Question

Assume adult IQ scores are normally distributed with a mean of 100 and a standard deviation of 15

a) What is the probability that a randomly selected adult has an IQ that is less than 115

b) Find the probability that an adult has an IQ greater than 131.5 (requirement to join MENSA)

c) Find the probability that a randomly selected adult has an IQ between 110 and 120

d) Find the IQ separating the top 15% from the others e) Find the IQ score separating the bottom 10% from the others

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = = 15

(a)

P(x < 115) = P((x - ) / < (115 - 100) / 15)

= P(z < 1)

= 0.8413

Probability = 0.8413

(b)

P(x > 131.5) = 1 - P(x < 131.5)

= 1 - P((x - ) / < (131.5 - 100) / 15)

= 1 - P(z < 2.1)

= 1 - 0.9821

= 0.0179

Probability = 0.0179

(c)

P(110 < x < 120) = P((110 - 100)/ 15) < (x - ) /  < (120 - 100) / 15) )

= P(0.67 < z < 1.33)

= P(z < 1.33) - P(z < 0.67)

= 0.9082 - 0.7486

= 0.1596

Probability = 0.1596

(d)

P(Z > z) = 15%

1 - P(Z < z) = 0.15

P(Z < z) = 1 - 0.15 = 0.85

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 15 +100 = 115.6

Score = 115.6

(e)

P(Z < z) = 10%

P(Z < 1.28) = 0.10

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 15 +100 = 119.2

Score = 119.2