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. Annual canned-food consumption $X\sim N(\mu=250.0\text{ kg},\ \sigma^2=40.0\text{ kg}^2)$, so $\sigma=\sqrt{40}=6.325$ kg. $M$ is the sample mean of $n=16$ families; $T$ is the sum of $n=25$ families.

Find. (a) $P(X<270)$; (b) the 90th percentile (upper decile) and the 25th percentile (lower quartile) of $X$; (c) the pdf of $M$ and $P(240<M<260)$; (d) the pdf of $T$ and $P(T<6{,}400)$.

Approach. Standardize each variable with its own mean and standard deviation — shrinking the standard error by $\sqrt n$ for the sample mean $M$, and scaling mean/variance by $n$ for the sum $T$ — then read the area (or invert it) from the standard normal table.

1. (a) Write the pdf of X and standardize 270. $f(x)=\dfrac{1}{6.325\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-250)^2}{2(6.325)^2}\right]$, $-\infty<x<\infty$. $z=\dfrac{270-250}{6.325}=3.16$. From the table, $\Phi(3.16)\approx0.5+0.4992=0.9992$, so $\boxed{P(X<270)=0.9992}$. The small variance (only $6.3$ kg about a mean of $250$ kg) means a 20–kg excess is an extremely rare, almost-certain-to-be-exceeded event on the low side — the shaded area in Fig. 1a covers essentially the whole curve.
2. (b) Percentiles via the inverse standard normal. The upper decile is the 90th percentile: $z_{0.90}=1.2816$, so $x_{0.90}=250+1.2816(6.325)=258.11$ kg. The lower quartile is the 25th percentile: $z_{0.25}=-0.6745$, so $x_{0.25}=250-0.6745(6.325)=245.73$ kg. $\boxed{\text{Upper decile}=258.11\text{ kg},\ \text{Lower quartile}=245.73\text{ kg}}$.
3. (c) M's distribution and P(240<M<260). By the sampling-distribution-of-the-mean result, $M\sim N\!\left(\mu,\ \sigma^2/n\right)=N\!\left(250,\ \dfrac{40}{16}\right)$, i.e. $\sigma_M=\dfrac{6.325}{4}=1.581$ kg, so $f_M(m)=\dfrac{1}{1.581\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-250)^2}{2(1.581)^2}\right]$. Standardizing: $z_1=\dfrac{240-250}{1.581}=-6.32$, $z_2=\dfrac{260-250}{1.581}=6.32$, so $P(240<M<260)=\Phi(6.32)-\Phi(-6.32)\approx 1.0000$, giving $\boxed{P(240<M<260)\approx1.0000}$. Overlaying $X$ and $M$ (Fig. 1c) shows $M$'s distribution is so tightly concentrated about $\mu=250$ that essentially all of its mass already sits inside $(240,260)$ — the CLT effect of averaging 16 families.
4. (d) T's distribution and P(T<6,400). The sum of $n$ i.i.d. normals is itself normal with mean $n\mu$ and variance $n\sigma^2$: $T\sim N\!\left(25(250),\ 25(40)\right)=N(6{,}250,\ 1{,}000)$, so $\sigma_T=\sqrt{1000}=31.62$ kg and $f_T(t)=\dfrac{1}{31.62\sqrt{2\pi}}\exp\!\left[-\dfrac{(t-6250)^2}{2(31.62)^2}\right]$. Standardizing: $z=\dfrac{6{,}400-6{,}250}{31.62}=4.74$, so $P(T<6{,}400)=\Phi(4.74)\approx0.99999\approx1.0000$, giving $\boxed{P(T<6{,}400)\approx1.0000}$.