Results 1 to 5 of 5

Math Help - Normal distribution

  1. #1
    Junior Member
    Joined
    Jan 2011
    Posts
    58

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by theodds View Post
    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 ....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    Quote Originally Posted by mr fantastic View Post
    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

    <br />
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)
    \approx P(D_1 + D_2 > 286.5) + 0 - 0.<br />

    D_1 would have to be about 25 standard deviations above its mean for it to exceed 286.5.
    Last edited by theodds; January 24th 2011 at 06:55 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by theodds View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 27th 2011, 01:08 PM
  2. normal distribution prior and posterior distribution proof
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 9th 2011, 06:12 PM
  3. Replies: 2
    Last Post: March 29th 2010, 02:05 PM
  4. Replies: 2
    Last Post: August 25th 2009, 10:39 PM
  5. Replies: 1
    Last Post: April 20th 2008, 06:35 PM

Search Tags


/mathhelpforum @mathhelpforum