Answer to Question #103581 in Statistics and Probability for eric

Question

The income of families in a slum have a normal distribution with a mean of $ 52 per year and a standard deviation of $ 5 . find (10mks) i)Percentage of families with an income of more than $50 per year ii). Percentage of families with an income of between $ 48 and $ 54 per year iii) .In a sample of 500 families in this slum, how many have an income of more than $ 50 per year?

Answer 1

Let X=the income of families in a slum. $X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0, 1)$$

Given that $\mu=52, \sigma=5$

i)

$$P(X>50)=1-P(X\leq50)=$$

$$=1-P(Z\leq{50-52 \over 5})=1-P(Z\leq-0.4)\approx$$

$$\approx1-0.3446\approx0.6554$$

$65.54\%$

ii)

$$P(48<x<54)=P(X<54)-P(X\leq48)=$$

$$=P(Z<{54-52 \over 5})-P(Z\leq{48-52 \over 5})=$$

$$=P(Z<0.4)-P(Z\leq-0.8)\approx0.65542-0.21186\approx$$

$$\approx0.4436$$

$44.36\%$

iii)

Given that $n=500$

$$P(X>50)\approx0.6554$$

$$500\cdot0.6554\approx328$$

Approximately 328 families in this slum have an income of more than $ 50 per year.