Question

Assume that family incomes are normally distributed with mean P30,000 and standard deviation of P10,000. If the poverty level is P10,000, find the percentage of the population that lies in poverty.

Answer 1

Let $X=$ family income: $X\sim N(\mu, \sigma^2).$

Given $\mu=30000, \sigma=10000.$

$$P(X<10000)=P(Z<\dfrac{10000-30000}{10000})$$

$$=P(Z<-2)\approx0.02275$$

$2.275\%$ of the population lies in poverty.