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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
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](https://tex.z-dn.net/?f=n%20%3D%202%2A6%20%3D%2012)
Sale on 30% of these calls, so ![p = 0.3](https://tex.z-dn.net/?f=p%20%3D%200.3)
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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B12%2C4%7D.%280.3%29%5E%7B4%7D.%280.7%29%5E%7B8%7D%20%3D%200.2311)
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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B12%2C0%7D.%280.3%29%5E%7B0%7D.%280.7%29%5E%7B12%7D%20%3D%200.0138)
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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B12%2C2%7D.%280.3%29%5E%7B2%7D.%280.7%29%5E%7B10%7D%20%3D%200.1678)
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](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
So
![E(X) = 12*0.3 = 3.6](https://tex.z-dn.net/?f=E%28X%29%20%3D%2012%2A0.3%20%3D%203.6)
The mean number of sales in the two-hour period is 3.6.