Question

If p(A) = 0.6 and k is the number of successes of A in n trials

(A) Show that p {550<= k <= 650} = 0.999 , for n = 1000

(B) Find n such that p{0.59n <= k <= 0.61n} = 0.95

Answer 1

$n=1000\\p=0.6\\q=1-p=1-0.6=0.4$

To solve this question, we shall use the Normal approximation to Binomial.

$\mu=E(x)=np=0.6\times1000=600$

Variance, $\sigma^2=npq=0.6\times0.4\times1000=240$

This implies that the standard deviation $\sigma=\sqrt{\sigma^2}=\sqrt{240}=15.4919334$

$a)$

$p(550\lt k\lt650)=p({550-\mu-0.5\over\sigma}\lt Z\lt{650-\mu+0.5\over\sigma}=p({550-600-0.5\over15.49}\lt Z\lt{650-600+0.5\over15.49})=p(-3.26\lt Z\lt 3.26)=\phi(3.26)-\phi(-3.26)=0.9994-0.0006=0.9988\approx0.999$

as required.

0.5 is the correction factor.

$b)$

$p=0.6\\q=1-p=1-0.6=0.4$

$error=\epsilon={1-0.95\over2}=0.025$

Now,

$p(0.59n \le k \le0.61n) = 2\phi({\epsilon\sqrt{{n\over pq}}})-1=0.95$

This implies that,

$2\phi({\epsilon\sqrt{{n\over pq}}})-1=0.95\implies\phi(0.025\sqrt{n\over0.24})=0.975$

So,

$(0.025\sqrt{n\over0.24})=Z_{0.975}$ where $Z_{0.975}$ is the table value associated with 0.975. From the tables, $Z_{0.975}=1.96$

Now we have that

$(0.025\sqrt{n\over0.24})=1.96\implies\sqrt{n\over0.24}=78.4\implies n=1475.1744\approx 1475$

For $p(0.59n \leq k \le 0.61n) = 0.95$ to hold, the value of $n$ must be greater than 1475. That is $n\gt1475$.