Question

A multichannel microwave link is to provide telephone communication to a remote community having 12 subscribers, each of whom uses the link 20% of the time during pick hours. How many channels are needed to make the link available during pick hours to:
(a) Eighty percent of the subscribers all of the time?
(b) All of the subscribers 80% of the time?
(c) All of the subscribers 95% of the time?

Answer 1

Answer on Question #55862 – Math – Statistics and Probability

A multichannel microwave link is to provide telephone communication to a remote community having 12 subscribers, each of whom uses the link 20% of the time during pick hours. How many channels are needed to make the link available during pick hours to:

(a) Eighty percent of the subscribers all of the time?

(b) All of the subscribers 80% of the time?

(c) All of the subscribers 95% of the time?

Solution

$$P(busy) = \frac{20}{100} = 0.2, \quad P(not busy) = 1 - 0.2 = 0.8.$$

(a)

$$N = 12 \cdot \frac{80\%}{100\%} = 9.6 \text{ round up to } 10.$$

(b) We work with the binomial distribution with $p = 0.2$ and $n = 12$.

Consider

$$P(X \leq N) = \text{binomialcdf}(N; 12; 0.2) = \sum_{k=0}^{N} \frac{12!}{k! (12 - k)!} \cdot 0.2^k (1 - 0.2)^{12 - k} > 0.8.$$

In Excel it is given by

$$= \text{BINOMDIST}(N; 12; 0.2; \text{TRUE})$$

where N can take on any value from the list $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

We calculate all these values and put them into the table:

Therefore, at $N = 4$ links from 12 users will be satisfied 92.7% of the time. At this level, 4 channels would be sufficient to guarantee access more than 80% of the time to all subscribers.

(c)

$$P(X \leq N) = \text{binomialcdf}(N; 12; 0.2) = \sum_{k=0}^{N} \frac{12!}{k! \, (12 - k)!} \cdot 0.2^k \, (1 - 0.2)^{12 - k} > 0.95.$$

From above, at $N = 5$ links users will be satisfied $98.1\%$ of the time. At this level, 5 channels would be sufficient to guarantee access $95\%$ of the time to all subscribers.

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