Question

In the first quarter of 2017, the average mortgage for first time buyers was R910,000. Assuming a normal distribution and a standard deviation of R50,000. What proportion of mortgages were between R890,000 and R990,000?

Answer 1

Given:

`\begin{gathered} \mu=R910,000 \\ \sigma=R50,000 \\ X_1=R890,000 \\ X_2=R990,000 \end{gathered}`

The formula for Z-score is,

`Z=(X-\mu)/(\sigma)`

Therefore,

`\begin{gathered} Z_1=(890000-910000)/(50000)=(-20000)/(50000)=-0.4 \\ \therefore Z_1=-0.4 \end{gathered}`
`\begin{gathered} Z_2=(990000-910000)/(50000)=(80000)/(50000)=\:1.6 \\ \therefore Z_2=1.6 \end{gathered}`

Hence, the proportion of the mortgages will be

`P(Z_1<strong>Therefore, the answer is</strong>[tex]0.60062\text{ or 60.062}\%`