Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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
[tex]n = 2*6 = 12[/tex]
Sale on 30% of these calls, so [tex]p = 0.3[/tex]
a. The probability of making exactly four sales.
This is P(X = 4).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311[/tex]
23.11% probability of making exactly four sales.
b. The probability of making no sales.
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138[/tex]
1.38% probability of making no sales.
c. The probability of making exactly two sales.
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678[/tex]
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
[tex]E(X) = np[/tex]
So
[tex]E(X) = 12*0.3 = 3.6[/tex]
The mean number of sales in the two-hour period is 3.6.
