04-BS-2, May 2015, Q1 | Nivaar Exam Prep

Question 1 (20 marks)

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. Bottle volume $V \sim N(\mu, \sigma^2)$ with $\mu = 4{,}430$ mL, $\sigma = 60$ mL.

Find. (a) $P(V>4400)$; (b) 25th and 90th percentiles of $V$; (c) distribution of $M=\bar V$ for $n=4$ and $P(M<4415)$; (d) distribution of $T=\sum_{i=1}^{16}V_i$ and $P(T>71000)$.

Approach. $V$ is already Normal, so every linear function of it (a sample mean $M$, a sample sum $T$) is Normal too — only the mean and standard deviation change, via $\sigma_M=\sigma/\sqrt{n}$ and $\sigma_T=\sigma\sqrt{n}$; every probability is then read off the standard Normal table after standardizing.

(a) Standardize and read the upper-tail area. $$Z=\frac{V-\mu}{\sigma}=\frac{4400-4430}{60}=\frac{-30}{60}=-0.50$$ $$P(V>4400)=P(Z>-0.50)=0.5+\Phi(0.50)=0.5+0.3413=\boxed{0.6915}$$ The pdf of $V$ is $f(v)=\dfrac{1}{60\sqrt{2\pi}}\exp\!\left[-\dfrac{(v-4430)^2}{2(60)^2}\right]$, $-\infty<v<\infty$.

pdf of V ~ N(4430, 60²); shaded area = P(V > 4400) = 0.6915

(b) Invert the Normal cdf at the required tail areas. The lower quartile satisfies $P(V<Q_1)=0.25$, i.e. $z_{0.25}=-0.6745$; the 90th percentile satisfies $P(V<P_{90})=0.90$, i.e. $z_{0.90}=1.2816$. $$Q_1 = 4430 + (-0.6745)(60) = \boxed{4389.53\text{ mL}}$$ $$P_{90} = 4430 + (1.2816)(60) = \boxed{4506.89\text{ mL}}$$ Meaning: 25% of bottles hold less than 4,389.53 mL of oil, and 90% of bottles hold less than 4,506.89 mL — only the top 10% of bottles exceed 4,506.89 mL.
(c)(i)–(ii) Distribution of the sample mean. For a sample of $n=4$ bottles drawn from a Normal population, $M=\bar V$ is exactly Normal with $$\mu_M=\mu=4430\text{ mL}, \qquad \sigma_M=\frac{\sigma}{\sqrt{n}}=\frac{60}{\sqrt{4}}=\boxed{30\text{ mL}}$$ so $M\sim N(4430,\,30^2)$ with pdf $f(m)=\dfrac{1}{30\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-4430)^2}{2(30)^2}\right]$.

pdf of V (solid, σ=60) and M (dashed, σ=30) — M is more concentrated about the shared mean 4,430 mL

(c)(iv) Probability for the sample mean. $$Z=\frac{4415-4430}{30}=\frac{-15}{30}=-0.50 \qquad P(M<4415)=P(Z<-0.50)=1-0.6915=\boxed{0.3085}$$
(d) Distribution of the sum of 16 bottles. For an i.i.d. Normal sample, $T=\sum_{i=1}^{16}V_i$ is Normal with $$\mu_T=n\mu=16(4430)=\boxed{70{,}880\text{ mL}}, \qquad \sigma_T^2=n\sigma^2=16(60)^2=\boxed{57{,}600\text{ mL}^2}\ \ (\sigma_T=240\text{ mL})$$ $$Z=\frac{71000-70880}{240}=\frac{120}{240}=0.50 \qquad P(T>71000)=P(Z>0.50)=\boxed{0.3085}$$

pdf of T ~ N(70,880, 240²); shaded area = P(T > 71,000) = 0.3085

Question 1 — final results
Quantity	Value
$P(V>4400)$	0.6915
Lower quartile of $V$	4,389.53 mL
90th percentile of $V$	4,506.89 mL
$M\sim N(\mu_M,\sigma_M^2)$	$\mu_M=4430$ mL, $\sigma_M=30$ mL
$P(M<4415)$	0.3085
$T\sim N(\mu_T,\sigma_T^2)$	$\mu_T=70{,}880$ mL, $\sigma_T^2=57{,}600$ mL$^2$
$P(T>71000)$	0.3085

Topic: Sampling distributions derived from a Normal population.

Key relations: If $X\sim N(\mu,\sigma^2)$ then for a random sample of size $n$: $\bar X\sim N\!\left(\mu,\ \sigma^2/n\right)$ and $\sum X_i\sim N(n\mu,\ n\sigma^2)$; any Normal variable is standardized via $Z=(X-\mu)/\sigma$, and a percentile is recovered via $x_p=\mu+z_p\sigma$.

Why this works: A linear combination of independent Normal variables is itself Normal (the Normal family is closed under linear combinations), so only the first two moments need to be recomputed — the shape never changes. This is what lets the same standard Normal table answer probability, percentile and sampling-distribution questions alike.

Common pitfall: Using $\sigma$ instead of $\sigma/\sqrt{n}$ when standardizing a sample mean (or $\sigma\sqrt n$ for a sum), and reading a lower-tail z-value as positive instead of negative.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 6–8 (Sampling Distributions).

Question 1 of 8

Question 1 (20 marks)