Results 1 to 5 of 5

Math Help - Adding two independent normal random variables

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    6
    Awards
    1

    Adding two independent normal random variables

    Hi guys,
    I couldn't figure this one out:
    Bulbs fail with normal distribution: mean = 400 days, st. dev = 100 days.

    Bulbs are replaced every 365 days. What is the probability that a bulb and its replacement will fail in that period?

    There's a hint though: The sum of two iid normal random variables is ______

    The book also says this: The sum of independent random variables whose distributions are normal is a random variable having the normal distrbution.

    I cant figure out neither the excerpt, the hint, nor the answer. Can someone please explain what it means?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by indian View Post
    Hi guys,
    I couldn't figure this one out:
    Bulbs fail with normal distribution: mean = 400 days, st. dev = 100 days.

    Bulbs are replaced every 365 days. What is the probability that a bulb and its replacement will fail in that period?

    There's a hint though: The sum of two iid normal random variables is ______

    The book also says this: The sum of independent random variables whose distributions are normal is a random variable having the normal distrbution.

    I cant figure out neither the excerpt, the hint, nor the answer. Can someone please explain what it means?
    Let X_1 and X_2 be two independent normal random variables:

    X_1 ~ Normal (\mu_1, \, \sigma_1)

    X_2 ~ Normal (\mu_2, \, \sigma_2)

    Then Y = X_1 + X_2 is a normal random variable:

    Y ~ Normal \left( \mu = \mu_1 + \mu_2, \, \sigma = \sqrt{\sigma_1^2 + \sigma_2^2} \right).


    So let Y = X + X be the random variable lifetime of bulb plus replacement bulb.

    Y ~ Normal \left( \mu = 400 + 400 = 800, \, \sigma = \sqrt{100^2 + 100^2} = 100 \sqrt{2}\right).

    You need to calculate Pr(Y < 365).

    I get 0.001049 (correct to six decimal places)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    6
    Awards
    1
    Thanks a LOT! That is AWESOME! Could you please help me out with a further sub-part? Basically, now they're saying that a college has 120 such bulbs, and they want to stock up spare ones. They generally change bulbs once a year.

    How many spare bulbs should they stock up to have P = 0.1 that they run out on replacements?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by indian View Post
    Thanks a LOT! That is AWESOME! Could you please help me out with a further sub-part? Basically, now they're saying that a college has 120 such bulbs, and they want to stock up spare ones. They generally change bulbs once a year.

    How many spare bulbs should they stock up to have P = 0.1 that they run out on replacements?
    I'm not sure I quite understand the question but here's something to chew on:

    The probability that a single bulb fails in less than 365 days is Pr(X < 365) = 0.36317 (correct to five decimal places).

    Let B be the random variable number of bulbs that fail in less than 365 days.

    Then B ~ Binomial(n = 120, p = 0.36317).

    You need to find the smallest of a such that Pr( B > a) < 0.1.

    This value of a will be the number of spare bulbs that should be kept in stock. The probability that a replacement bulb will fail before the year is finished is so small as to not be worth worrying about.

    Finding the value of a is straightforward using standard technology: I get

    Pr(B > 50) = 0.1310
    Pr(B > 51) = 0.0954

    I'd be inclined to give 51 spare bulbs as the answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2008
    Posts
    6
    Awards
    1
    ohhh! I got it now! Thanks a LOT!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sup of independent random variables
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 23rd 2010, 02:34 PM
  2. Independent random variables and their PDF
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 25th 2010, 09:16 AM
  3. independent random variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 15th 2009, 02:55 PM
  4. Independent Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: January 17th 2009, 01:24 PM
  5. Independent Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 28th 2008, 05:39 PM

Search Tags


/mathhelpforum @mathhelpforum