Question 1 of 8: Snow Tire Life (Normal Distribution and Sampling)

Question 1: Snow Tire Life (Normal Distribution and Sampling) (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. Tire life X ~ Normal with μ = 165,000 km, σ = 10,000 km.

Find. (a) P(X<180,000). (b) P(|X−165,000|>16,000). (c) the distribution of M = mean of n=16 tires and P(M>160,000). (d) the distribution of T = sum of 36 tires and P(T>6,000,000).

Approach. Standardize X (and its derived sums/means M, T, which are themselves normal by the linear-combination/CLT property of the normal family) and read tail probabilities from the standard normal table.

1. (a) P(X<180,000). The pdf of X is $$f(x)=\dfrac{1}{10000\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{x-165000}{10000}\right)^{2}}$$ Standardizing, $z=\dfrac{180000-165000}{10000}=1.50$. From the normal table, $P(Z\lt 1.50)=0.5+0.4332=0.9332$. $$\boxed{P(X\lt 180{,}000)=0.9332}$$
Fig. 1 — pdf of X ~ N(165,000, 10,000²); shaded = P(X<180,000) = 0.9332.
2. (b) P(|X−165,000|>16,000). This is $P(X\lt 149{,}000)+P(X\gt 181{,}000)$. Standardizing, $z=\dfrac{16000}{10000}=1.60$, so each tail has area $0.5-0.4452=0.0548$. $$\boxed{P(|X-165{,}000|\gt 16{,}000)=2(0.0548)=0.1096}$$
Fig. 2 — pdf of X; shaded (two tails) = P(|X−165,000|>16,000) = 0.1096.
3. (c)(i)–(ii) Distribution of the sample mean M. For a sample of n=16 tires drawn from the normal population, $M\sim N\!\left(\mu,\ \dfrac{\sigma^{2}}{n}\right)$, so $$\mu_M=165{,}000\text{ km}, \qquad \sigma_M=\dfrac{10{,}000}{\sqrt{16}}=2{,}500\text{ km}$$ and the pdf of M is $$f(m)=\dfrac{1}{2500\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{m-165000}{2500}\right)^{2}}$$ (iii) The figure below overlays the (wider, flatter) pdf of X and the (narrower, taller) pdf of M — both centred at 165,000 but M's spread is 1/4 that of X since $\sigma_M=\sigma/\sqrt{16}$.
Fig. 3 — pdfs of X ~ N(165,000,10,000²) (dashed) and M ~ N(165,000,2,500²) (solid); shaded = P(M>160,000) = 0.9772.
4. (c)(iv) P(M>160,000). Standardizing with M's own parameters, $z=\dfrac{160000-165000}{2500}=-2.00$, so $P(M\gt 160{,}000)=P(Z\gt -2.00)=0.5+0.4772=0.9772$. $$\boxed{P(M\gt 160{,}000)=0.9772}$$
5. (d)(i)–(ii) Distribution and tail probability of T. T is the sum of the lives of n=36 independently drawn tires, so $T\sim N(n\mu,\ n\sigma^{2})$: $$\mu_T=36(165{,}000)=5{,}940{,}000\text{ km}, \qquad \sigma_T=10{,}000\sqrt{36}=60{,}000\text{ km}$$ with pdf $$f(t)=\dfrac{1}{60000\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{t-5940000}{60000}\right)^{2}}$$ Standardizing, $z=\dfrac{6{,}000{,}000-5{,}940{,}000}{60{,}000}=1.00$, so $P(T\gt 6{,}000{,}000)=0.5-0.3413=0.1587$. $$\boxed{P(T\gt 6{,}000{,}000)=0.1587}$$ (The "twenty limousines" detail is a red herring for the calculation — only the 36-tire sample size matters.)