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 :-
[tex]\mu=np=300(0.25)=75[/tex]
[tex]\sigma=\sqrt{p(1-p)n}=\sqrt{(0.25)(1-0.25)(300)}\\\\=\sqrt{56.25}=7.5[/tex]
Now, using [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 70 :-
[tex]z=\dfrac{70-75}{7.5}\approx-0.67[/tex]
The z-value corresponds to x= 90 :-
[tex]z=\dfrac{90-75}{7.5}\approx2[/tex]
By using the standard normal distribution table for z, the probability that he will make more than 70 sales but less than 90 sales:-
[tex]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[/tex]
Hence, the probability that he will make more than 70 sales but less than 90 sales= 0.7258
