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Statistics (z-scores) question?

  • 18-12-2012 5:53pm
    #1
    skyscraperblue
    Registered Users, Registered Users 2 Posts: 747 ✭✭✭


    I'm having trouble with comparing two z-scores where I don't have any actual values to insert for x. I have two normal distributions: call one A, which has one mean and S.D, and the other B has a different mean and S.D. and I'm asked for the probability that a random selection from A will be greater than one from B. I just seem to have too many unknowns no matter what I do!

    That probably didn't make sense so I'm going to type the question (I'll insert the information from previous sections to avoid making it too long)

    Okay a mixed (6 type A, 5 type B) box of chocolates has mean weight 226.9g and standard deviation 4.46g. A box of chocolates consisting of ten type B chocolates has mean weight 227g and S.D. 4.74g. My question is:
    What is the probability that the mass of the contents of a mixed box will exceed that of a box containing 10 type B chocolates? (Assume random selection.)


Comments

  • #2
    hivizman
    Registered Users, Registered Users 2 Posts: 1,163 ✭✭✭hivizman


    Intuitively, the probability you are after will be slightly less than 0.5, because a random mixed box will on average weigh a little less than a random type B box.

    What I think you need to do is define a new random variable, which is equal to the difference between the two random variables A and B. This means that you have only one z-score to work with rather than two.

    Let A be the random variable represented by the weight of the mixed box, then A is distributed normally with mean E(A) = 226.9 and s.d. 4.46 (i.e. variance 4.46^2 = 19.8916 = var(A)).

    Let B be the random variable represented by the weight of the type B box, then B is distributed normally with mean E(B) = 227 and s.d. 4.74 (i.e. variance 4.74^2 = 22.4676 = var(B)).

    Define C = A - B, then, because A and B are independent, C will also be normally distributed.

    The mean of C will be equal to E(A) - E(B) = 226.9 - 227 = -0.1

    The variance of C will be equal to var(A) + var(B) = 19.8916 + 22.4676 = 42.3952, so s.d.(C) will be 6.51 (note, because we have squared the s.d. to get the variance, we add the two variances even though we subtract the means).

    You want to calculate the probability that the weight of a random mixed box is greater than the weight of a random type B box, which is equivalent to the probability that C>0. You can work out the z-value (as you now have only two parameters rather than the original four) and look this up in the relevant table for the normal distribution.
    Given the information above, your z-value will be 0.1/6.51 = 0.01536, and your required probability p(C>0) will be 0.493872. I used the NORMDIST function on Excel to calculate this - the function gives the cumulative probability (i.e. p(C<0)) as 0.506128, so p(C>0) = 1 - p(C<0) = 0.493872.


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