Answer to Question #289401 in Statistics and Probability for anne

Question

The probability of a telesales representative making a sale on a customer call is 0.15. Assuming this percentage holds true for future attempts, answer the following questions.

i. Find the probability that in the next 12 calls, the number of sales made is less than 2.

ii. Find the probability that in the next 15 calls, the number of sales made is more than 3.

Answer 1

Let the random variable $X$ represent the number of sales made on a customer call, then $X\sim Binom(n,p)$ where $p=0.15$.

$a)$

$n=12$.

The probability to be determined is, $p(x\lt2)=p(x=0)+p(x=1)$

$p(x=0)=\binom{12}{0}(0.15)^0(0.85)^{12}=(0.85)^{12}=0.142241757$ and

$p(x=1)=\binom{12}{1}(0.15)^1(0.85)^{11}=12\times0.15\times0.85^{11}=0.301217838$

Now we have,

$p(x\lt2)=0.1422+0.3012=0.4434596$

Therefore, the probability that in the next 12 calls, the number of sales made is less than 2 is 0.4434596.

$b)$

$n=15$

We determine the following probability.

$p(x\gt3)=1-p(x\leq3)$.

Now,

$p(x\leq3)=p(x=0)+p(x=1)+p(x=2)+p(x=3)$

So,

$p(x=0)=\binom{15}{0}(0.15)^0(0.85)^{15}=(0.85)^{15}=0.08735422$

$p(x=1)=\binom{15}{1}(0.15)^1(0.85)^{14}=15\times0.15\times0.85^{14}=0.23123176$

$p(x=2)=\binom{15}{2}(0.15)^2(0.85)^{13}=105\times0.0225\times0.85^{13}=0.28563922$

$p(x=3)=\binom{15}{3}(0.15)^3(0.85)^{12}=445\times0.003375\times0.85^{12}=0.21362934$

Now we have,

$p(x\leq3)=0.08735422+0.23123176+0.28563922+0.21362934=0.81785454$

Therefore,

$p(x\gt3)=1-0.81785454=0.18214546$

Therefore, the probability that in the next 15 calls, the number of sales made is more than 3 is 0.18214546.