Question

Suppose family incomes in a town are normally distributed with a mean of P25000 and a standard deviation of P6000 per month. What is the probability that a family has an income between P14,000 and P32,500?

Answer 1

The formula for the Z-score is,

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

Given:

`\begin{gathered} x_1=P14,000,x_2=P32,500 \\ \mu=P25,000,\sigma=P6000 \end{gathered}`

Therefore,

`\begin{gathered} Z_1=(14000-25000)/(6000)=-1.83333 \\ Z_2=(32500-25000)/(6000)=1.25 \end{gathered}`

Hence, the probability will be

`P(Z_1<strong>Therefore, the answer is</strong>[tex]\begin{equation*} 0.86097 \end{equation*}`