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Homework Help: Statistics: normal distribution

  1. Jun 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that X is normally distributed with mean 95 and standard deviation 17.

    A. What is the probability that is greater than 126.79?

    B. What value of X does only the top 16% exceed?



    2. Relevant equations

    z= (X-(mean of x))/standard deviation



    3. The attempt at a solution

    using the equation, I found z= (126.79-95)/17=1.87

    then using a statistical table, the value for P(z greater than 1.87) = 1 - P(z=1.87) = 0.0307

    But this is wrong..

    For B)

    I considered that 84% must be the value which the top 16% must exceed.. then 84% of 95 is 88.42

    Also wrong (well I wasn't really too sure about part B)

    Any help would be appreciated!
     
  2. jcsd
  3. Jun 27, 2011 #2
    The tabular value for above that z-score seems correct. Your z-score seems correct to the given number of decimal places. How do you know that the answer under a) is "wrong"?

    For B), you should find such a [itex]Z_{0}[/itex] that:

    [tex]
    P(Z > Z_{0}) = 0.16 \Rightarrow P(Z < Z_{0}) = 0.84
    [/tex]

    Using that value for [itex]Z_{0}[/itex], solve for X in the expression:
    [tex]
    Z = \frac{X - \mu}{\sigma}
    [/tex]
    where [itex]\mu[/itex] is the mean and [itex]\sigma[/itex] is the standard deviation.
     
  4. Jun 27, 2011 #3

    HallsofIvy

    User Avatar
    Science Advisor

    No, that is correct. Why do you say it is wrong?

    95 is the mean- 50% are larger than 95!

    Looking at a table of the normal distribution, N(x< a)= 0.84 for a= 1.0 (approximately).

    [tex]\frac{z- 95}{17}= 1.0[/tex]
    solve for z.
     
  5. Jun 27, 2011 #4
    Thank you both for the help with part B. As for part A, since its a webwork assignment, it tells me when I enter an incorrect answer. I guess it's possible that there might be some sort of problem with the webwork.
     
  6. Jun 28, 2011 #5
    maybe you need to use the correct number of significant figures.
     
  7. Jun 28, 2011 #6
    I put all the numbers (0.0307).. The chart I have for the z-score only goes upto 4 decimal places so I'm not sure how much it's actually supposed to go up to
     
  8. Jun 29, 2011 #7
    Well, assuming the mean and the standard deviation are known to the same number of significant figures as 126.79, then the z-score is:
    [tex]
    \frac{126.79 - 95}{17} = \frac{31.79}{17} = 1.870
    [/tex]
    to 4 significant figures (actually the quotient is exact). For this, the probability in question is:

    1 - 0.9693 = 0.0307

    So, it can't be that.

    Are you sure this is the standard deviation and not the variance?
     
  9. Jun 29, 2011 #8
    Yes it is the standard deviation. I copied the question word for word..
    In any case it's too late, the homework set is closed and apparently the accepted answer was 0307419079428789. I think I was supposed to use excel to find the exact value to the correct amount of significant figures..
     
  10. Jun 30, 2011 #9
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