Question

According to psychologists, IQs are normally distributed, with a mean of 100 and a standard deviation of 18.a. What percentage of the population has IQs between 82 and 100?

Answer 1

To solve this question, we will have to find the Z score.

`Z=\frac{x-\bar{x}}{s.d}`

Where x is the value of the IQ

X-bar is the mean.

s.d is the standard deviation

`\begin{gathered} Z_1=(82-100)/(18) \\ =-(18)/(18) \\ =-1 \end{gathered}`
`Z_2=(100-100)/(18)`
`\begin{gathered} Z_2=(0)/(18) \\ =0 \end{gathered}`

The percentage of the population with IQs between 82 and 100 will be calculated thus:

`\begin{gathered} P(-1-1) \\ =0.5-0.1587 \\ =0.34134 \\ \text{The percentage is}\colon \\ =0.34134*100=34.134\text{ \%} \end{gathered}`