91Ó°ÊÓ

a) Create a random value of $X$.

Representing the lifespan of a chosen randomly automotive tire.

That is given to us $X~\mathcal{N}\left(34000,{4000}^2\right)$

$\begin{array}{r}P(X>40000)=1−P(X≤40000)\\ \end{array}$

$\begin{array}{r}=1−P\left(\frac{X−34000}{4000}≤\frac{40000−34000}{4000}\right)\\ \end{array}$

$=1−\mathrm{Φ}\left(1.5\right)$

$=1-0.93319$

$=0.06681$

b) The given information is $X~\mathcal{N}\left(34000,{4000}^2\right)$

$P(30000≤X≤35000)=P\left(\frac{30000−34000}{4000}≤\frac{X−34000}{4000}≤\frac{35000−34000}{4000}\right)$

localid="1649488038646" role="math" $=P\left(−1≤\frac{X−34000}{4000}≤0.25\right)$

$=\mathrm{Φ}\left(0.25\right)−\mathrm{Φ}(−1)$

role="math" localid="1649488073471" $=0.59871−0.15866$

$=0.44$

c). Given $X~\mathcal{N}\left(34000,{4000}^2\right)$

$P(Xâ‰\yen 40000âˆ\pounds Xâ‰\yen 30000)=\frac{P(Xâ‰\yen 40000,Xâ‰\yen 30000)}{P(Xâ‰\yen 30000)}$

$=\frac{P(Xâ‰\yen 40000)}{P(Xâ‰\yen 30000)}$

The preceding expression is equivalent to

$\frac{P(Xâ‰\yen 40000)}{P(Xâ‰\yen 30000)}=\frac{1−P\left(\frac{X−34000}{4000}≤\frac{4000034000}{4000}\right)}{1−P\left(\frac{X34000}{4000}≤\frac{30000−34000}{4000}\right)}$

$=\frac{1−\mathrm{Φ}\left(1.5\right)}{1−\mathrm{Φ}(−1)}$

$=\frac{1−0.93319}{1−0.15866}$

$=0.0794$