Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help on normal distribution question please

  1. Nov 10, 2004 #1
    " A contractor has recently purchased a new bulldozer. On previous jobs, 15 out of a total of 50 bulldozers have broken down before the end of the job.

    What is the mean and standard deviation of the probability distribution describing the probability of failure of a bulldozer?

    Note: assume a normal approximation is applicable."

    Mean of the probablity distribution seems obvious (0.3), but how does one find the standard deviation?? :confused:

    Thanks for any help that can be offered. :approve:
     
  2. jcsd
  3. Nov 11, 2004 #2
    [tex]\sqrt{\frac{1}{n-1}\sum_{i=1}^n ({x_i}-{\bar{x}})^2}[/tex]
    Hope this helps.

    Paden Roder
     
  4. Nov 11, 2004 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    In the standard dev. expression, the mean is 0.3, as you noted. The xi are either 0 (35 terms) or 1 (15 terms), where n=50.
     
  5. Nov 12, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is a binomial distribution (I'm not sure why they mention a "normal approximation"- if you were asked for specific probablities of, say, 17 breaking down, you might want to use an approximation but it isn't necessary just to find the mean and standard deviation).

    In a binomial distribution with n occurances, probability p of "sucess" on any one and (1-p) of "failure", the mean is np and the standard deviation is √(np(1-p)).

    The probability of a single bulldozer breaking down, here, is p= 15/50= 0.3 and you are asked about 1 bulldozer so, yes, the mean is 0.3. The standard deviation is
    √(0.3*0.7)= √(0.21).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help on normal distribution question please
Loading...