Answer to Question #227081 in Statistics and Probability for Mike

Question

Suppose a census was conducted to determine the impact of COVID-19 on the 

operations of local vendors and it found that the mean and variance of the daily 

income of all local vendors is N$ 600 and 156250 (N$ squared), respectively. Assume 

a sample of 100 local vendors is selected, what is the probability that their sample 

mean daily income is within N$ 100 of the population mean daily income.

Answer 1

Let $X=$ sample mean daily income: $X\sim N(\mu, \sigma^2/n).$  

Given $\mu=600, \sigma^2=156250, n=100$

$P(\mu-100<X<\mu+100)$

$=P(X<\mu+100)-P(X\leq \mu-100)$

$=P(Z<\dfrac{100}{\sqrt{\dfrac{156250}{100}}})-P(Z\leq\dfrac{-100}{\sqrt{\dfrac{156250}{100}}})$

$\approx P(Z<2.5298)-P(Z\leq -2.5298)$

$\approx0.99429-0.00571\approx0.9886$