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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

A doctor knows from experience that \(10 \%\) of the patients to whom she gives a certain drug will have undesirable side effects. Find the probabilities that among the 10 patients to whom she gives the drug: a. At most two will have undesirable side effects. b. At least two will have undesirable side effects.

What does it mean for the trials to be independent in a binomial experiment?

Four cards are selected, one at a time, from a standard deck of 52 playing cards. Let \(x\) represent the number of aces drawn in the set of four cards. a. If this experiment is completed without replacement, explain why \(x\) is not a binomial random variable. b. If this experiment is completed with replacement, explain why \(x\) is a binomial random variable.

a. Explain how the various values of \(x\) in a probability distribution form a set of mutually exclusive events. b. Explain how the various values of \(x\) in a probability distribution form a set of "all-inclusive" events.

Bill has completed a 10-question multiple-choice test on which he answered 7 questions correctly. Each question had one correct answer to be chosen from five alternatives. Bill says that he answered the test by randomly guessing the answers without reading the questions or answers. a. Define the random variable \(x\) to be the number of correct answers on this test, and construct the probability distribution if the answers were obtained by random guessing. b. What is the probability that Bill guessed 7 of the 10 answers correctly? c. What is the probability that anybody can guess six or more answers correctly? d. Do you believe that Bill actually randomly guessed as he claims? Explain.
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