Answer to Question #210487 in Statistics and Probability for Vickie

Question

The daily income of a clerk works in a road construction sites is normally distributed with a mean of 2800 per day and a standard deviation of 300.what is the probability that a clerk of works selected at random will earn

Less than 2400 per week

Between 2500 and 3500 per week

If 2000 clerks were employed in a given day ,how many earned less than 2400 that day

Answer 1

Let $X=$the daily income of a clerk: $X\sim N(\mu, \sigma^2)$

Given $\mu=2800, \sigma=300$

Less than 2400 per day

$$P(X<2400)=P(Z<\dfrac{2400-2800}{300})$$

$$\approx P(Z<-1.333333)\approx0.091211$$

Between 2500 and 3500 per week

$$P(2500<X<3500)=P(X<3500)-P(X\leq 2500)$$

$$=P(Z<\dfrac{3500-2800}{300})-P(Z\leq\dfrac{2500-2800}{300})$$

$$\approx P(Z<2.333333)-P(Z\leq-1)$$

$$\approx0.9901847-0.1586553\approx0.831529$$

$$0.091211(2000)=182$$

182 clerks earned less than 2400.