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

Step-by-step 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 :-

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

$\sigma=\sqrt{p(1-p)n}=\sqrt{(0.25)(1-0.25)(300)}\\\\=\sqrt{56.25}=7.5$

Now, using $z=\dfrac{x-\mu}{\sigma}$, the z-value corresponds to x= 70 :-

$z=\dfrac{70-75}{7.5}\approx-0.67$

The z-value corresponds to x= 90 :-

$z=\dfrac{90-75}{7.5}\approx2$

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

$P(-0.67$

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