Question 1 of 8

Question 1 (20 marks: (a) 5, (b) 5, (c) 5, (d) 5)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. X, the travel time in minutes, is normally distributed with μ = 52 and σ = 4.

Find. Four tail/interval probabilities on X, on the sample mean M of n = 4, and on the sample sum T of n = 144, plus the pdfs of X, M, and the graphs showing the requested areas.

Approach. Standardize each event with the Normal distribution, using the Central Limit Theorem exactly (X is itself Normal, so M and T are Normal for any n).

1. (a) Write f(x) and find P(X > 57). The pdf is $$f(x) = \frac{1}{4\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{x-52}{4}\right)^2\right], \quad -\infty < x < \infty.$$ Standardizing, $Z = (57-52)/4 = 1.25$, so $$P(X>57) = P(Z>1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = \boxed{0.1056}.$$
2. (b) Find P(|X−52| < 3). The event is $49 < X < 55$, giving $Z$-bounds $\pm 0.75$: $$P(|X-52|<3) = \Phi(0.75) - \Phi(-0.75) = 0.7734 - 0.2266 = \boxed{0.5467}.$$
3. (c)(i)–(ii) Distribution of M, the mean of n = 4. Since X is Normal, $M = \bar{X}$ is exactly Normal with $$\mu_M = \mu = 52, \qquad \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{4}} = \boxed{2 \text{ min}}.$$ So $M \sim N(52, 2^2)$, with pdf $g(m) = \dfrac{1}{2\sqrt{2\pi}}\exp\left[-\tfrac12\left(\tfrac{m-52}{2}\right)^2\right]$.
4. (c)(iii) Overlay the two curves. M's curve is the same bell shape as X's but narrower (smaller σ), centred at the same mean — see the figure below.
5. (c)(iv) Compute P(M > 50). $$Z = \frac{50-52}{2} = -1.00, \qquad P(M>50) = 1-\Phi(-1.00) = \Phi(1.00) = \boxed{0.8413}.$$
6. (d)(i) Distribution of T, the sum of n = 144. $$E(T) = n\mu = 144(52) = \boxed{7{,}488 \text{ min}}, \qquad \text{Var}(T) = n\sigma^2 = 144(16) = 2{,}304, \quad \sigma_T = \sqrt{2304} = \boxed{48 \text{ min}}.$$
7. (d)(ii) Compute P(T > 7,500). $$Z = \frac{7500-7488}{48} = 0.25, \qquad P(T>7500) = 1-\Phi(0.25) = 1-0.5987 = \boxed{0.4013}.$$