Question

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

Answer 1

Answer:

1c

 $n = 33$

1d

 $n = 19$

Step-by-step explanation:

From the question we are told that

   The  probability of telesales representative making a sale on a customer call is  $p = 0.15$

     The mean is  $\mu = 5$

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

i.e  

         $X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial  distribution is  

      $P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

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

     $\mu = n* p$

=>  $5= n * 0.15$

=>  $n = 33$

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

      $P( X \ge 1) = 1 - P( X < 1 ) > 0.95$

=>    $P( X \ge 1) = 1 - P( X =0 ) > 0.95$

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

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

=>    $[(0.85)^{n}] > 0.05$

taking natural  log of both sides

$n = \frac{ln(0.05)}{ln(0.85)}$

=>  $n = 19$