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Chapter 5: Problem 31

The random variable \(A\) has the following probability distribution: $$\begin{array}{l|ccccc}\hline \mathbf{A} & 1 & 2 & 3 & 4 & 5 \\\P(\boldsymbol{A}) & 0.6 & 0.1 & 0.1 & 0.1 & 0.1 \\\\\hline\end{array}$$ a. Find the mean and standard deviation of \(A .\) b. How much of the probability distribution is within 2 standard deviations of the mean? c. What is the probability that \(A\) is between \(\mu-2 \sigma\) and \(\mu+2 \sigma ?\)

Short Answer

Expert verified
The mean of \(A\) is 1.8. Its standard deviation is 1.25. The total probability of the distribution within 2 standard deviations of the mean is 1, or 100%. The probability of \(A\) being between \(\mu-2\sigma\) and \(\mu+2\sigma\) is also 1, or 100%.

Step by step solution

01

Calculate the Mean (\(\mu\))

The mean or expected value of a random variable is given by: \(\mu = \sum x * P(x)\), where \(x\) is the value of the variable and \(P(x)\) is its probability. For variable \(A\), this would be: \[ \mu = 1*0.6 + 2*0.1 + 3*0.1 + 4*0.1 + 5*0.1 = 1.8 \]
02

Calculate the Variance (\(\sigma^2\))

The variance is the average of the squared differences from the mean. It is computed by: \(\sigma^2 = \sum (x - \mu)^2 * P(x)\). For variable \(A\), this would be: \[\sigma^2 = (1 - 1.8)^2*0.6 + (2 - 1.8)^2*0.1 + (3 - 1.8)^2*0.1 + (4 - 1.8)^2*0.1 + (5 - 1.8)^2*0.1 = 1.56 \]
03

Calculate the Standard Deviation (\(\sigma\))

The standard deviation is simply the square root of the variance: \(\sigma = \sqrt{\sigma^2}\). Therefore, \(\sigma = \sqrt{1.56} = 1.25\)
04

Calculate Probability Within 2 Standard Deviations from the Mean

To find out how much of the probability distribution is within 2 standard deviations of the mean, calculate the probability of \(A\) being between \(\mu-2\sigma\) and \(\mu+2\sigma\). Here, this implies the values of \(A\) being between 1.8-2*1.25 and 1.8+2*1.25, which simplifies to -0.7 and 4.3, respectively. All the given values of \(A\) lie within this range, so the probability is 1 or 100%
05

Determine the Probability that A is Between \(\mu-2\sigma\) and \(\mu+2\sigma\)

We've already computed this in the previous step. The probability of \(A\) falling between these values is 1

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Key Concepts

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

Mean Calculation
Calculating the mean of a probability distribution is fundamentally about finding its expected value. This process involves multiplying each possible outcome of a random variable by its probability, and then summing up all these products.
For example, if you have a random variable, denoted as \( A \), with a set of probabilities linked to its outcomes, you can compute its mean \( \mu \) using the formula: \[ \mu = \sum (x \cdot P(x)) \]
  • Each \( x \) represents an outcome of the variable.
  • Each \( P(x) \) represents the probability of that outcome.

In our exercise, the computation of the mean for \( A \) yielded a mean \( \mu \) of 1.8.
Variance
Variance is a measure of how much the values of a probability distribution are spread out from the mean. It provides an average of the squared differences from the mean. To find variance, you follow these steps:
  • First, calculate the difference between each outcome and the mean.
  • Then, square these differences to remove negative signs and highlight larger discrepancies.
  • Finally, multiply each squared difference by its respective probability and sum these values.

The formula looks like this: \[ \sigma^2 = \sum ((x - \mu)^2 \cdot P(x)) \]
In our example, this calculation showed that the variance of \( A \) was 1.56.
Standard Deviation
Standard deviation provides a practical way to understand the dispersion of a dataset. It is simply the square root of the variance, offering a value in the same units as the random variable, thus making it more interpretable.
By taking the square root of the variance, you eliminate the squared units and can better gauge how much the data tends to differ from the mean on average.
In mathematical terms, the formula is \( \sigma = \sqrt{\sigma^2} \). With respect to our previous calculations, for the variable \( A \), the standard deviation calculated was approximately 1.25.
Random Variable
A random variable is a way to describe uncertain events in mathematics and statistics. In probability theory, it represents outcomes of a probabilistic event in numerical form.
For discrete random variables, like \( A \) in this exercise, they take on a countable number of distinct values. Each value occurs with some probability. These probabilities are part of what we call a probability distribution, which visualizes all possible outcomes and their probabilities.

Random variables help in formulating and solving problems in probability and statistics, providing a base for more advanced concepts like expectation, variance, and standard deviation.

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Most popular questions from this chapter

In another germination experiment involving old seed, 50 rows of seeds were planted. The number of seeds germinating in each row were recorded in the following table (each row contained the same number of seeds). $$\begin{array}{cc|cc}\begin{array}{c}\text { Number } \\\\\text { Germinating }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} & \begin{array}{c}\text { Number } \\\\\text { Germinating }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} \\\\\hline 0 & 17 & 3 & 2 \\\1 & 20 & 4 & 1 \\\2 & 10 & 5 \text { or more } & 0 \\\\\hline\end{array}$$ a. What probability distribution (or function) would be helpful in modeling the variable "number of seeds germinating per row"? Justify your choice. b. What information is needed in order to apply the probability distribution you chose in part a? c. Based on the information you do have, what is the highest or lowest rate of germination that you can estimate for these seeds? Explain.

Above-average hot weather extended over the northwest on August \(3,2009 .\) The day's forecasted high temperatures in four cities in the affected area were: $$\begin{array}{lc} \text { City } & \text { Temperature } \\ \hline \text { Boise, } 1 \mathrm{D} & 100^{\circ} \\ \text { Spokane, WA } & 95^{\circ} \\\ \text { Portland, OR } & 91^{\circ} \\ \text { Helena, } \mathrm{MT} & 91^{\circ} \\ \hline \end{array}$$ a. What is the random variable involved in this study? b. Is the random variable discrete or continuous? Explain.

The probability of success on a single trial of a binomial experiment is known to be \(\frac{1}{4} .\) The random variable \(x\), number of successes, has a mean value of \(80 .\) Find the number of trials involved in this experiment and the standard deviation of \(x.\)

According to United Mileage Plus Visa (November22,2004)\(, 41 \%\) of passengers say they "put on the earphones" to avoid being bothered by their seatmates during flights. To show how important, or not important, the earphones are to people, consider the variable \(x\) to be the number of people in a sample of 12 who say they "put on the earphones" to avoid their seatmates. Assume the\(41 \%\) is true for the whole population of airline travelers and that a random sample is selected. a. Is \(x\) a binomial random variable? Justify your answer. b. Find the probability that \(x=4\) or 5. c. Find the mean and standard deviation of \(x .\) d. Draw a histogram of the distribution of \(x:\) label it completely, highlight the area representing \(x=4\) and \(x=5,\) draw a vertical line at the value of the mean, and mark the location of \(x\) that is 1 standard deviation larger than the mean.

If \(x\) is a binomial random variable, calculate the probability of \(x\) for each case. a. \(\quad n=4, x=1, p=0.3\) b. \(\quad n=3, x=2, p=0.8\) c. \(\quad n=2, x=0, p=\frac{1}{4}\) d. \(\quad n=5, x=2, p=\frac{1}{3}\) e. \(\quad n=4, x=2, p=0.5\) f. \(\quad n=3, x=3, p=\frac{1}{6}\)
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