04-BS-2, May 2016, Q1 | Nivaar Exam Prep

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. Service-call duration $X\sim N(\mu=1{,}500\text{ s},\ \sigma=250\text{ s})$. $M$ is the sample mean of $n=25$ calls; $T$ is the sum of $n=100$ calls.

Find. (a) $P(X<1200)$; (b) $P(1300<X<1700)$; (c) the pdf of $M$ and $P(M>1450)$; (d) the pdf of $T$ and $P(T<155{,}000)$.

Approach. Standardize each variable using its own mean and standard deviation (with the standard error shrunk by $\sqrt n$ for $M$, and mean/variance scaled by $n$ for the sum $T$), then read the area from the standard normal table.

(a) Write the pdf of X and standardize 1,200. $f(x)=\dfrac{1}{250\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-1500)^2}{2(250)^2}\right]$, $-\infty<x<\infty$. $z=\dfrac{1200-1500}{250}=-1.20$. From the table, $\Phi(-1.20)=1-\Phi(1.20)=1-0.8849=0.1151$. So $\boxed{P(X<1200)=0.1151}$.
(b) Standardize both endpoints. $z_1=\dfrac{1300-1500}{250}=-0.80$, $z_2=\dfrac{1700-1500}{250}=0.80$. $P(1300<X<1700)=\Phi(0.80)-\Phi(-0.80)=0.7881-0.2119=0.5762$, so $\boxed{P(1300<X<1700)=0.5762}$.
(c) M's distribution and P(M>1450). By the sampling-distribution-of-the-mean result, $M\sim N\!\left(\mu,\ \sigma^2/n\right)=N\!\left(1500,\ \dfrac{250^2}{25}\right)$, i.e. $\sigma_M=\dfrac{250}{5}=50$ s, so $f_M(m)=\dfrac{1}{50\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-1500)^2}{2(50)^2}\right]$. Standardizing: $z=\dfrac{1450-1500}{50}=-1.00$, so $P(M>1450)=1-\Phi(-1.00)=\Phi(1.00)=0.8413$, giving $\boxed{P(M>1450)=0.8413}$. Overlaying $X$ and $M$ on one diagram (Fig. 1) shows $M$'s distribution is far more concentrated about the same mean — the CLT effect of averaging 25 calls.
(d) T's distribution and P(T<155,000). The sum of $n$ i.i.d. normals is itself normal with mean $n\mu$ and variance $n\sigma^2$: $T\sim N\!\left(100(1500),\ 100(250)^2\right)=N(150{,}000,\ 6{,}250{,}000)$, so $\sigma_T=2{,}500$ s and $f_T(t)=\dfrac{1}{2500\sqrt{2\pi}}\exp\!\left[-\dfrac{(t-150000)^2}{2(2500)^2}\right]$. Standardizing: $z=\dfrac{155{,}000-150{,}000}{2{,}500}=2.00$, so $P(T<155{,}000)=\Phi(2.00)=0.9772$, giving $\boxed{P(T<155{,}000)=0.9772}$.

Fig. 1a — pdf of X, shaded area = P(X<1200).

Fig. 1b — pdf of X, shaded area = P(1300<X<1700).

Fig. 1c — pdf of X (broad, dashed) and pdf of M (narrow, n=25), shaded area = P(M>1450).

Question 1 — final results
Part	Quantity	Value
(a)	P(X < 1,200 s)	0.1151
(b)	P(1,300 < X < 1,700 s)	0.5762
(c)	σ_M (n=25); P(M > 1,450 s)	50 s; 0.8413
(d)	σ_T (n=100); P(T < 155,000 s)	2,500 s; 0.9772

Topic: Normal distribution and the sampling distributions of the mean and the sum.

Key relations: $X\sim N(\mu,\sigma^2)\Rightarrow \bar X_n\sim N(\mu,\sigma^2/n)$ and $T_n=\sum X_i\sim N(n\mu,n\sigma^2)$; standardize with $z=(x-\mu)/\sigma$ (or the appropriate $\sigma_{\bar X}$/$\sigma_T$) before reading $\Phi(z)$.

Why this works: a linear combination of independent normal variables is itself normal, with mean and variance obtained by the usual expectation/variance algebra; averaging divides the variance by $n$ (concentrating $M$ around $\mu$), while summing multiplies it by $n$ (spreading $T$ out in absolute terms even though it is tight relative to its own much larger mean).

Common pitfall: using $\sigma$ instead of $\sigma/\sqrt n$ when standardizing $M$, or forgetting that $T$'s standard deviation is $\sqrt n\,\sigma$ (not $n\sigma$).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 6 (Sampling Distributions).

Question 1 of 8

Question 1 (20 marks)