# A problem on normal distribution.

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?

• What have you tried? You might also want to check the question numbers for typos – Henry Jul 18 '15 at 12:02
• The mean should be 510 or 5100? – Satish Ramanathan Jul 18 '15 at 12:04
• Perhaps the answer to (i) is about $0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999767$ – Henry Jul 18 '15 at 12:06
• @henry, Out of curiosity. How did you get that number from EXCEL? – Satish Ramanathan Jul 18 '15 at 12:22
• @satishramanathan =NORM.DIST(500,5100,200,TRUE) gives 2.33E-117 which I then subtracted from $1$, though in fact I used R and pnorm(500,mean=5100,sd=200) which gave the same thing – Henry Jul 18 '15 at 12:46

Answer: Part I $P(X\ge500) = 1-P(X\le500) = 1-P(\frac{X-5100}{200}\le \frac{500-5100}{200})$

$1 - P(z\le -23) \approx 1$

Part II

$P(X\ge x) = 0.98 => 1-P(X\le x) = 0.98 =>P(X\le x) = 0.02$

$\frac{X-5100}{200} = -2.053748911 => X = 4689$

$x \sim N(\mu,\sigma^2)$. 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.