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)

To find the minimum number of telephone lines, we need to calculate the number of lines required to ensure that at most 1% of the callers in town A will fail to have immediate access to a telephone line to town B.

First, let's calculate the average number of calls made during the busiest hour of the day. Each subscriber in town A requires a telephone connection to town B for an average of 2 minutes. Since there are 2000 telephones in town A, the total number of minutes of calls during the busiest hour is 2000 * 2 = 4000 minutes.

Next, let's calculate the standard deviation of the distribution of calls. We can approximate the distribution by a Gaussian distribution. The standard deviation (σ) of the Gaussian distribution can be calculated using the formula:

σ = √(n * p * (1 - p))

where n is the average number of calls and p is the probability of a single call occurring. In this case, n = 4000 and p = 1/(60/2) = 1/30 (since there are 60 minutes in an hour and each call lasts for 2 minutes).

σ = √(4000 * 1/30 * (1 - 1/30))
≈ √(133.33)
≈ 11.55

Now, let's calculate the number of lines required to ensure at most a 1% failure rate. We can use the 99th percentile of the Gaussian distribution, which is approximately 2.33 standard deviations above the mean.

Number of lines (M) = n + (z * σ)

where z is the number of standard deviations above the mean.

M = 4000 + (2.33 * 11.55)
≈ 4257

Therefore, the minimum number of telephone lines required to ensure that at most 1% of the callers in town A will fail to have immediate access to a telephone line to town B is approximately 4257.