Answer to Question #171750 in Statistics and Probability for subh

Question

The probability that a marketing representative make a sale on a customer’s online order is 0.15. Find the probability in case of no sales made in ten online orders and in case of more than three sales made in twenty online orders. The marketing representatives are required to achieve a mean of at least five sales each day. Find the least number of online orders each day that a marketing representative should get to achieve this requirement. Which distribution is suitable for solving the above problems? Discuss the features of this distribution in detail

Answer 1

Solution:

Let $X$ be the random variable denoting a telesales representative making a sale on a customer call.

(a): Given, $p=0.15,q=0.85,n=10$

$X\sim\text{Binomial}(10,0.15)$

No sales made means $X=0$

$P(X=0)=^{10}C_0(0.15)^0(0.85)^{10}\approx0.1969$

(b): Given, $p=0.15,q=0.85,n=20$

$X\sim\text{Binomial}(10,0.15)$

More than three sales made means $X>3$

$P(X\le3)=^{20}C_0(0.15)^0(0.85)^{20}+^{20}C_1(0.15)^1(0.85)^{19}+ \\^{20}C_2(0.15)^2(0.85)^{18}+^{20}C_3(0.15)^3(0.85)^{17}\approx0.6477$

Then, $P(X\ge3)=1-P(X>3)=1-0.6477=0.3523$

(c): Mean$=np=0.15n$

$Mean\ge5$

$\Rightarrow 0.15n\ge5$

$\Rightarrow n\ge 33.333...$

So, at least 34 orders online orders should be there to achieve the requirement.

(d): Binomial is the best distribution in this case as values are discrete and low.