Question 1 of 8

Question 1 (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. Bottle volume $X\sim N(\mu=10{,}100\text{ mL},\ \sigma=200\text{ mL})$. $M$ is the sample mean of $n=4$ bottles; $T$ is the sum of $n=64$ bottles.

Find. (a) $P(X>10{,}000)$; (b) the lower and upper quartiles of $X$; (c) the mean, sd and pdf of $M$, and $P(M>9{,}900)$; (d) the mean, sd of $T$, and $P(T>645{,}000)$.

Approach. Standardize each variable with its own mean and standard deviation ($\sigma/\sqrt n$ for the mean $M$, $\sqrt n\,\sigma$ for the sum $T$), then read the area from the standard normal table (quartiles use the inverse table, $z=\pm0.6745$).

1. (a) Write the pdf and standardize 10,000. $f(x)=\dfrac{1}{200\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-10100)^2}{2(200)^2}\right]$, $-\infty<x<\infty$. $z=\dfrac{10000-10100}{200}=-0.50$. $P(X>10{,}000)=1-\Phi(-0.50)=\Phi(0.50)=0.6915$, so $\boxed{P(X>10{,}000)=0.6915}$.
2. (b) Quartiles from the inverse normal table. The lower quartile satisfies $P(X<Q_1)=0.25$, i.e. $z_{0.25}=-0.6745$; the upper quartile satisfies $P(X<Q_3)=0.75$, i.e. $z_{0.75}=+0.6745$. $Q_1=10{,}100-0.6745(200)=10{,}100-134.9=\boxed{9{,}965.1\text{ mL}}$; $Q_3=10{,}100+0.6745(200)=10{,}100+134.9=\boxed{10{,}234.9\text{ mL}}$. $Q_1$ is the volume below which 25% of bottles fall (the 25th percentile) and $Q_3$ is the volume below which 75% fall (equivalently, above which only 25% of bottles fall) — together they bracket the middle 50% (the interquartile range) of the bottle-volume distribution.
3. (c) M's distribution and P(M>9,900). By the sampling-distribution-of-the-mean result, $M\sim N\!\left(\mu,\ \sigma^2/n\right)=N\!\left(10{,}100,\ \dfrac{200^2}{4}\right)$, so $\sigma_M=\dfrac{200}{\sqrt4}=100$ mL, giving mean $=10{,}100$ mL and $\boxed{\sigma_M=100\text{ mL}}$. The pdf of $M$ is $f_M(m)=\dfrac{1}{100\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-10100)^2}{2(100)^2}\right]$. Standardizing: $z=\dfrac{9900-10100}{100}=-2.00$, so $P(M>9{,}900)=1-\Phi(-2.00)=\Phi(2.00)=\boxed{0.9772}$. Overlaying $X$ ($\sigma=200$) and $M$ ($\sigma=100$) on one diagram (Fig. 1c) shows $M$'s curve is exactly half as spread, the CLT effect of averaging 4 bottles.
4. (d) T's distribution and P(T>645,000). The sum of $n$ i.i.d. normals is itself normal with mean $n\mu$ and variance $n\sigma^2$: $T\sim N\!\left(64(10{,}100),\ 64(200)^2\right)=N(646{,}400,\ 2{,}560{,}000)$, so mean $=\boxed{646{,}400\text{ mL}}$ and $\sigma_T=200\sqrt{64}=\boxed{1{,}600\text{ mL}}$. Standardizing: $z=\dfrac{645{,}000-646{,}400}{1{,}600}=-0.875$, so $P(T>645{,}000)=1-\Phi(-0.875)=\Phi(0.875)=\boxed{0.8092}$.