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Consider two discrete probability distribution with the same sample space and the same expected value. Are the standard deviations of the two distributions necessarily equal? Explain.

Short Answer

Expert verified
No, they are not necessarily equal. The same expected value does not imply equal dispersion or standard deviation.

Step by step solution

01

Understanding Expected Value

The expected value of a probability distribution is a measure of the center of the distribution, calculated as the weighted average of all possible outcomes. Both distributions having the same expected value means that their central tendency is the same.
02

Defining Standard Deviation

The standard deviation measures the spread or dispersion of a distribution around its expected value. It is calculated as the square root of the variance, which is the average of the squared deviations from the expected value.
03

Link Between Expected Value and Standard Deviation

While the expected value indicates where the center of the distribution lies, the standard deviation provides insights into how the values are distributed around this center. Two distributions can have the same expected value but different spreads around this center.
04

Conclusion: Expected Value vs. Standard Deviation

Since the expected value does not account for how data is spread around this central value, two distributions with the same expected value can indeed have different standard deviations. Variations in spread (dispersion) are independent of the central value.

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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 denoted as \( E(X) \), of a probability distribution is essentially a weighted average of all possible outcomes. It provides us with a central measure, giving us an idea of what the average result would be if we could repeat the random experiment infinite times.
This is calculated by multiplying each possible outcome by its probability and then summing all these products. It's like figuring out a destination, even if the routes there vary. In our context, two distributions may have the same expected value. Picture it as both paths leading to the same endpoint, even if they take different twists and turns to get there.
Understanding the expected value helps us anticipate the average outcome, but it doesn't tell us about the variability or possible deviation in outcomes from this central point.
Standard Deviation
While the expected value gives us a point to focus on, the standard deviation paints a richer picture by illustrating how far the values lie from this average point. It's a measure of dispersion in a probability distribution. Calculated as the square root of the variance, the standard deviation is expressed in the same units as the data, making it very intuitive to interpret.
When analyzing distributions, a smaller standard deviation indicates that the data points cluster tightly around the mean, while a larger one suggests that the data is more spread out. Even if two distributions share the same expected value, they may have vastly different standard deviations. Imagine two archers aiming for the center of a target: one archer may consistently hit closer to the bullseye (lower standard deviation), while the other might have a broader spread of arrows (higher standard deviation), though both have the same average score.
Variance
Variance is an essential concept underpinning standard deviation and offers insights into the distribution's variability. Given by the symbol \( \sigma^2 \), it measures how much the values in the distribution spread out from the expected value (or mean).
Calculating variance involves determining the average squared deviation from the mean. This squaring step is key because it amplifies larger deviations, providing a more pronounced picture of the distribution's dispersion.
While variance itself is not expressed in the original data units due to the squaring, it remains a fundamental step in calculating the standard deviation. In context, two distributions can have identical expected values and still have differing variances, highlighting disparities in how the data is spread. Variance helps identify these differences, even when the central tendencies remain aligned.

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Most popular questions from this chapter

Consider a binomial experiment with \(n=7\) trials where the probability of success on a single trial is \(p=0.60\) (a) Find \(P(r=7)\) (b) Find \(P(r \leq 6)\) by using the complement rule.

The Denver Post reported that a recent audit of Los Angeles 911 calls showed that \(85 \%\) were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls. (a) What is the probability that all four calls are, in fact, emergencies? (b) What is the probability that three or more calls are not emergencies? (c) How many calls \(n\) would the 911 operators need to answer to be \(96 \%\) (or more) sure that at least one call is, in fact, an emergency?

Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies' Liesi't Lies\%' The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

Insurance Risk Insurance companies know the risk of insurance is greatly reduced if the company insures not just one person, but many people. How does this work? Let \(x\) be a random variable representing the expectation of life in years for a 25 -year-old male (i.e., number of years until death). Then the mean and standard deviation of \(x\) are \(\mu=50.2\) years and \(\sigma=11.5\) years (Vital Statistics Section of the Statistical Abstract of the United States, 116th edition). Suppose Big Rock Insurance Company has sold life insurance policies to Joel and David. Both are 25 years old, unrelated, live in different states, and have about the same health record. Let \(x_{1}\) and \(x_{2}\) be random variables representing Joel's and David's life expectancies. It is reasonable to assume \(x_{1}\) and \(x_{2}\) are independent. $$\begin{aligned} &\text { Joel, } x_{1}: \mu_{1}=50.2 ; \sigma_{1}=11.5\\\ &\text { David, } x_{2}: \mu_{2}=50.2 ; \sigma_{2}=11.5 \end{aligned}$$ If life expectancy can be predicted with more accuracy, Big Rock will have less risk in its insurance business. Risk in this case is measured by \(\sigma\) (larger \(\sigma\) means more risk). (a) The average life expectancy for Joel and David is \(W=0.5 x_{1}+0.5 x_{2}\) Compute the mean, variance, and standard deviation of \(W\). (b) Compare the mean life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W).\) (c) Compare the standard deviation of the life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W).\) (d) The mean life expectancy is the same for a single policy \(\left(x_{1}\right)\) as it is for two policies \((W),\) but the standard deviation is smaller for two policies. What happens to the mean life expectancy and the standard deviation when we include more policies issued to people whose life expectancies have the same mean and standard deviation (i.e., 25 -year-old males)? For instance, for three policies, \(W=(\mu+\mu+\mu) / 3=\mu\) and \(\sigma_{W}^{2}=(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}=\) \((1 / 3)^{2}\left(3 \sigma^{2}\right)=(1 / 3) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{3}} \sigma .\) Likewise, for \(n\) such policies, \(W=\mu\) and \(\sigma_{W}^{2}=(1 / n) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{n}} \sigma .\) Looking at the general result, is it appropriate to say that when we increase the number of policies to \(n\), the risk decreases by a factor of \(\sigma_{W}=\frac{1}{\sqrt{n}} ?\)

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