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91Ó°ÊÓ

If a finite random variable has an expected value of 10 and a standard deviation of 0, what must its probability distribution be?

Short Answer

Expert verified
The probability distribution of this finite random variable is such that it only has one possible value, 10, with an associated probability of 1, since the standard deviation is 0 and there is no spread or dispersion in the values. Therefore, \(P(X = 10) = 1\).

Step by step solution

01

Understand the properties of expected value and standard deviation

The expected value (mean) of a random variable is the weighted average of its possible values, with the probabilities as weights. The standard deviation measures the dispersion or spread of the values around this mean. Given that the standard deviation is 0, this indicates that all the values of the random variable are equal to the mean since there is no dispersion. In this case, the mean is 10.
02

Establish the probability distribution

Since the standard deviation is 0, it means that the random variable takes only one possible value, which is the expected value (mean), in this case, 10. The associated probability for this single outcome will be 1, as it is the only possible outcome. Therefore, the probability distribution for this finite random variable can be described as: \(P(X = 10) = 1\)
03

Conclusion

The probability distribution of this finite random variable is such that it only has one possible value, 10, with an associated probability of 1. The standard deviation of 0 confirms that there is no spread or dispersion in the values of this random variable.

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Key Concepts

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

Expected Value
The expected value, often referred to as the mean, is a foundational concept in probability and statistics. It represents the average outcome you would anticipate if you could repeat an experiment or process an infinite number of times. Imagine rolling a die. Each face has an equal probability of 1/6, so the expected value of this die roll is the average of all possible outcomes, taking into account these probabilities.
For a discrete random variable, the expected value is found using the formula:
  • \( E(X) = \sum{[x_i \times P(x_i)]} \)
This means you multiply each possible value \( x_i \) by its probability \( P(x_i) \), and then sum those products.
In the context of our original exercise, the random variable has an expected value of 10. This signifies that, as a weighted average of its outcomes, the mean or expected long-term value when considering all possible outcomes and their probabilities is 10.
Standard Deviation
Standard deviation is a statistical measure that illustrates the spread or variability of a set of values. A low standard deviation means that values tend to be close to the mean, while a high standard deviation indicates that they are spread out over a wider range.
For a finite random variable, you'll calculate the standard deviation by first determining the variance, which is the expected value of the squared deviations from the mean:
  • \( \text{Variance} = E[(X - E(X))^2] = \sum{[(x_i - E(X))^2 \times P(x_i)]} \)
  • \( \text{Standard Deviation} = \sqrt{\text{Variance}} \)
In the exercise presented, the standard deviation is 0. This implies there is absolutely no variability; all outcomes are the same as the expected value. Therefore, if you calculate the variance, it results in zero—showing no dispersion of the values around the mean.
Finite Random Variable
A finite random variable is one that has a limited set of distinct outcomes it can take. Since real-world scenarios often have finite outcomes, understanding this concept is crucial.
Such a random variable could represent, for example, the result of drawing a card from a deck, the number of heads in a series of coin flips, or the specific case in our exercise that results in only one value, 10.
  • The probability distribution for this kind of random variable is a complete list of possible outcomes along with their likelihoods.
  • For instance, if only one outcome consistently occurs, as in the exercise where \( P(X = 10) = 1 \), it indicates no other values occur.
With a standard deviation of 0, the finite random variable can only equal the expected value, as all probabilities consolidate into a single, certain outcome. This ensures all realizations of the random variable equate to that single specific outcome.

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Most popular questions from this chapter

The Acme Insurance Company is launching a drive to generate greater profits, and it decides to insure racetrack drivers against wrecking their cars. The company's research shows that, on average, a racetrack driver races four times a year and has a 1 in 10 chance of wrecking a vehicle, worth an average of \(\$ 100,000\), in every race. The annual premium is \(\$ 5,000\), and Acme automatically drops any driver who is involved in an accident (after paying for a new car), but does not refund the premium. How much profit (or loss) can the company expect to earn from a typical driver in a year? HINT [Use a tree diagram to compute the probabilities of the various outcomes.]

The Blue Sky Flight Insurance Company insures passengers against air disasters, charging a prospective passenger \(\$ 20\) for coverage on a single plane ride. In the event of a fatal air disaster, it pays out \(\$ 100,000\) to the named beneficiary. In the event of a nonfatal disaster, it pays out an average of \(\$ 25,000\) for hospital expenses. Given that the probability of a plane's crashing on a single trip is \(.00000087,{ }^{32}\) and that a passenger involved in a plane crash has a \(.9\) chance of being killed, determine the profit (or loss) per passenger that the insurance company expects to make on each trip. HINT [Use a tree to compute the probabilities of the various outcomes.]

Must the expected number of times you hit a bull's-eye after 50 attempts always be a whole number? Explain.

I \(\vee\) Supermarkets A survey of supermarkets in the United States yielded the following relative frequency table, where \(X\) is the number of checkout lanes at a randomly chosen supermarket: \(^{49}\) \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \(\boldsymbol{x}\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \(\boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x})\) & \(.01\) & \(.04\) & \(.04\) & \(.08\) & \(.10\) & \(.15\) & \(.25\) & \(.20\) & \(.08\) & \(.05\) \\ \hline \end{tabular} a. Compute the mean, variance, and standard deviation (accurate to one decimal place). b. As financial planning manager at Express Lane Mart, you wish to install a number of checkout lanes that is in the range of at least \(75 \%\) of all supermarkets. What is this range according to Chebyshev's inequality? What is the least number of checkout lanes you should install so as to fall within this range?

Based on data from Nielsen Research, there is a \(15 \%\) chance that any television that is turned on during the time of the evening newscasts will be tuned to \(\mathrm{ABC}\) 's evening news show. \({ }^{58}\) Your company wishes to advertise on a small local station carrying \(\mathrm{ABC}\) that serves a community with 2,500 households that regularly tune in during this time slot. Find the approximate probability that at least 400 households will be tuned in to the show. HINT [See Example 4.]

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