Question

A distribution of values is normal with a mean of 90 and a standard deviation of 8.

Find the interval containing the middle-most 32% of scores:

Answer 1

Solution:-

Given that,

mean = = 90

standard deviation = = 8

Using standard normal table,

P( -z < Z < z) = 32%

= P(Z < z) - P(Z <-z ) = 0.32

= 2P(Z < z) - 1 = 0.32

= 2P(Z < z) = 1 + 0.32

= P(Z < z) = 1.32 / 2

= P(Z < z) = 0.66

= P(Z < 0.4125) = 0.66

= z  ± 0.4125

Using z-score formula,

x = z * +

x = - 0.4125 * 8 + 90

x = 86.7

Using z-score formula,

x = z * +

x = 0.4125 * 8 + 90

x = 93.3

The middle 32 % are from 86.7 to 93.3