1. The probability of telesales representative making a sale on a customer call is 0.15.

Representatives are required to achieve a mean of at least 5 sales each day.
(C) Find the least number oftcalls each day a representative should make to achieve this
requirement

(d) Calculate the least number of calls that need to be made by a representative for the
probability of at least 1 sale to exceed 0.95

Respuesta :

Answer:

1c

 [tex]n = 33[/tex]

1d

 [tex]n = 19[/tex]

Step-by-step explanation:

From the question we are told that

   The  probability of telesales representative making a sale on a customer call is  [tex]p = 0.15[/tex]

     The mean is  [tex]\mu = 5[/tex]

Generally the distribution of sales call  made by a  telesales representative follows a binomial distribution  

i.e  

         [tex]X  \~ \ \ \  B(n , p)[/tex]

and the probability distribution function for binomial  distribution is  

      [tex]P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}[/tex]

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

Generally the mean is mathematically represented as

     [tex]\mu = n* p[/tex]

=>  [tex]5= n * 0.15[/tex]

=>  [tex]n = 33[/tex]

Generally the least number of calls that need to be made by a representative for the  probability of at least 1 sale to exceed 0.95 is mathematically represented as

      [tex]P( X \ge 1) = 1 - P( X < 1 ) > 0.95[/tex]

=>    [tex]P( X \ge 1) = 1 - P( X =0 ) > 0.95[/tex]

=>    [tex]P( X \ge 1) = 1 - [ ^{n}C_0 *  (0.15 )^0 *  (1- 0.15)^{n-0}] > 0.95[/tex]

=>    [tex] 1 - [1  *  1*  (0.85)^{n}] > 0.95[/tex]

=>    [tex]  [(0.85)^{n}] > 0.05[/tex]

taking natural  log of both sides

[tex]n = \frac{ln(0.05)}{ln(0.85)}[/tex]

=>  [tex]n = 19[/tex]