Question

Use the Normal model N(95,14​) for the IQs of sample participants.

​a) What IQ represents the 22nd percentile?

​b) What IQ represents the 99th percentile?

​c) What's the IQR of the​ IQs?

Answer 1

Solution:-

Given that,

mean = $\mu$ = 95

standard deviation = $\sigma$ = 14

a) Using standard normal table,

P(Z < z) = 22%

= P(Z < z) = 0.22  

= P(Z < -0.77) = 0.22

z = -0.77

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.77 * 14 + 95

x = 84.22

IQ = 84

b) Using standard normal table,

P(Z < z) = 99%

= P(Z < z) = 0.99

= P(Z < 2.33) = 0.99  

z = 2.33

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 2.33 * 14 + 95

x = 127.62

IQ = 128

c) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.6745 * 14 +95

x = 85.56

First quartile =Q1 = 86

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 0.6745 * 14 + 95

x = 104.44

Third quartile =Q3 = 104

IQR = Q3 - Q1

IQR = 104 - 86 = 18