Question 1 of 8: Normal Distribution — Cement Block Weight

Question 1: Normal Distribution — Cement Block 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. X ∼ N(μ = 45.2 kg, σ = 0.4 kg); a sample mean M of n = 16 blocks, and a sample sum T of n = 9 blocks.

Find. (a) P(X > 45.1); (b) P(|X−45.2| < 0.1); (c) the sampling distribution of M and P(M < 45.0); (d) E(T), Var(T) and P(T > 408.0).

Approach. Standardize each Normal variable with Z=(X−μ)/σ, using 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) Right-tail probability for a single block. The pdf is $$f(x)=\dfrac{1}{0.4\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-45.2)^2}{2(0.4)^2}\right],\quad -\infty<x<\infty.$$ Standardizing, $Z=\dfrac{45.1-45.2}{0.4}=-0.25$, so $$P(X>45.1)=P(Z>-0.25)=\Phi(0.25)=\boxed{0.5987}.$$
   pdf of X ∼ N(45.2, 0.4²) kg²; the shaded region (right of 45.1 kg) is P(X > 45.1).
2. (b) Two-tailed deviation from the mean. "Differs from the mean by less than 0.1 kg" means $|X-45.2|<0.1$, i.e. $45.1<X<45.3$. With $Z=0.1/0.4=0.25$, $$P(|X-45.2|<0.1)=2\Phi(0.25)-1=2(0.5987)-1=\boxed{0.1974}.$$
   pdf of X; the shaded central band (45.1 to 45.3 kg) is P(|X−45.2| < 0.1).
3. (c) Sampling distribution of the mean of n = 16 blocks. By the sampling-distribution theorem for a Normal population, $M=\bar X$ is itself Normal with $$\mu_M=\mu=45.2\ \text{kg},\qquad \sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.4}{\sqrt{16}}=\boxed{0.1\ \text{kg}}.$$ (i)/(ii) So $M\sim N(45.2,\,0.1^2)$ with pdf $f(m)=\dfrac{1}{0.1\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-45.2)^2}{2(0.1)^2}\right]$ — the same bell shape as X's pdf but four times narrower (σ_M=σ/4 here, since n=16). (iv) $Z=\dfrac{45.0-45.2}{0.1}=-2.00$, so $$P(M<45.0)=P(Z<-2.00)=1-\Phi(2.00)=1-0.9772=\boxed{0.0228}.$$
   (iii) pdfs of X (dashed, wide) and M (solid, narrow, σ_M=σ/4=0.1) on the same axes, both centred at 45.2 kg.
4. (d) Sum of n = 9 blocks. For the sum of n i.i.d. Normal draws, $E(T)=n\mu=9(45.2)=\boxed{406.8\ \text{kg}}$ and $\mathrm{Var}(T)=n\sigma^2=9(0.4)^2=\boxed{1.44\ \text{kg}^2}$, so $\sigma_T=\sqrt{1.44}=1.2\ \text{kg}$ and $T\sim N(406.8,\,1.2^2)$. With $Z=\dfrac{408.0-406.8}{1.2}=1.00$, $$P(T>408.0)=1-\Phi(1.00)=1-0.8413=\boxed{0.1587}.$$