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.

a. X ~ _____(_____,_____)

b. Find the probability that the person has an IQ greater than 120.

A. 0.05            B. 0.08           C. 0.09           D. 0.06           E. 0.07

c. Find the probability that the person has an IQ between 90 and 115.

A. 0.589            B. 0.664           C. 0.732           D. 0.531           E. 0.469

d. Find the 60^th percentile

A. 98.6              B. 100.0             C. 103.8            D. 105.4           E. 108.5

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = = 15

a.

X N (100 , 15)

b.

P(x > 120) = 1 - P(x < 120)

= 1 - P[(x - ) / < (120 - 100) / 15)

= 1 - P(z < 1.33)

= 1 - 0.9082

= 0.09

Probability = 0.09

c.

P( 10< x < 115) = P[(90 - 100)/ 15) < (x - ) /  < (115 - 100) / 15) ]

= P(-0.67 < z < 1)

= P(z < 1) - P(z < -0.67)

= 0.8413 - 0.2514

= 0.589

Probability = 0.589

d.

Using standard normal table ,

P(Z < z) = 60%

P(Z < 0.25) = 0.6

z = 0.25

Using z-score formula,

x = z * +

x = 0.25 * 15 + 100 = 103.8

the 60th percentile = 103.8