Question

The CEO of a clothing company estimates that 52% of customers will make a purchase. Part A: How many customers should a salesperson expect until she finds a customer that makes a purchase? Part B: What is the probability that a salesperson helps 3 customers until she finds the first person to make a purchase?

Answer 1

Answer:

(a) The expected number of should a salesperson expect until she finds a customer that makes a purchase is 0.9231.

(b) The probability that a salesperson helps 3 customers until she finds the first person to make a purchase is 0.058.

Step-by-step explanation:

Let the random variable X be defined as the number of customers the salesperson assists before a customer makes a purchase.

The probability that a customer makes a purchase is, p = 0.52.

The random variable X follows a Geometric distribution since it describes the distribution of the number of trials before the first success.

The probability mass function of X is:

$P(X=x)=(1-p)^{x}p$

The expected value of a Geometric distribution is:

$E(X)=\frac{1-p}{p}$

(a)

Compute the expected number of should a salesperson expect until she finds a customer that makes a purchase as follows:

$E(X)=\frac{1-p}{p}$

         $=\frac{1-0.52}{0.52}\\=0.9231$

This, the expected number of should a salesperson expect until she finds a customer that makes a purchase is 0.9231.

(b)

Compute the probability that a salesperson helps 3 customers until she finds the first person to make a purchase as follows:

$P(X=3)=(1-0.52)^{3}\times0.52\\=0.110592\times 0.52\\=0.05750784\\\approx 0.058$

Thus, the probability that a salesperson helps 3 customers until she finds the first person to make a purchase is 0.058.