A study of data collected at a tyre factory shows that a batch of $8000$ tyres have a mean wear life of $35000$ km with a standard deviation of $7000$ km. Assuming a Normal Distribution, estimate:
(i) How many tyres will wear out before $30000$ km?
(ii) How many tyres will continue for more than $45000$ km before wearing out?
$X = 8,000\\ U= 35,000\\ SD= 7,000\\ \frac{X-\mu}\sigma$
(i) First we find the $Z$ score$$\frac{35,000-30,0000}{7,000}=0.71$$ $Z$-score$= .2580$
Tyres to wear out$$8000(.5-.2580)=1936$$ (ii) Tiers to continue more than $45,000$
This is where I am stuck as I am not sure how to go about
I am not sure if I just repeat the same steps as before and just change the values $$\frac{45,000-35,000}{7000}=1.42$$
$Z$-score $= .4207$
Answer is$$8000(.5-.4207)=634.3$$
Should I subtract the first answer from $X~(8000-1936)$ first as we already know that these tiers will fail?
$(1)~P(X\le30,000)=P\left(Z\le\frac{30,000-35,000}{7,000}=\color{red}-\frac57\right)=P(Z\ge5/7)=0.5-P(0\le Z\le5/7)$
$5/7\approx0.71$ and $P(0\le Z\le0.71)$ is $0.2611$ from the table. Thus, the number of tyres is $8,000(0.5-0.2611)\approx1,911$. The difference in our answers is because I had access to a more refined value of $P(0\le Z\le5/7)$.
$(2)~P(X\ge45,000)=P\left(Z\ge\frac{45,000-35,000}{7,000}=\frac{10}7\right)=0.5-P(0\le Z\le10/7)$
$10/7\approx1.43$ and $P(0\le Z\le1.43)$ is $0.4236$ from the table. Thus, the number of tyres is $8,000(0.5-0.4236)\approx611$. The difference in our answers is because I had access to a more refined value of $P(0\le Z\le10/7)$.
There is no requirement to subtract the above answer from $8,000$, as $8,000-1,911$ will give the number of tyres that wear out after $30,000\text{ km}$ which is not required to be found.