A change is made to one product page on the retail companies’ web site. To determine if the change does improve the efficiency of that product page, data must be collected on the proportion of visitors to the new page that ultimately purchase the product. It is known that 3.2% of visitors, to the original page, make purchases. Assume that this proportion holds for the next 500 visitors to the new page. Use the normal distribution to approximate the probability that, among these 500 visitors, the number who purchase will be (a) 11 or fewer. (b) 21 or more.

$n=500\\p=0.032$

If we use the normal approximation then, $\mu=np=500\times 0.032=16$ and $\sigma=\sqrt{npq}=\sqrt{500\times0.032\times 0.968}=\sqrt{15.488}=3.93547964$

Let the random variable X represent the number of purchases

a)

$p(X\le 11)=p({X-\mu\over \sigma}\le{11-\mu\over \sigma})=p(Z\le {11-16\over 3.93547964})=p(Z\le -1.27)=\phi(-1.27)=0.1020$

The probability that there are 11 or fewer purchases is 0.1020

b)

$p(X\ge 21)=p({X-\mu\over \sigma}\ge{21-\mu\over \sigma})=p(Z\ge {21-16\over 3.93547964})=p(Z\ge 1.27)=1-p(Z\le 1.27)=1-\phi(1.27)=1-0.8980=0.1020$

The probability that there are 21 or more purchases is 0.1020