Question 1 of 8: Normal Distribution of Bag Weight

Question 1: Normal Distribution of Bag Weight (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. W (weight of one bag) is Normal with mean μ = 39.80 kg and standard deviation σ = 0.30 kg.

Find. (a) P(W>39.50); the pdf and its shaded area. (b) the 25th percentile and 90th percentile of W. (c) the sampling distribution of M = mean of n=9 bags, and P(|M−39.80|<0.05). (d) the mean, variance of T = total of 100 bags, and P(T>4000).

Approach. Standardize W (and its derived variables M, T) with the Central Limit Theorem/normal-sum properties, then read tail and quantile probabilities from the standard-normal table.

1. (a) Probability weight exceeds 39.50 kg. The pdf is $$f(w)=\dfrac{1}{0.30\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{w-39.80}{0.30}\right)^{2}}$$ Standardizing, $z=\dfrac{39.50-39.80}{0.30}=-1.00$. From the normal table, $P(Z\gt -1.00)=0.5+0.3413=0.8413$. So $$\boxed{P(W\gt 39.50)=0.8413}$$ The figure below plots f(w) with the required area (w ≥ 39.50) shaded.
Fig. 1 — pdf of W ~ N(39.80, 0.30²); shaded area = P(W>39.50) = 0.8413.
2. (b) Lower quartile and upper decile. The lower quartile has cumulative probability 0.25, so $z_{0.25}=-0.6745$ (upper tail area 0.25 below/above the median at the corresponding $z=0.6745$): $$w_{0.25}=39.80+(-0.6745)(0.30)=39.598\text{ kg}$$ The upper decile has cumulative probability 0.90, so $z_{0.90}=1.2816$: $$\boxed{w_{0.25}=39.598\text{ kg}, \qquad w_{0.90}=40.184\text{ kg}}$$ Meaning: 25% of bags weigh less than 39.598 kg (and 75% weigh more); 90% of bags weigh less than 40.184 kg (only the heaviest 10% of bags exceed this).
3. (c)(i)–(ii) Distribution of the sample mean M. By the sampling-distribution result for a normal population, $M=\bar{W}$ over $n=9$ bags is itself Normal with $$\mu_M=39.80\text{ kg}, \qquad \sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.30}{\sqrt{9}}=0.10\text{ kg}$$ so the pdf of M is $$f(m)=\dfrac{1}{0.10\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{m-39.80}{0.10}\right)^{2}}$$ $$\boxed{\mu_M=39.80\text{ kg}, \ \sigma_M=0.10\text{ kg}}$$
Fig. 2 — pdf of a single bag W (σ=0.30, dashed) overlaid with the pdf of the 9-bag sample mean M (σ=0.10, solid) — same mean, M is much more concentrated.
4. (c)(iv) Probability M is within 0.05 kg of the mean. $z=\dfrac{0.05}{0.10}=0.50$. $$P(|M-39.80|\lt 0.05)=P(-0.50\lt Z\lt 0.50)=2(0.1915)=0.3829$$ $$\boxed{P(|M-39.80|\lt 0.05)=0.3829}$$
5. (d) Total weight of a lot of 100 bags. $T=\sum_{i=1}^{100}W_i$ is Normal (sum of independent normals) with $$\mu_T=100(39.80)=3980\text{ kg}, \qquad \sigma_T^2=100(0.30)^2=9.0\text{ kg}^2 \ (\sigma_T=3.0\text{ kg})$$ Standardizing, $z=\dfrac{4000-3980}{3.0}=6.667$. This is far beyond the tabulated range (table stops at $z=3.0$, where the tail is already only 0.0013); the exact tail probability is $$\boxed{\mu_T=3980\text{ kg}, \ \sigma_T^2=9.0\text{ kg}^2, \ P(T\gt 4000)\approx1.3\times10^{-11}\ (\text{effectively zero})}$$