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  • Continuity Correction

    I took the practice exam 1 from TIA and I ran into a question with continuity correction. The question stated:

    A fair 6-sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability that the number of 3's that are rolled is greater than 150 and less than 180.

    Fairly simple question:

    I calculated:
    mean: 166.67 (1000/6)
    Sx: 11.78 [sqrt (1000 *5/36)]

    The continuity part however was the part I got "wrong".

    The problem is, I believe I am right in that: to calculate the prob of rolling greater than 150 w/ a continuity correction is 149.5 where as the solution wrote 150.5. And less than 180 w/ a continuity correction as 180.5 instead of the solution of 179.5.

    The entire point of the continuity correction is to include the number (in this case: 150 and 180) while calculating the probability because the normal distribution only calc. probability of < the specified number.

    Please explain whether I am wrong or the solution? And if I am wrong...why?
    :laugh:

  • #2
    Originally posted by plastictaxi420 View Post
    I took the practice exam 1 from TIA and I ran into a question with continuity correction. The question stated:

    A fair 6-sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability that the number of 3's that are rolled is greater than 150 and less than 180.

    Fairly simple question:

    I calculated:
    mean: 166.67 (1000/6)
    Sx: 11.78 [sqrt (1000 *5/36)]

    The continuity part however was the part I got "wrong".

    The problem is, I believe I am right in that: to calculate the prob of rolling greater than 150 w/ a continuity correction is 149.5 where as the solution wrote 150.5. And less than 180 w/ a continuity correction as 180.5 instead of the solution of 179.5.

    The entire point of the continuity correction is to include the number (in this case: 150 and 180) while calculating the probability because the normal distribution only calc. probability of < the specified number.

    Please explain whether I am wrong or the solution? And if I am wrong...why?
    Strictly greater than and strictly less than.

    NOT-- greater than or equal to and less than or equal to.

    x>150 is equivalent to x=>151. You get it now?

    150 is not included in x>150 and 180 is not included in x<180. Make sense?
    (/

    Comment


    • #3
      this is a dumb question, but how did you get the mean and var?

      i was thinking its a n*mu and n*sigma squared, but wouldnt this be a uniform distribution so the mean of a uniform with n=6 would be 7/2? (n+1)/2? Maybe not, im really having a problem here

      Comment


      • #4
        Its central limit theorem with n= 1000.

        Its the version with a TOTAL number of 3's. If n is large, then a sequence of identically distributed "occurrences" is approximately normally distributed with E(x) = n * the probability of a number on a die = expected value of any of the 6 numbers being rolled in 1000 rolls.

        The new standard deviation is the original deviation divided by the sqrt of n.

        Use the new found mu and standard dev. and find the probability (with correction factor of course because the occurence of a 3 is not continuous) of this occurence.

        Think of throwing a die 1000 times. The chart/graph/ or distribution of the amount of 3's that would appear would look approximately like a bell curve (normal).

        If that helps.

        Comment


        • #5
          I'm still not getting it. It makes sense but CLT says that when the number of trials is greater than 30, then the distribution is N(nmu, nsigmasquared). Isn't mu for a die rolling equal to (n+1)/2 (because a die is uniform distribution?) wouldn't that be 7/2?

          Comment


          • #6
            So your saying P(x<150) with a continuity correction is equivilant to P(x<150.5) because you WANT to include the 150?

            To Dystrother - well yes you would use the normal approximation, however in order to use it you need the mean and standard deviation for it. The mean and variance for any distribution has its respective formulas. For rolling a die multiple times and obtaining x-3s, we want to use a binomial distribution with mean (np) and sqrt(variance (np(1-p))). The mean would be 1000*(1/6) and variance would be 1000 * (1/6) * (5/6)
            :laugh:

            Comment


            • #7
              Originally posted by plastictaxi420 View Post
              So your saying P(x<150) with a continuity correction is equivilant to P(x<150.5) because you WANT to include the 150?
              Wrong.

              I meant P(x>150) is equivalent with CC to P(x>150.5). You are not including 150 with x>150.

              You have the inequality backwards.

              Bottom line.......

              P(a<x<b) ==using CC==> P(a+1/2<=x<=b-1/2)

              P(a<=x<b) ==using CC==> P(a-1/2<=x<=b-1/2)

              P(a<=x<=b) ==using CC==> P(a-1/2<=x<=b+1/2)
              (/

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              • #8
                Thanks man. Il memorize it and eventualy understand it.
                :laugh:

                Comment


                • #9
                  Ok yea that definitely makes sense. When would you use a Uniform distribution then with respect to a die problem?

                  Comment


                  • #10
                    Originally posted by dystrother View Post
                    Ok yea that definitely makes sense. When would you use a Uniform distribution then with respect to a die problem?
                    You can use it here to get the exact answer. Don't just blindly use CLT, think about what it means and why we use it. It's saying when you have a LOT of trials (define a LOT), then your uniform, binomial, poisson, etc. converge to normal and you can just figure out the moments and then the Z score etc. But using a cont. distribution to estimate a discrete one isn't completely kosher, you have to use a continuity correction to adjust for that change. So for this problem, you could've used the uniform to arrive at an answer though I wager it would've taken some time.

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