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,

$Z=\frac{X-\mu}{\sigma}$

Therefore,

$\begin{gathered} Z_1=\frac{890000-910000}{50000}=\frac{-20000}{50000}=-0.4 \\ \therefore Z_1=-0.4 \end{gathered}$$\begin{gathered} Z_2=\frac{990000-910000}{50000}=\frac{80000}{50000}=\:1.6 \\ \therefore Z_2=1.6 \end{gathered}$

Hence, the proportion of the mortgages will be

$P(Z_1Therefore, the answer is[tex]0.60062\text{ or 60.062}\%$