Question

16. A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours, find: a. The probability of making exactly four sales. b. The probability of making no sales. c. The probability of making exactly two sales. d. The mean number of sales in the two-hour period. Lind, Douglas. Basic Statistics for Business and Economics (p. 181). McGraw-Hill Education. Kindle Edition.

Answer 1

Answer:

a) 23.11% probability of making exactly four sales.

b) 1.38% probability of making no sales.

c) 16.78% probability of making exactly two sales.

d) The mean number of sales in the two-hour period is 3.6.

Step-by-step explanation:

For each phone call, there are only two possible outcomes. Either a sale is made, or it is not. The probability of a sale being made in a call is independent from other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours, find:

Six calls per hour, 2 hours. So

$n = 2*6 = 12$

Sale on 30% of these calls, so $p = 0.3$

a. The probability of making exactly four sales.

This is P(X = 4).

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

$P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311$

23.11% probability of making exactly four sales.

b. The probability of making no sales.

This is P(X = 0).

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

$P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138$

1.38% probability of making no sales.

c. The probability of making exactly two sales.

This is P(X = 2).

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

$P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678$

16.78% probability of making exactly two sales.

d. The mean number of sales in the two-hour period.

The mean of the binomia distribution is

$E(X) = np$

So

$E(X) = 12*0.3 = 3.6$

The mean number of sales in the two-hour period is 3.6.