Approximate Distribution of Y-X | P(Y-X>13) | Binomial & Normal Distributions

[drawar]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

X~B(20,0.4)~N(8,4.8)
Y~B(30,0.6)~(18,7.2)
Y-X~N(10,12)
P(Y-X>13)=P(Z>(13-10)/sqrt(12))=P(Z>0.866)=0.193 (This is the correct answer)

My question is why don't we use continuity correction in this case?

 

[Ray Vickson]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

X~B(20,0.4)~N(8,4.8)
Y~B(30,0.6)~(18,7.2)
Y-X~N(10,12)
P(Y-X>13)=P(Z>(13-10)/sqrt(12))=P(Z>0.866)=0.193 (This is the correct answer)

My question is why don't we use continuity correction in this case?

You should use the continuity correction; it improves the accuracy a lot. The exact answer, using the actual binomial distributions, is 0.1560906047 ~ 0.156. The normal approximation (without the continuity correction) is 0.1932381154 ~ 0.193 (as you said), while the normal approximation with the continuity correction is 0.1561607109 ~ 0.156.

BTW: the easiest way to do this question is to recognize that 20-X ~ B(20,0.6), so Y-X+20~B(30,0.6) + B(20,0.6) = B(50,0.6), giving {Y-X > 13} = {Y + 20 -X > 33} = {B(50,0.6) > 33}.

RGV

 

[drawar]

20-X ~ B(20,0.6)

Can you explain the reasoning to this please?

 

[Ray Vickson]

Can you explain the reasoning to this please?

If X has the distribution B(N,p), it counts the number of successes in n trials, with success probability p per trial. That means that n-X is the number of failures in n trials, with failure probability (1-p) per trial, so it is binomial with parameters n and 1-p.

RGV

 

[drawar]

Got it now, thank you very much!