# Thread: Normal distribution

1. ## Normal distribution

Have got the following question was fine on the first section just not sure on the others, not sure whether the route to go down is to find the probability greater than or equal to 286.5 hours for the d2 component. I tried this just jusing the standard normal distribution functions as in part a however the numbers seem to large....
(c) The life lengths (times to failure, in hours) of two types of electronic component,
D1 and

D
2, have distributions N(139.6, 36) and N(145, 4) respectively.

(i) Which type of component should be used to maximize the probability of running for
at least 148 hours without failure.

(ii) A type
D1 component is used. When it fails, it is replaced by a type D2 component.
What is the probability that the second component is still running after 286.5 hours?

(iii) The two types of device are run in parallel. What is the probability that one device

fails more than 5.4 hours before the other?

Many thanks in advance.

2. For (ii) you want (if I'm interpreting the question correctly, it's a little vague) P(D_1 + D_2 > 286.5). D_1 and D_2 are presumably independent so their sum is a normal where you sum the means and variances of each.

For (iii) You want P(D_1 > D_2 + 5.4 OR D_2 > D_1 + 5.4), which is the sum of the two probabilities. You'll use the distribution of D_1 - D_2.

3. Originally Posted by theodds
For (ii) you want (if I'm interpreting the question correctly, it's a little vague) P(D_1 + D_2 > 286.5). D_1 and D_2 are presumably independent so their sum is a normal where you sum the means and variances of each.

For (iii) You want P(D_1 > D_2 + 5.4 OR D_2 > D_1 + 5.4), which is the sum of the two probabilities. You'll use the distribution of D_1 - D_2.
Yes, I initially thought that for (ii) as well. The only problem is that D1 has to fail before D2 gets used. I have no time to think about it, but it seems to me that Pr(D1 < x), where x < 286.5, has to be considered along with Pr(D2 > 286.5 - x) and an integral over 0 < x < 286.5 set up ....

4. Originally Posted by mr fantastic
Yes, I initially thought that for (ii) as well. The only problem is that D1 has to fail before D2 gets used. I have no time to think about it, but it seems to me that Pr(D1 < x), where x < 286.5, has to be considered along with Pr(D2 > 286.5 - x) and an integral over 0 < x < 286.5 set up ....
The thing is, the way the question is worded doesn't make it clear whether or we consider the event to have happened if D_1 hasn't failed at 286.5. The question seems to assume that D_1 will have failed at that point. If you want to go with your interpretation, we want P(D_1 + D_2 > 286.5 and D_1 < 286.5)

Regardless of your interpretation, however, the result is going to be the same. P(D_1 > 286.5) is 0 to within any reasonable rounding error, so

$\displaystyle P(D_1 + D_2 > 286.5 \cap D_1 < 286.5) = P(D_1 + D_2 > 286.5) + P(D_1 < 286.5) - P(D_1 + D_2 > 286.5 \cup D_1 < 286.5)$
$\displaystyle \approx P(D_1 + D_2 > 286.5) + 0 - 0.$

D_1 would have to be about 25 standard deviations above its mean for it to exceed 286.5.

5. Originally Posted by theodds
The thing is, the way the question is worded doesn't make it clear whether or we consider the event to have happened if D_1 hasn't failed at 286.5. The question seems to assume that D_1 will have failed at that point.

[snip]
Yes, I agree the wording is open to several interpretations.