Answer to Question #145821 in Statistics and Probability for Ray

Question

1. The probability of telesales representative making a sale on a customer call is 0.15. Find the probability that
(a) no sales are made in 10 calls, (2 marks) (b) more than 1 sales are made in 20 calls. (3 marks)
Representatives are required to achieve a mean of at least 5 sales each day.
(c) Find the least number of calls each day a representative should make to achieve this requirement. (1 mark) (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.

Answer 1

$X=$ a telesales representative making a sale on a customer call

(a) $X\sim Bin(10,0.15)$

$$P(X=0)=\binom{10}{0}(0.15)^0(1-0.15)^{10}=0.1969$$

The probability that No sales are made in 10 calls 0.1969

(b) $X\sim Bin(20,0.15)$

$$P(X>1)=1-P(X=0)-P(X=1)=$$ $$=1-\binom{20}{0}(0.15)^0(1-0.15)^{20}-$$ $$-\binom{20}{1}(0.15)^1(1-0.15)^{19} =$$

$$=1-0.03875953108-0.136798345=$$

$$=0.82444212391$$

The probability that more than 1 sales are made in 20 calls is 0.8244.

(c) $\mu=np\geq5$

$$0.15n\geq5=>n\geq34$$

The least number of calls each day is 34.

(d)

$$P(X\geq1)=1-P(X=0)>0.95$$

$$1-(1-0.15)^n>0.95$$

$$(0.85)^n<0.05$$

$$n\geq19$$

The least number of calls is 19.