Answer:
Given:
The population mean $(μ) = 83$
The population standard deviation $(σ) = 13$
Solution:
(a) The probability of completing the exam in one hour or less (i.e. 60 or less) :
$$\therefore \text{P}(x \leq 60) = \text{P}\left( \frac{x – \mu}{\sigma} \leq \frac{60 – 83}{13} \right)$$ $$= \text{P}(z \leq -1.769)$$ $$= 0.0384$$
(b) The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes :
$$\text{P}(60 < x < 75) = \text{P}\left( \frac{60 – 83}{13} < \frac{x – \mu}{\sigma} < \frac{75 – 83}{13} \right)$$ $$= \text{P}(-1.769 < z < -0.615)$$ $$= \text{P}(z < -0.615) – \text{P}(z < -1.769)$$ $$= 0.2693 – 0.0384$$ $$= 0.2309$$
(c)
The total number of students in class $(N) = 60$
The duration of examination period is 90 minutes.
→ The probability that a randomly selected student will be unable to complete the exam in the allotted time (i.e. more than 90 minutes) :
$$\text{P}(x > 90) = \text{P}\left( \frac{x – \mu}{\sigma} > \frac{90 – 83}{13} \right)$$ $$= \text{P}(z > 0.538)$$ $$= 1 – \text{P}(z < 0.538)$$ $$= 1 – 0.7047$$ $$= 0.2953$$
