Is continuity correction needed for this qn? (H2 maths)

• Mad Hat

  27 May 13, 17:18

  This question is from A-levels N2003/II/26:

  The random variable X has the binomial distribution B(20,0.4) and the independent random variable Y has the binomial distribution B(30,0.6). State the approximate distribution of Y−X, and hence find an approximate value for P(Y−X>13).

  Here, X ~ N(8, 4.8) approx., since both np and nq > 5.
  Likewise, Y ~ N(18, 7.2) approx.
  

  So, Y−X ~ N(10, 12) approx.

  So far, okay. Now they are asking for P(Y−X>13). Since we have switched from binomial to normal, it seems that a cc. is required. So we should calculate P(Y−X>13.5), and the answer we get is 0.156.

  Problem is that the TYS Dyna answer is 0.193. And I noticed that you can get this answer if you DON'T use cc. Just calculate P(Y−X>13) directly, and get 0.193.

  So is the Dyna answer wrong? (This is Q10 under "Normal distribution" in my TYS Dyna, Nov 2002-2011 edition.) Or is there some reason why you should NOT use cc for this example? Because the question is unusual and involves the difference Y−X. So maybe the cc for Y somehow subtracts the cc for X, ending up with no cc. Is there such a thing??

  Is the correct answer 0.156 (with cc) or 0.193 (no cc)?

• wee_ws

  27 May 13, 18:01

  Hi,

    Continuity correction is required, this type of question has been asked again in 2011 and 2012. Thanks!

  Cheers,
  Wen Shih
  
  P.S. Sorry for the mistake :P

• wee_ws

  27 May 13, 18:16

  Hi,

    Interestingly, if X, Y are binomial random variables with sample sizes m, n and with the same probability of success p, then X + Y is a binomial random variable with sample size m + n and probability of success p.

    Thanks!

  Cheers,
  Wen Shih

• Mad Hat

  27 May 13, 20:53

  WS, thanks for clarifying! No prob about the typo, small thing. I haven't looked closely at the 2011 and 2012 papers yet, but ok, will look out for this question again.

  As for X+Y being binomial, yup, I think I can see that. Because you have m+n trials, and prob of success in each trial is constant at p. So X+Y should be binomial too. But I didn't notice it before.