Question 1 of 8: Normal Distribution — Fabric Softener Volume

Question 1: Normal Distribution — Fabric Softener Volume (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. X ∼ N(μ = 4,850 mL, σ = 40 mL); samples of n = 4 (mean M) and n = 9 (sum T) bottles are drawn.

Find. (a) P(X < 4,900); (b) P(|X−4,850| > 60); (c) the sampling distribution of M and P(M > 4,840); (d) E(T), Var(T) and P(T > 43,680).

Approach. Standardize each Normal variable with Z=(X−μ)/σ, using μ_M=μ, σ_M=σ/√n for the sample mean and E(T)=nμ, Var(T)=nσ² for the sample sum (exact, since X is already Normal).

1. (a) Left-tail probability for a single bottle. The pdf is $$f(x)=\dfrac{1}{40\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-4850)^2}{2(40)^2}\right],\quad -\infty<x<\infty.$$ Standardizing, $Z=\dfrac{4900-4850}{40}=1.25$, so $$P(X<4900)=P(Z<1.25)=\Phi(1.25)=\boxed{0.8944}.$$
   pdf of X ~ N(4,850, 40&sup2;) mL; the shaded region (left of 4,900 mL) is P(X &lt; 4,900).
2. (b) Two-tailed deviation from the mean. "Differs from the mean by more than 60 mL" means $|X-4850|>60$, i.e. $X<4790$ or $X>4910$. With $Z=60/40=1.50$, $$P(|X-4850|>60)=2\big[1-\Phi(1.50)\big]=2(1-0.9332)=\boxed{0.1336}.$$
   pdf of X; the two shaded tails (X &lt; 4,790 or X &gt; 4,910) are P(|X&minus;4,850| &gt; 60).
3. (c) Sampling distribution of the mean of n = 4 bottles. By the sampling-distribution theorem for a Normal population, $M=\bar X$ is itself Normal with $$\mu_M=\mu=4850\ \text{mL},\qquad \sigma_M^2=\dfrac{\sigma^2}{n}=\dfrac{40^2}{4}=400\ \Rightarrow\ \sigma_M=\boxed{20\ \text{mL}}.$$ So $M\sim N(4850,\,20^2)$ with pdf $f(m)=\dfrac{1}{20\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-4850)^2}{2(20)^2}\right]$ — the same bell shape as X's pdf but five times narrower (σ_M=σ/2 here, since n=4). For part (iv), $Z=\dfrac{4840-4850}{20}=-0.50$, so $$P(M>4840)=P(Z>-0.50)=\Phi(0.50)=\boxed{0.6915}.$$
   pdf of M (sample mean, n=4) &mdash; narrower than the pdf of a single bottle X since &sigma;_M=&sigma;/&radic;4=20; shaded region is P(M &gt; 4,840).
4. (d) Sum of n = 9 bottles. For the sum of n i.i.d. Normal draws, $E(T)=n\mu=9(4850)=\boxed{43{,}650\ \text{mL}}$ and $\mathrm{Var}(T)=n\sigma^2=9(40)^2=\boxed{14{,}400\ \text{mL}^2}$, so $\sigma_T=\sqrt{14400}=120\ \text{mL}$ and $T\sim N(43650,\,120^2)$. With $Z=\dfrac{43680-43650}{120}=0.25$, $$P(T>43680)=1-\Phi(0.25)=1-0.5987=\boxed{0.4013}.$$
   pdf of T (sum of n=9 bottles), T ~ N(43,650, 120&sup2;); shaded region is P(T &gt; 43,680).