Question

A telemarketer makes a sale on 25% of his calls. If he makes 300 calls in a night, what is the probability that he will make more than 70 sales but less than 90 sales?

Answer 1

Answer: 0.7258

Explanation:

Given : A telemarketer makes a sale on 25% of his calls.

i.e. p=0.25

He makes 300 calls in a night, i.e. n=300

Let x be a random variable that represents the number of calls make in night.

To convert the given binomial distribution to normal distribution we have :-

`\mu=np=300(0.25)=75`

`\sigma=√(p(1-p)n)=√((0.25)(1-0.25)(300))\\\\=√(56.25)=7.5`

Now, using
`z=(x-\mu)/(\sigma)`, the z-value corresponds to x= 70 :-

`z=(70-75)/(7.5)\approx-0.67`

The z-value corresponds to x= 90 :-

`z=(90-75)/(7.5)\approx2`

By using the standard normal distribution table for z, the probability that he will make more than 70 sales but less than 90 sales:-

`P(-0.67<z<2)=P(z<2)-P(z<-0.67)\\\\=P(z<2)-(1-P(z<0.67))\\\\=0.9772-(1-0.7486)\\\\=0.9772-0.2514=0.7258`

Hence, the probability that he will make more than 70 sales but less than 90 sales= 0.7258