Question

1. What is the first quartile of a normal distribution with mean 40, and standard deviation 7? Round to three decimal places.

2. What is the third quartile of a normal distribution with mean 40, and standard deviation 7? Round to three decimal places.

3. What is the IQR of a normal distribution with mean 40, and standard deviation 7? Round to three decimal places.

Answer 1

Solution:-

Given that,

mean = = 40

standard deviation = = 7

Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.67 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 7+40

x = 35.310

First quartile =Q1 = 35.310

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.67 ) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 7+40

x = 44.690

Third quartile =Q3 =44.690

IQR = Q3 - Q1=44.690-35.310=9.380