Answer to Question #301062 in Statistics and Probability for kavya

Question

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.

Answer 1

$n=500\\p=0.032$

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