Question

Critical-part failures in NASCAR vehicles. Refer to The Sport Journal (Winter 2007) analysis of critical-part failures at NASCAR races, Exercise 4.144 (p. 277). Recall that researchers found that the time x (in hours) until the first critical-part failure is exponentially distributed with $\mathbf{\mu }$= .10 and s = .10. Now consider a random sample of n = 50 NASCAR races and let $\overline{\mathbf{\chi }}$ represent the sample meantime until the first critical-part failure.

a) Find $\mathbf{E}\left(\overline{\mathbf{\chi }}\right)$ and $\mathbf{Var}\left(\overline{\mathbf{\chi }}\right)$

b) Although x has an exponential distribution, the sampling distribution of x is approximately normal. Why?

c) Find the probability that the sample meantime until the first critical-part failure exceeds .13 hour.

Short answer

It is frequently used to simulate the time spent among occurrences. Continuous probability distributions are used to calculate the incidence of occurrences.

Step-by-step solution

 Step 1: (a) Given

The time $\chi$ is exponentially distributed with the mean:

$\begin{array}{c}\mu =10\\ \sigma \text{=0.10}\\ \text{E}\left(\text{χ}\right)=\mu =0.10\end{array}$

$\begin{array}{c}\text{V}\left(\text{χ}\right)={\sigma }^2={\left(0.1\right)}^2\\ =0.01\end{array}$

The calculation is given below:

Now, $\chi$represents the sample mean period for n = 50

localid="1651469168306" $\stackrel{-}{\chi }=\frac{{\text{χ}}_{\text{1}}+{\text{χ}}_2+\mathrm{.......}+{\text{χ}}_{50}}{50}$

$\begin{array}{c}\text{E}\left(\stackrel{-}{\chi }\right)=\frac{1}{50}\text{E}({\text{χ}}_{\text{1}}+{\text{χ}}_2+\mathrm{.......}+{\text{χ}}_{50})\\ =\frac{1}{50}(\text{E}\left({\text{χ}}_{\text{1}}\right)+\text{E}\left({\text{χ}}_2\right)+\mathrm{.......}+\text{E}\left({\text{χ}}_{50}\right))\\ =\frac{1}{50}(\mu +\mu +50\text{times+}\mu )\\ =\frac{50\mu }{50}\end{array}$

localid="1651469270820" $=0\mu$

$\text{E}\left(\stackrel{-}{\chi }\right)=0.1$

localid="1651469378610" $\begin{array}{c}\text{V}\left(\stackrel{-}{\chi }\right)=\text{V}\left(\frac{{\displaystyle \sum _{i\ge 1}^{50}\times \text{i}}}{50}\right)\\ =\frac{1}{{\left(50\right)}^2}\text{V}(\sum _{i\ge 1}^{50}\times \text{i})\\ =\frac{1}{2500}\sum _{i\ge 1}^{50}\times \text{V}\left(\text{χi}\right)\\ =\frac{1}{2500}\sum _{i\ge 1}^{50}{\sigma }^2\end{array}$

localid="1651469435234" $\begin{array}{c}=\frac{1}{2500}\times 500{\sigma }^2\\ =\frac{{\sigma }^2}{50}\\ =\frac{0.01}{50}\end{array}$

$\text{V}\left(\stackrel{-}{\chi }\right)=0.0002$

(b) Explanation

The $\text{E}\left(\stackrel{-}{\chi }\right)=0.1$or$\mu \left(\stackrel{-}{\chi }\right)andV\left(\stackrel{-}{\chi }\right)=0.0002=\sigma {\text{χ}}^{\text{2}}$. Then using the control limit theorem $\stackrel{-}{\chi }~\text{N}\left(\mu {\text{χ}}_{\text{1}}\sigma {\text{χ}}_2\right)$that is$\stackrel{-}{\chi }~\text{N}(0.001,0.002).$It is since n =50 that is the higher size of the sample $(>30)$.

(c) Explanation

The calculation is given below:

$\begin{array}{c}\text{P}\left(\text{Samplemeantime>0.13}\right)=\text{P}\left(\text{.13}\right)\\ =\text{P}(\frac{-0.01}{\sqrt{0.002}}>\frac{0.13-0.01}{\sqrt{0.002}})=\text{P}(z>8.49)\\ =0\end{array}$