Question

Long distance telephone calls are normally distributed with a mean of 8 minutes and a

     standard deviation of 2 minutes.

        

1. If random samples of 25 calls were selected, what is the probability that telephone calls would be between 7.83 and 8.2 minutes?
2. If random samples of 25 calls were selected, what is the probability that telephone calls would be at most 8.2 minutes?

      c.   If random samples 100 calls were selected, what is the probability that       telephone calls would be between 8.35 and 8.42 minutes?

Answer 1

Given,

=8

=2

a) if n = 25 calculate p(7.83 < x < 8.2)

We know that z = (x-​​​​​​) /(/n)

p(7.83 < x < 8.2) = p(7.83-8/(2/25)<z<8.2-8/(2/25))

= p(-0.43 < z < 0.50)

= p(z<0.43) + p(z < 0.50)

= 0.1664 + 0.1915

p(7.83 < z < 8.2) = 0.3579

Therefore probability that telephone calls would be between 7.83 and 8.2 minutes is 0.3579.

b) p(x < 8.2) = 0.5 + p(0 < z < 8.2-8/(2/25))

= 0.5 + p(0 < z < 0.50)

= 0.5 + 0.1915

p(x < 8.2) = 0.6915

Therefore probability that telephone calls would be at most 8.2 minute is 0.6915.

c) if n= 100

p(8.35<x<8.42) =p(8.42-8/(2/100)<z<8.35-8/(2/100))

= p(z < 2.10) - p(z < 1.75)

= 0.4821 - 0.4599

p(8.35< x < 8.42) = 0.0222

Therefore probability that telephone calls would be between 8.35 and 8.42 is 0.0222.