Question

suppose that during any hour in a large department store, the average number of shoppers is 448, with a standard deviation of 21 shoppers. what is the probability that a random sample of 49 different shopping hours will yield a sample mean between 441 and 446 shoppers?

Answer 1

$$P(441<X<446)=P(Z<\dfrac{446-448}{21/\sqrt{49}})$$

$$-P(Z\le\dfrac{441-448}{21/\sqrt{49}})$$

$$=P(Z<-0.6667)-P(Z\le -2.3333)$$

$$\approx0.25249-0.00982\approx0.2427$$