Question

Assuming that in some part of Quebec the probability is 0.00007 that a child will develop cancer. Therefore, the number Z among 28, 572 children that will develop cancer follows a Binomial distribution with parameters p = 0.00007 and n = 28, 572. We would like to use the Poisson distribution to approximate these binomial probabilities.

(a) What is the adequate value of the variance of the Poisson distribution to use in order to approximate the precedent binomial distribution?

(b) Find the probabilities of the 11 first possible values of Z (i.e. Z = 0, 1, . . . , 10) using both the formulas for the binomial distribution and then the Poisson approximation. Plot the two histograms and make a comparison. Is this approximation close enough? Justify!

(c) Use the Poisson probabilities to approximate the binomial probabilities that among 28, 572 children

i) None will develop cancer ii) At most two will develop cancer.

(d) Using Poisson approximation, calculate the probability that at least seven in a sample of ten children will not develop cancer. Is it a good approximation? Justify!

Answer 1

Given:

Assuming that in some part of Quebec the probability is 0.00007 that a child will develop cancer. Therefore, the number Z among 28, 572 children that will develop cancer follows a Binomial distribution with parameters p = 0.00007 and n = 28, 572.

a) The adequate value of the variance of the Poisson distribution to use in order to approximate the precedent binomial distribution :

For Binomial distribution , variance is given by

Variance = np(1-p)

Here p is very small so 1-p is approximately 1.

Therefore the variance of the Poisson distribution to approximate Binomial distribution is

np = 28572 * 0.00007 = 2.00004 = 2

So the variance is 2

b) The probabilities of the 11 first possible values of Z (i.e. Z = 0, 1, . . . , 10) using both the formulas for the binomial distribution and then the Poisson approximation :

The probability density function of Binomial distribution is given by

P(X=X) = (nCx) * p^x (1-p)^n-x

So P(X=0) = (28572C0) * (0.00007)^0 * (1-0.00007)^28572-0

= 0.1353

Similarly we can find remaining probability.

The probability density function of Poisson distribution is given by

P(X=X) = e^- * ()^x /X!

Where = variance = 2

So P(X=0) = e^-2 * (2)^0 / 0!

= 0.1353

Similarly we can find remaining probability.

Z	Binomial	Poisson
0	0.1353	0.1353
1	0.2707	0.2707
2	0.2707	0.2707
3	0.1804	0.1804
4	0.0902	0.0902
5	0.0361	0.0361
6	0.0120	0.0120
7	0.0034	0.0034
8	0.0009	0.0009
9	0.0002	0.0002
10	3.82*10^-5	3.82*10^-5

The two histograms and make a comparison :

c) The Poisson probabilities to approximate the binomial probabilities that among 28, 572 children :

i)The probability that bone will develop cancer :

P(X=X) = e^- * ()^x /X!

Where = variance = 2

So P(X=0) = e^-2 * (2)^0 / 0!

= 0.1353

So the probability that none will develop a cancer is 0.1353

ii)The probability that at most two will develop cancer:

P(X<=2) = P(X=0) + P(X=1) + P(X=2)

= 0.1353 + 0.2707 + 0.2707

= 0.6767

So the probability that at most 2 will develop a cancer is 0.6767

d) The probability that at least seven in a sample of ten children will not develop cancer = The probability that at the most 3 out of 10 will develop cancer

For n = 10

= np = 0.00007*10 = 0.0007

P(X <= 3) = e^- 0.0007* (0.0007)^x/x!

= 1