula,

x = z * +

x = 0.84 * 15 + 100

x = 112.6

x₂ = 113

The middle 60% are from 87 to 113