Answer to Question #145821 in Statistics and Probability for Ray

Question #145821
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.
1
Expert's answer
2020-11-24T04:28:32-0500

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

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

P(X=0)=(100)(0.15)0(10.15)10=0.1969P(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) XBin(20,0.15)X\sim Bin(20,0.15)

P(X>1)=1P(X=0)P(X=1)=P(X>1)=1-P(X=0)-P(X=1)==1(200)(0.15)0(10.15)20=1-\binom{20}{0}(0.15)^0(1-0.15)^{20}-(201)(0.15)1(10.15)19=-\binom{20}{1}(0.15)^1(1-0.15)^{19} =

=10.038759531080.136798345==1-0.03875953108-0.136798345=

=0.82444212391=0.82444212391

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


(c) μ=np5\mu=np\geq5


0.15n5=>n340.15n\geq5=>n\geq34

The least number of calls each day is 34.

(d)



P(X1)=1P(X=0)>0.95P(X\geq1)=1-P(X=0)>0.95

1(10.15)n>0.951-(1-0.15)^n>0.95

(0.85)n<0.05(0.85)^n<0.05

n19n\geq19

The least number of calls is 19.



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