Question 1 of 8: Normal Distribution of Parcel Weight
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations May 2019 (course code 04-IE-2 / 04-BS-2, shared statistics paper) — Probability and Statistics, 2 hours, closed book except one hand-written information sheet and an approved calculator. Any 5 of 8 questions constitute a complete paper; all 8 are solved below.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists (Pearson) — used throughout for distribution theory, estimation and hypothesis-testing formulas.
Question 1: Normal Distribution of Parcel Weight (20 marks)
Given. W (weight of one parcel) is Normal with mean μ = 5,000.0 gr and standard deviation σ = 400.0 gr.
Find. (a) P(W>5,300); the pdf and its shaded area. (b) P(|W−5,000|<500). (c) the sampling distribution of M = mean of n=16 parcels, and P(M>4,900). (d) the mean, variance of T = sum of 25 parcels, and P(T>126,000).
Approach. Standardize W (and its derived variables M, T) using the Central Limit Theorem / normal-sum properties, then read tail probabilities from the standard-normal table.
(a) Probability a parcel weighs more than 5,300 gr. The pdf is $$f(w)=\dfrac{1}{400\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{w-5000}{400}\right)^{2}}$$ Standardizing, $z=\dfrac{5300-5000}{400}=0.75$. From the normal table, $P(Z\gt 0.75)=0.5-0.2734=0.2266$. So $$\boxed{P(W\gt 5{,}300)=0.2266}$$ The figure below plots f(w) with the required area (w ≥ 5,300) shaded.
Fig. 1a — pdf of W with the shaded tail P(W>5,300).
(b) Probability weight differs from mean by less than 500 gr. This is $P(4{,}500\lt W\lt 5{,}500)$. Standardizing both bounds, $z=\dfrac{500}{400}=1.25$. From the table, $P(-1.25\lt Z\lt 1.25)=2(0.3944)=0.7887$. So $$\boxed{P(|W-5{,}000|\lt 500)=0.7887}$$
Fig. 1b — pdf of W with the shaded central region P(|W−5,000|<500).
(c) Sampling distribution of M, mean of n=16 parcels. (i) Since W is Normal, $M\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)$, so $\mu_M=5{,}000$ gr and $\sigma_M=\dfrac{400}{\sqrt{16}}=100$ gr. (ii) The pdf is $$f(m)=\dfrac{1}{100\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{m-5000}{100}\right)^{2}}$$ (iii) M's pdf (red) is four times taller and narrower than W's pdf (blue) about the shared mean 5,000 — see the figure below. (iv) Standardizing, $z=\dfrac{4{,}900-5{,}000}{100}=-1.00$, so $P(M\gt 4{,}900)=P(Z\gt -1.00)=0.5+0.3413=0.8413$. $$\boxed{P(M\gt 4{,}900)=0.8413}$$
Fig. 1c — pdf of W (σ=400, blue) and M (σ=100, red) on the same axes.
(d) Distribution of T, sum of 25 parcels. For an i.i.d. sum, $\mu_T=n\mu=25(5{,}000)=125{,}000$ gr and $\sigma_T^2=n\sigma^2=25(400)^2=4{,}000{,}000\ \text{gr}^2$ (so $\sigma_T=2{,}000$ gr). Standardizing, $z=\dfrac{126{,}000-125{,}000}{2{,}000}=0.50$, so $P(T\gt 126{,}000)=P(Z\gt 0.50)=0.5-0.1915=0.3085$. $$\boxed{P(T\gt 126{,}000)=0.3085}$$
