Question

1. Suppose an online retailer has determined that if a potential customer spends more than 5 minutes on their site, that the probability of them making a purchase is 0.40. If 12 customers who spent more than 5 minutes on the site were randomly selected, what is the probability that more than 7 will make a purchase

Answer 1

Using the binomial distribution, it is found that there is a 0.056 = 5.6% probability that more than 7 will make a purchase.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• There are 12 customers, hence n = 12.

• The probability of any of them making a purchase is of p = 0.4.

The probability that more than 7 will make a purchase is given by:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12).

Hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.4)^{8}.(0.6)^{4} = 0.042$

$P(X = 9) = C_{12,9}.(0.4)^{9}.(0.6)^{3} = 0.012$

$P(X = 10) = C_{12,10}.(0.4)^{10}.(0.6)^{2} = 0.02$

$P(X = 11) = C_{12,11}.(0.4)^{11}.(0.6)^{1} \approx 0$

$P(X = 12) = C_{12,12}.(0.4)^{12}.(0.6)^{0} \approx 0$

Then:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.042 + 0.012 + 0.02 + 0 + 0 = 0.056.

0.056 = 5.6% probability that more than 7 will make a purchase.

More can be learned about the binomial distribution at brainly.com/question/24863377