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a. Form the probability distribution table for \(P(x)=\frac{x}{6},\) for \(x=1,2,3.\) b. Find the extensions \(x P(x)\) and \(x^{2} P(x)\) for each \(x.\) c. \(\quad\) Find \(\Sigma[x P(x)]\) and \(\Sigma\left[x^{2} P(x)\right].\) d. Find the mean for \(P(x)=\frac{x}{6},\) for \(x=1,2,3.\) e. Find the variance for \(P(x)=\frac{x}{6},\) for \(x=1,2,3.\) f. Find the standard deviation for \(P(x)=\frac{x}{6},\) for \(x=1,2,3.\)

Short Answer

Expert verified
The probability distribution table for \(P(x)=\frac{x}{6}\), \(x=1,2,3\), is formed with values \(\frac{1}{6}\), \(\frac{2}{6}\), and \(\frac{3}{6}\) respectively. The extensions \(x P(x)=\frac{1}{6}, \frac{4}{6}, \frac{9}{6}\) and \(x^{2} P(x)=\frac{1}{6}, \frac{4}{6}, \frac{9}{6}\). Summations \(\Sigma[x P(x)] = \frac{14}{6}\) and \(\Sigma[x^{2} P(x)] = \frac{28}{6}\). The mean, variance and standard deviation for \(P(x)=\frac{x}{6}\), for \(x=1,2,3\), equal \(\frac{14}{6}\), \(\frac{28}{6} - (\frac{14}{6})^{2}\), and \(\sqrt{\mathrm{Var}(x)}\) respectively.

Step by step solution

01

Forming the probability distribution table

We create a table with values ranging from 1 to 3. Each probability \(P(x)\) is computed by substituting the \(x\) value into the probability function \(P(x)=\frac{x}{6}\). So, for \(P(1)\), \(P(2)\), and \(P(3)\) it gives us \(\frac{1}{6}\), \(\frac{2}{6}\), and \(\frac{3}{6}\) respectively.
02

Find the extensions \(x P(x)\) and \(x^{2} P(x)\)

The extensions are calculated by multiplying each \(x\) value in the probability distribution table with the respective \(P(x)\), we got \(1*\frac{1}{6} = \frac{1}{6}\), \(2*\frac{2}{6} = \frac{4}{6}\), and \(3*\frac{3}{6} = \frac{9}{6}\). To find \(x^{2} P(x)\), we multiply each \(x^{2}\) with its respective \(P(x)\), leading us to these values \((1^{2}*\frac{1}{6} = \frac{1}{6}\), \(2^{2}*\frac{2}{6} = \frac{4}{6}\), \(3^{2}*\frac{3}{6} = \frac{9}{6}\).
03

Find \(\Sigma[x P(x)]\) and \(\Sigma[x^{2} P(x)]\)

The sum \(\Sigma[x P(x)]\) is calculated by adding all values in the \(x P(x)\) column. The same is done for \(\Sigma[x^{2} P(x)]\), resulting in the sums \(\frac{14}{6}\) and \(\frac{28}{6}\) respectively.
04

Finding the mean for \(P(x)\)

The mean (expected value) of a discrete probability distribution is calculated by the formula \(E[x] = \Sigma[x P(x)]\), thus utilizing the result from step 3. Thus, the mean is \(\Sigma[x P(x)] = \frac{14}{6}\).
05

Find the variance for \(P(x)\)

The variance is calculated using the formula \(\mathrm{Var}(x) = E[x^{2}] - (E[x])^{2}\). As we already found \(E[x]\) in the previous step and \(E[x^{2}]=\Sigma[x^{2} P(x)]\) in step 3, we just substitute these values into the formula. This results in \(\mathrm{Var}(x) = \frac{28}{6} - (\frac{14}{6})^{2}\)
06

Find the standard deviation for \(P(x)\)

The standard deviation is the square root of variance. The square root of the variance value computed in step 5 gives the standard deviation. That is \(SD = \sqrt{\mathrm{Var}(x)}\).

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Key Concepts

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

Mean
The mean, often referred to as the average or expected value, is a key concept in probability distributions. It represents a central value of the probability distribution, where the sum of all the possible values of a random variable, each weighted by their respective probability of occurrence, is calculated. In simpler terms, it's where you expect the outcomes to converge if the experiment is repeated many times.

To compute the mean of a discrete probability distribution, you use the formula:
  • E[x] = \( \Sigma [x P(x)] \)
This involves taking each outcome \(x\), multiplying it by its probability \(P(x)\), and then summing these products together. In our example, we found that the mean for the probability distribution \(P(x)=\frac{x}{6}\) for \(x=1,2,3\) was \(\frac{14}{6}\).

