Answer to Question #171753 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

Count of sales from total customer’s online orders count (n) is Binomial random variable X with parameters p=0.15 and n.

Distribution function:

$Pr(X=k) = C_n^kp^k(1-p)^{n-k}$

The probability in case of no sales made in ten online orders (n = 10):

$Pr(X = 0) = C_{10}^0*0.85^{10} = 0.196$

The probability in case of more than three sales made in twenty online orders (n = 20):

$P(X > 3) = 1 - Pr(X <4) = 1 - (C_{20}^3*0.15^3*0.85^{17} + C_{20}^2*0.15^2*0.85^{18} + C_{20}^1*0.15^1*0.85^{19} + C_{20}^0*0.15^0*0.85^{20}) = 0.352$

Mathematical expectation of X:

$EX = n p \geq E_0 = 5 => n_{min} = E_0 / p = 5 / 0.15 = 34$