A set of telephone lines is to be installed so as to connect between town A and town B. The town A has 2000 telephones. If each of the telephone users of A were to be guaranteed instant access to make calls to B, 2000 telephone lines would be needed. This would be rather extravagant. Suppose that during the busiest hour of the day, each supscriber in A requires, on the average, a telephone connection to B for two minutes, and that these telephone calls are made at random. Find the minimum number M of telephone lines to B which must be installed so that at most, only 1% of the callers of town A will fail to have immediate access to a telephone line to B. (Suggestion!: approximate the distribution by a Gaussian distribution to facilitate the arithmetic)"

Answer 1

This question was posted twice in a row. I may look at it later, if someone else doesn't do it first.

Answer 2

Yes please. I think I hit the refresh button while posting. Sorry. Would love to work with you on getting some hints on getting this done.

Answer 3

If I'm interpreting this problem correctly and since the suggestion is to approximate the distribution by using the Gaussian (or normal) distribution, I would first find the z-score equated to 1% using a z-table. That would be -2.33 (below the mean of the distribution). Next, you will need to find mean and standard deviation. Once you have those values, substitute into the z-score formula, then solve for x.
Mean = np = (2000)(.5) = ?
Standard deviation = √npq = √(2000)(.5)(.5) = ?
Note: I am using .5 since no value for p is stated. Also, q = 1 - p.

Formula for z-score:
z = (x - mean)/sd

Substitute the values for mean and standard deviation when you finish the above calculations:
-2.33 = (x - mean)/sd

Now solve for x.

I hope this will help and is what the problem was asking.