The mean provides valuable insight into where you can "expect" the center of your distribution to be located. It's especially useful for analyzing and comparing different probability distributions.
Variance
Variance measures how much the values of a random variable differ from the mean. Essentially, it quantifies the spread or dispersion of the distribution. A higher variance indicates that the data points are more spread out from the mean, while a lower variance means that they are closer to the mean.

The variance for a discrete probability distribution is calculated using the formula:
  • \( \mathrm{Var}(x) = E[x^2] - (E[x])^2 \)
Where:
  • \( E[x^2] \) is the expected value of the square of \(x\)
  • \( (E[x])^2 \) is the square of the expected value of \(x\)
In our given distribution, variance is computed as \( \frac{28}{6} - (\frac{14}{6})^2 \).

Knowing the variance helps us understand the variability of the data and predict the probability of different outcomes more accurately.
Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion of a set of values. It's essentially the square root of the variance and is expressed in the same units as the data, making it easier to interpret relative to the mean.

To find the standard deviation of a probability distribution, you calculate:
  • \( SD = \sqrt{\mathrm{Var}(x)} \)
In the case of our example, taking the square root of the variance determined in the earlier step yields the standard deviation. This helps to better visualize the spread of the data in the context of the mean. A small standard deviation means that most of the numbers are close to the average, while a large standard deviation indicates that the numbers are more spread out.
Expected Value
The expected value is a fundamental concept in probability, representing the average or mean value that you anticipate when repeating an experiment. It serves as a central tendency, predicting where outcomes should land on average.

The expected value, which is similar to the mean, is computed using the same formula:
  • \( E[x] = \Sigma[x P(x)] \)
Each outcome \(x\) of the random variable is multiplied by its respective probability \(P(x)\) and summed over all possible outcomes. For the probability distribution given by \(P(x) = \frac{x}{6}\) for \(x=1,2,3\), the expected value is the same as the mean, calculated to be \(\frac{14}{6}\).

This value is crucial for probabilistic assessments in decision-making, finances, and various other fields, providing a reliable indication of what to expect over time.

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Most popular questions from this chapter

A January 2005 survey of bikers, commissioned by the Progressive Group of Insurance Companies, showed that \(40 \%\) of bikers have body art, such as tattoos and piercings. A group of 10 bikers are in the process of buying motorcycle insurance. a. What is the probability that none of the 10 has any body art? b. What is the probability that exactly 3 have some body art? c. What is the probability that at least 4 have some body art? d. What is the probability that no more than 2 have some body art?

As reported in the chapter opener "USA and Its Automobiles," Americans are in love with the automobile - the majority have more than one vehicle per household. In fact, the national average is 2.28 vehicles per household. The number of vehicles per household in the United States can be described as follows: $$\begin{array}{cc}\text { Vehicles, } \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\\\\hline 1 & 0.34 \\\2 & 0.31 \\\3 & 0.22 \\\4 & 0.06 \\\5 & 0.03 \\\6 & 0.02 \\\7 & 0.01 \\\8 \text { or more } & 0.01 \\\\\hline\end{array}$$ a. \(\quad\) Replacing the category "8 or more" with exactly "8," find the mean and standard deviation of the number of vehicles per household in the United States. b. How does the mean calculated in part a correspond to the national average of \(2.28 ?\) c. Explain the effect that replacing the category "8 or more" with "8" has on the mean and standard deviation.

Census data are often used to obtain probability distributions for various random variables. Census data for families in a particular state with a combined income of 50,000 dollar or more show that \(20 \%\) of these families have no children, \(30 \%\) have one child, \(40 \%\) have two children, and \(10 \%\) have three children. From this information, construct the probability distribution for \(x,\) where \(x\) represents the number of children per family for this income group.

a. Explain the difference and the relationship between a probability distribution and a probability function. b. Explain the difference and the relationship between a probability distribution and a frequency distribution, and explain how they relate to a population and a sample.

According to the article "Season's Cleaning," the U.S. Department of Energy reports that \(25 \%\) of people with two-car garages don't have room to park any cars inside. Assuming this to be true, what is the probability of the following? a. Exactly 3 two-car-garage households of a random sample of 5 two-car-garage households do not have room to park any cars inside. b. Exactly 7 two-car-garage households of a random sample of 15 two-car-garage households do not have room to park any cars inside. c. Exactly 20 two-car-garage households of a random sample of 30 two-car- garage households do not have room to park any cars inside.

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