The lifetime of a component in a computer is advertised to last for $500$ hours. It is known that the lifetime follows a normal distribution with mean $5100$ hours and standard deviation $200$ hours.
(i) Calculate the probability that a randomly chosen component will last longer than the advertised hours.
(ii) If a dealer wants to be sure that $98\%$ of all the components for sale lasts longer than the advertised figure, what figure should be advertised?
$x \sim N(\mu,\sigma^2)$.
Now one can calculate z as:
$z = (x-\mu) / \sigma $
(i) in this case we can have z score as $z = (500 - 510)/200 = -0.05 $ now we can see the z-table to see the probability below or equal to the z value. Then one can calculate the probability greater than z value as 1-p(z).
(ii) in this case you are provided with the $\alpha = 0.98$ so you need to calculate the z value.And then calculate X value for it correspondingly.