Question

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

Short answer

The probabilities of the events are as follows: \n a. 0.70, \n b. 0.45, \n c. 0.55, \n d. 0.71, \n e. 0.45, \n f. 0.45 \n These are calculated by adding up the corresponding probabilities from the given probability distribution.

Step-by-step solution

Translate Statements

First, each event needs to be translated into terms of \(x\). In each statement, the terms 'at most', 'fewer than', 'at least' and 'between...inclusive' refers to ranges of x-values. Thus, these events corresponds to the following x-values: a. 'At most three lines are in use' implies \(x = 0, 1, 2, 3\). b. 'Fewer than three lines are in use' implies \(x = 0, 1, 2\).c. 'At least three lines are in use' implies \(x = 3, 4, 5, 6\).d. 'Between two and five lines are in use' implies \(x = 2, 3, 4, 5\).e. 'Between two and four lines are not in use' implies \(x = 2, 3, 4\).f. 'At least four lines are not in use' implies \(x = 0, 1, 2\).

Calculate Probabilities

Now, to calculate the probability of each event occurring, the associated probabilities from given distribution should be added up for all corresponding x-values. Here are the probability calculations for each event using given distribution:a. \(P(x \leq 3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\).b. \(P(x < 3) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).c. \(P(x \geq 3) = P(x=3) + P(x=4) + P(x=5) + P(x=6) = 0.25 + 0.20 + 0.06 + 0.04 = 0.55\).d. \(P(2 \leq x \leq 5) = P(x=2) + P(x=3) + P(x=4) + P(x=5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\).e. \(P(2 \leq x \leq 4) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).f. \(P(x \leq 2) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).