1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability of normally distributed life lengths

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A standard light bulb is claimed to have a lifelength of 8000 hours with standard deviation of 800 hours.
    A newly designed light bulb is claimed to have a life length of 9200 hours with a standard deviation of 600 hours

    2. Relevant equations

    If 36 standard and 36 newly designed bulbs are tested what is the probability that the mean lifelength of the standard bulbs will be greater than the mean lifelength of the newly designed bulbs.


    [itex]
    z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}
    [/itex]
    3. The attempt at a solution

    Since the sample size is greater than 30, the central limit theorem can be used

    [itex]
    z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}
    [/itex]

    We have [itex]\sigma [/itex] = 600
    n = 36
    [itex]\mu [/itex]= 9200 ( the mean of the newly designed bulbs.)
    X = 8000 ( the mean of the standard bulbs.)

    Putting those into the formula we get:

    [itex]
    z = \frac{8000-9200}{\frac{\600}{\sqrt{36}}}
    [/itex]


    [itex]
    z = \frac{8000-9200}{\frac{400}{3}}
    [/itex]

    = -1

    which -1 standard deviations from mean so equates to aprroximately .85 on the normal distribution.

    So the probability that the mean lifelength of the standard bulbs will be greater than the mean lifelength of the newly designed bulbs.

    equals 1-0.85 = .15

    How does that look ?
     
  2. jcsd
  3. Jun 8, 2009 #2

    Mark44

    Staff: Mentor

    I don't think your work is correct at all. You are treating the problem as if you had one population of light bulbs, but clearly there are two, each with its own std. dev., and possibly its own mean (which is what you are testing).

    This are the hypotheses I believe you need to test:

    H0: [itex]\mu_Y = \mu_X[/itex]
    H1: [itex]\mu_Y > \mu_X[/itex]

    where X represents the lifetimes of the older bulbs, and Y represents the lifetimes of the redesigned bulbs.

    The statistic to work with is
    [tex]
    Z = \frac{\overline{Y} - \overline{X} - (\mu_Y - \mu_X)}{\sqrt{\frac{\sigma_Y^2}{n} + \frac{\sigma_X^2}{m}}}
    [/tex]
     
  4. Jun 8, 2009 #3
    In the formula you gave

    [tex]Z = \frac{\overline{Y} - \overline{X} - (\mu_Y - \mu_X)}{\sqrt{\frac{\sigma_Y^2}{n} + \frac{\sigma_X^2}{m}}}[/tex]

    You are using the sample mean of X and Y [itex] \overline{Y}\text{ and } \overline{X}[/itex]

    How would I calculate those values?

    I have
    [itex]
    \mu Y = 9200 [/itex]

    [itex]\mu X = 8000[/itex]

    [itex]\sigma_Y^2 = 600^2[/itex]

    [itex]\sigma_X^2 = 800^2[/itex]

    [itex]n = 36[/itex]

    [itex]
    m = 36
    [/itex]
    regards
     
  5. Jun 8, 2009 #4

    Mark44

    Staff: Mentor

    You raise a good point. On second thought, I take back what I said, and agree with your approach, with this change:
    X = 9200, and [itex]\mu[/itex] = 8000. You should get a positive value for z.

    Is there some information about the confidence level given in the problem, that you didn't show?
     
  6. Jun 8, 2009 #5

    statdad

    User Avatar
    Homework Helper

    I don't think this is a hypothesis question at all.
    You can use the CLT to work out the distribution for the mean life of the standard bulb, and also for the mean life of the new bulb.
    You now have two new normal distributions - I'll call them S for the standard, N for the new.
    Since they correspond to two different populations, it is safe to assume they are independent: you need to calculate
    [tex]
    \Pr(S -N > 0)
    [/tex]

    ("what is the probability the mean life of the standard bulbs exceeds the mean life of the new bulbs?)

    Since S and N are independent, [tex] S - N [/tex] also is normally distributed, and you can find the desired probability quite easily.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability of normally distributed life lengths
Loading...