Normal Distribution — Parcel Weight: 04-BS-2, May 2013, Q1

Question 1: Normal Distribution — Parcel Weight (22 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. W ∼ N(μ = 6,000.0 g, σ = 500.0 g).

Find. (a) P(W > 6,600); (b) P(|W−6,000| < 300); (c) the sampling distribution of the sample mean M for n = 25 and P(M > 5,900); (d) E(T), Var(T) and P(T > 98,000) for the sum T of n = 16 parcels.

Approach. Standardize each Normal random variable with Z = (X−μ)/σ, using the sampling-distribution results μ_M=μ, σ_M=σ/√n for the sample mean and E(T)=nμ, Var(T)=nσ² for the sample sum (Central Limit Theorem, exact here since W is already Normal).

(a) Right-tail probability. The pdf is $$f(w)=\dfrac{1}{500\sqrt{2\pi}}\exp\!\left[-\dfrac{(w-6000)^2}{2(500)^2}\right],\quad -\infty<w<\infty.$$ Standardizing, $Z=\dfrac{6600-6000}{500}=1.20$, so $$P(W>6600)=P(Z>1.20)=1-\Phi(1.20)=1-0.8849=\boxed{0.1151}.$$ The shaded region below is the right tail of f(W) beyond w = 6,600.
Fig. 1(a) — f(W), shaded area = P(W > 6,600) = 0.1151.
(b) Two-sided probability within 300 g of the mean. $Z=\dfrac{300}{500}=0.60$, so $$P(|W-6000|<300)=P(-0.60<Z<0.60)=2\Phi(0.60)-1=2(0.7257)-1=\boxed{0.4515}.$$
Fig. 1(b) — f(W), shaded area = P(5,700 < W < 6,300) = 0.4515.
(c) Sampling distribution of the mean, n = 25. By the sampling-distribution theorem for a Normal population, $$\mu_M=\mu=6000.0\text{ g},\qquad \sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{500}{\sqrt{25}}=100.0\text{ g}.$$ So $M\sim N(6000,\,100^2)$ with pdf $$f(m)=\dfrac{1}{100\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-6000)^2}{2(100)^2}\right].$$ Both curves are shown below — M's density is narrower (smaller spread) because averaging 25 parcels cancels individual variation.
Fig. 1(c) — f(W) (solid, σ=500) versus f(M) (dashed, σ_M=100), both centred at 6,000 g.
Standardizing for (iv): $Z=\dfrac{5900-6000}{100}=-1.00$, so $$P(M>5900)=P(Z>-1.00)=\Phi(1.00)=\boxed{0.8413}.$$
(d) Sampling distribution of the sum, n = 16. $$E(T)=n\mu=16(6000)=\boxed{96{,}000\text{ g}},\qquad \mathrm{Var}(T)=n\sigma^2=16(500)^2=\boxed{4{,}000{,}000\text{ g}^2}\;(\sigma_T=2000\text{ g}).$$ $T\sim N(96000,\,2000^2)$. Standardizing, $Z=\dfrac{98000-96000}{2000}=1.00$, so $$P(T>98000)=1-\Phi(1.00)=1-0.8413=\boxed{0.1587}.$$
Fig. 1(d) — f(T), shaded area = P(T > 98,000) = 0.1587.

Question 1 — final results
Part	Quantity	Value
(a)	P(W > 6,600)	0.1151
(b)	P(\|W−6,000\| < 300)	0.4515
(c)(i)	μ_M, σ_M	6,000.0 g, 100.0 g
(c)(iv)	P(M > 5,900)	0.8413
(d)	E(T), Var(T)	96,000 g, 4,000,000 g²
(d)	P(T > 98,000)	0.1587

Topic: Normal distribution and sampling distributions of the mean and sum.

Key relations: $Z=(X-\mu)/\sigma$; for a Normal population, $\bar X\sim N(\mu,\sigma^2/n)$ exactly (any n), and $T=\sum X_i\sim N(n\mu,n\sigma^2)$.

Why this works: a linear combination of independent Normal random variables is itself Normal, so both the sample mean and the sample sum inherit Normality from the parent population without needing the Central Limit Theorem's large-n approximation — the results in (c) and (d) are exact for any sample size.

Common pitfall: using σ instead of σ/√n when standardizing the sample mean, or forgetting that Var(T) scales with n (not n²) so σ_T = σ√n.

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

Question 1 of 8: Normal Distribution — Parcel Weight

Question 1: Normal Distribution — Parcel Weight (22 marks)