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: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ a. Calculate the mean value and standard deviation of \(x\). b. What is the probability that the number of lines in use is farther than 3 standard deviations from the mean value?

Short answer

The mean number of telephone lines in use at any given time is approximately 2.74 with a standard deviation of about 1.32. The probability that the number of lines in use would be farther than 3 standard deviations from the mean value is 0.14, or 14%.

Step-by-step solution

Calculation of Mean

First, calculate the mean value. The mean of a probability distribution is calculated as \( mean = \sum x \cdot p(x) \), where \(x\) represents the number of lines in use, and \(p(x)\) is the probability of \(x\) their occurrence. So we will multiply each \(x\) with its corresponding \(p(x)\), and sum everything up: \(0 * 0.10 + 1 * 0.15 + 2 * 0.20 + 3 * 0.25 + 4 * 0.20 + 5 * 0.06 + 6 * 0.04 = 2.74\).This result means that on average, there are 2.74 phone lines being used at any given time.

Calculation of Standard Deviation

The standard deviation is a measure of how spread out numbers are. If we denote mean as \( \mu \), the standard deviation \(\sigma\) can be calculated as: \(\sigma = \sqrt{\sum (x - \mu)^2 \cdot p(x)}\). So, the next step is to plug the determined mean in this formula: \(\sqrt{(0 - 2.74)^2 * 0.10 + (1 - 2.74)^2 * 0.15 + (2 - 2.74)^2 * 0.20 + (3 - 2.74)^2 * 0.25 + (4 - 2.74)^2 * 0.20 + (5 - 2.74)^2 * 0.06 + (6 - 2.74)^2 * 0.04} = 1.32\).This calculated standard deviation about 1.32 allows us to understand the dispersion in the data, as in how far any given number of lines in use can be from the mean number of 2.74.

Calculation of Probability

Lastly, the exercise asks about the probability that the number of lines in use is farther than 3 standard deviations from the mean value. That's equivalent to asking for the percentage of values that lie outside the range \(\mu - 3\sigma\) to \(\mu + 3\sigma\). The range results in from -1.22 to 6.7. As this is a discrete distribution, we are only interested in whole numbers. Hence, we need to sum probabilities for x=0 and x=6, because only these values lie outside the defined range. This results in 0.10 + 0.04 = 0.14 or 14%.