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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

Expert verified
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

01

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.
02

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.
03

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%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
The mean is a statistical measure that helps us understand the average or expected value of a data set. In the context of a probability distribution, the mean is calculated by taking each possible value of the random variable, multiplying it by its probability, and summing all these products. This can be express as:
  • Formula: \( \text{Mean} = \sum x \cdot p(x) \)
In this exercise, we have several customer service lines with varying probabilities of being in use. Calculating the mean allows us to understand, on average, how many lines are being used at any given point in time.
The calculation is performed by multiplying each number of lines by the probability it occurs and summing these up. For example, the probability that no line is in use is 0.10, and for one line, it is 0.15, and so on, which eventually gives us an average of 2.74 lines in use.
Understanding the mean helps us visualize the central tendency of the probability distribution effectively.
Standard Deviation
Standard deviation is a crucial concept in statistics, revealing how much individual data points in a set differ from the mean. It measures the spread or variance of a data set. For any probability distribution, calculating the standard deviation involves taking the square root of the variance of the data set.
In this exercise, we use the formula:
  • Formula: \( \sigma = \sqrt{\sum (x - \mu)^2 \cdot p(x)} \)
Here, \( \mu \) is the mean we previously calculated.
This formula evaluates each data point's deviation from the mean, squares it to eliminate negative values, multiplies it by the probability, and aggregates these products before taking the square root. A standard deviation of 1.32 was computed. This relatively small spread implies that the usage of phone lines is quite consistent around the mean of 2.74 lines.
Understanding standard deviation is invaluable when examining how much variation exists within probability distributions or data sets.
Discrete Distribution
A discrete distribution is one in which the random variable can take on a specific set of distinct values. Unlike a continuous distribution, where data can take any value within a given range, a discrete distribution is concerned with specific outcomes like rolling a die or counting phone lines in use.
In our exercise, we see this concept in action with the phone lines, where the values might be 0 through 6, each with an associated probability.
These probabilities must sum up to 1, indicating the certainty that one of these values will occur.
  • The probabilities here are given as \( p(x) \), for each \( x \) which ranges from 0 to 6.
This kind of distribution is often used for countable data and allows us to analyze problems involving distinct, separate events.
Understanding discrete distributions is fundamental when assessing data that categorically falls into clear and distinct groups or events.
Probability Calculation
Probability calculation involves determining the likelihood that a particular outcome occurs within a probability distribution. In our exercise, part of the aim was to find the probability that the number of lines in use falls outside a particular range determined by the standard deviation from the mean.
The concept of calculating probability within specific bounds is essential, as it allows us to understand how likely different event regions are. For our specific question, this is expressed as calculating the range:
  • \( \mu - 3\sigma \) to \( \mu + 3\sigma \)
With the mean and standard deviation known, the boundary values were calculated as -1.22 and 6.7. Since the variable \( x \) is discrete, practical evaluation only occurs at whole number points.
In this distribution, only the probabilities for \( x = 0 \) and \( x = 6 \) fall outside these bounds, leading to a combined probability of 0.14 or 14%.
Probability calculation gives insights into how often we can expect certain segments of data from our distribution to occur.

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Most popular questions from this chapter

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

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