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A business has six customer service telephone lines. Consider the random variable \(x=\) number of lines in use at a randomly selected 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 value of \(x\) is 2.64 and the standard deviation is approximately 8.947. The probability that the number of lines in use is farther than 3 standard deviations from the mean value is 0.

Step by step solution

01

Calculate the mean value

To calculate the mean value of \(x\), we will use the formula for the mean of a discrete random variable: \[E(x)=\sum xP(x)\] This means that we will multiply each \(x\) value by its corresponding probability, and then sum up the results. E(x) = \(0\times0.10 + 1\times0.15 + 2\times0.20 + 3\times0.25 + 4\times0.20 + 5\times0.06 + 6\times0.04 = 0 + 0.15 + 0.4 + 0.75 + 0.8 + 0.3 + 0.24 = 2.64\) The mean value of \(x\) is 2.64.
02

Calculate the variance and standard deviation

To calculate the variance of \(x\), we will use the formula for the variance of a discrete random variable: \[Var(x) = E(x^2) - E(x)^2\] First, we need to find E(x^2): \[E(x^2) = \sum (x^2)P(x)\] E(x^2) = \(0^2\times0.10 + 1^2\times0.15 + 2^2\times0.20 + 3^2\times0.25 + 4^2\times0.20 + 5^2\times0.06 + 6^2\times0.04 = 87\) Now, we can calculate the variance: \[Var(x) = E(x^2) - E(x)^2 = 87 - (2.64)^2 = 87 - 6.9696 = 80.0304\] To find the standard deviation, we just need to take the square root of the variance: \[SD(x) = \sqrt{Var(x)} = \sqrt{80.0304} ≈ 8.947\] The standard deviation of \(x\) is approximately 8.947.
03

Calculate the probability that the number of lines in use is farther than 3 standard deviations from the mean value

To find the range of values within 3 standard deviations of the mean, we will calculate the lower and upper bounds: Lower bound: \(2.64 - 3\times8.947 ≈ -24.201\) Upper bound: \(2.64 + 3\times8.947 ≈ 29.481\) Since the random variable \(x\) is a discrete random variable representing the number of customer service lines in use, it cannot take negative values. So, the lower bound will be 0. Also, the highest possible value for \(x\) is 6. Therefore, the range of values we are interested in is \(0 \le x \le 6\). Since this range includes all possible values for the random variable \(x\), the probability that the number of lines in use is farther than 3 standard deviations from the mean value is 0.

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Key Concepts

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

Discrete Random Variables
When we talk about discrete random variables, we are referring to variables that can take on a countable number of distinct outcomes. In this exercise, the discrete random variable is the number of phone lines in use at any given time. The number of possible values for this variable ranges from 0 to 6, which corresponds to the total number of lines available. Each possible outcome is associated with a specific probability, indicating how likely it is to occur.

Discrete random variable outcomes can often be visualized with a probability distribution. This distribution displays all possible values the variable can assume along with their respective probabilities. In our example:
  • 0 lines in use has a probability of 0.10
  • 1 line in use has a probability of 0.15
  • 2 lines in use has a probability of 0.20
  • 3 lines in use has a probability of 0.25
  • 4 lines in use has a probability of 0.20
  • 5 lines in use has a probability of 0.06
  • 6 lines in use has a probability of 0.04
Understanding discrete random variables helps to model and forecast outcomes in various fields, from telecommunication to finance.
Mean and Standard Deviation
The mean and standard deviation provide a summary of our probability distribution. Let's first consider the mean, which is essentially the average expected value of our random variable. It is calculated by multiplying each possible value by its probability and then adding all the results together. For our distribution, this results in a mean, or expected value, of 2.64. This means on average, about 2 to 3 phone lines are in use at any time.

The standard deviation, on the other hand, tells us how much variation or dispersion exists from the mean. It's crucial to understanding the spread of our data around this average. We first need the variance as an intermediate step, which is derived from the mean of the squares of each value minus the square of the mean.
  • The variance, calculated as 80.0304, gives us an idea of the overall spread of the data.
  • The standard deviation is the square root of the variance, giving us approximately 8.947.
This higher standard deviation indicates a large spread of values, suggesting that the number of lines in use can vary significantly from the mean.
Variance Calculation
Variance calculation is an important part of statistical analysis as it shows how much the values of a random variable differ from its mean. In our example, we began by computing the expected value of the square of each outcome, \(E(x^2)\), which involves squaring each possible value, multiplying by its probability, and summing them.

This result is then used to find the variance using the formula: \[Var(x) = E(x^2) - E(x)^2\]
  • We calculated this to be 87 - 6.9696 = 80.0304.
The variance reflects the degree to which each probable value of the random variable deviates from the mean. A larger variance indicates that the data points are spread out over a wide range of values. In practical terms, this means there can be a significant fluctuation in the number of lines in use. By understanding variance, businesses can better anticipate and plan for peak usage times, potentially improving service efficiency and customer satisfaction. Variance is computed directly from probability formulas and provides a foundational measure for further statistical analysis, such as determining the probability of events concerning standard deviations.

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Most popular questions from this chapter

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass Index in Children" (International Journal of Cardiology [2009]: 1-7) described a study in which the left atrial size was measured for a large number of children ages 5 to 15 years. Based on these data, the authors concluded that for healthy children, left atrial diameter was approximately normally distributed with a mean of \(26.4 \mathrm{~mm}\) and a standard deviation of \(4.2 \mathrm{~mm}\). a. Approximately what proportion of healthy children have left atrial diameters less than \(24 \mathrm{~mm}\) ? b. Approximately what proportion of healthy children have left atrial diameters greater than \(32 \mathrm{~mm} ?\) c. Approximately what proportion of healthy children have left atrial diameters between 25 and \(30 \mathrm{~mm} ?\) d. For healthy children, what is the value for which only about \(20 \%\) have a larger left atrial diameter?

A point is randomly selected on the surface of a lake that has a maximum depth of 100 feet. Let \(x\) be the depth of the lake at the randomly chosen point. What are possible values of \(x\) ? Is \(x\) discrete or continuous?

An appliance dealer sells three different models of freezers having 13.5,15.9 , and 19.1 cubic feet of storage space. Consider the random variable \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: \(x\) \(\begin{array}{lll}13.5 & 15.9 & 19.1\end{array}\) \(p(x)\) \(\begin{array}{lll}0.2 & 0.5 & 0.3\end{array}\) a. Calculate the mean and standard deviation of \(x\). (Hint: See Example \(6.15 .\) ) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

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