The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean $\mu =8.6$ minutes and a standard deviation $\sigma =3.5$ minutes. If a random sample of 49 customers is observed, find the probability that their mean time at the teller’s window is:

(a) at most 7.6 minutes;
(b) more than 9.1 minutes;
(c) at least 8.6 minutes but less than 9.4 minutes.

Answer :

Given :

The population mean $(\mu )=8.6$

The population standard deviation $(\sigma )=3.5$

The sample size $(n)=49$

Solution :

(A) The probability that their mean time at the teller’s window is at most 7.6 minutes:

$$\therefore \text{P}(\overline{x}<7.6)=\text{P}(\frac{\overline{x}–\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{7.6–8.6}{\frac{3.5}{\sqrt{49}}})$$ $$=\text{P}(z<-2)$$ $$=0.0228$$

(B) The probability that their mean time at the teller’s window is more than 9.1 minutes:

$$\therefore \text{P}(\overline{x}>9.1)=\text{P}(\frac{\overline{x}–\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{9.1–8.6}{\frac{3.5}{\sqrt{49}}})$$ $$=\text{P}(z>1)$$ $$=0.1587$$

(C) The probability that their mean time at the teller’s window is at least 8.6 minutes but less than 9.4 minutes:

$$\therefore \text{P}(8.6<\overline{x}<9.4)=\text{P}(\frac{8.6–8.6}{\frac{3.5}{\sqrt{49}}}<\frac{\overline{x}–\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{9.4–8.6}{\frac{3.5}{\sqrt{49}}})$$ $$=\text{P}(0<z<1.6)$$ $$=\text{P}(z<1.6)–\text{P}(z<0)$$ $$=0.9452–0.5$$ $$=0.4452$$