(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 $(μ) = 8.6$
The population standard deviation $(σ) = 3.5$
The sample size $(n) = 49$
Solution :
(A) The probability that their mean time at the teller’s window is at most 7.6 minutes:
$$\therefore \text{P}\left(\bar{x} < 7.6\right) = \text{P}\left(\frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{7.6 – 8.6}{\frac{3.5}{\sqrt{49}}}\right)$$ $$= \text{P}(z < -2)$$ $$= 0.0228$$
(B) The probability that their mean time at the teller’s window is more than 9.1 minutes:
$$\therefore \text{P}\left(\bar{x} > 9.1\right) = \text{P}\left(\frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} > \frac{9.1 – 8.6}{\frac{3.5}{\sqrt{49}}}\right)$$ $$= \text{P}(z > 1)$$ $$= 0.1587$$
(C) The probability that their mean time at the teller’s window is at least 8.6 minutes but less than 9.4 minutes:
$$\therefore \text{P}(8.6 < \bar{x} < 9.4) = \text{P}\left(\frac{8.6 – 8.6}{\frac{3.5}{\sqrt{49}}} < \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{9.4 – 8.6}{\frac{3.5}{\sqrt{49}}}\right)$$ $$= \text{P}(0 < z < 1.6)$$ $$= \text{P}(z < 1.6) – \text{P}(z < 0)$$ $$= 0.9452 – 0.5$$ $$= 0.4452$$
