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The \(n\) candidates for a job have been ranked 1,2 , \(3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)=\left\\{\begin{array}{cc} 1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise } \end{array}\right. $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6\).]

Short Answer

Expert verified
\(E(X) = \frac{n+1}{2}\), \(V(X) = \frac{(n^2-1)}{12}\).

Step by step solution

01

Understanding the Distribution

The given problem mentions that the rank of a randomly selected candidate, denoted by \(X\), follows a discrete uniform distribution. This means that each candidate's rank from 1 to \(n\) is equally likely, and the probability mass function (pmf) is \(p(x) = \frac{1}{n}\) for \(x = 1, 2, \ldots, n\).
02

Calculating the Expected Value

To find the expected value \(E(X)\) for a discrete uniform distribution over \(n\) ranks, use the formula: \[ E(X) = \frac{n + 1}{2} \]. This formula is derived based on the uniform distribution properties and accounts for the equal probability of each rank.
03

Applying the Expected Value Formula

Substitute \(n\) into the formula: \(E(X) = \frac{n + 1}{2}\). For the discrete uniform distribution, this gives the mean rank. This is consistent with the arithmetic mean for integers from 1 to \(n\).
04

Calculating the Variance

For a discrete uniform distribution, the variance \(V(X)\) is given by the formula: \[ V(X) = \frac{(n^2 - 1)}{12} \]. This formula accounts for the spread of the ranks around the mean. It is derived from the basic variance formula considering the properties of uniform distribution.
05

Applying the Variance Formula

Plug \(n\) into the variance formula: \(V(X) = \frac{(n^2 - 1)}{12}\). This tells us about the variability or spread of the ranks for a random selection. The formula is simplified using the formula for the sum of squares of integers, which is provided in the hint.

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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 \(E(X)\), is a fundamental concept in probability theory that gives us the mean or average of a random variable. In the context of the discrete uniform distribution, the expected value can be thought of as the "center" of all possible outcomes. For a distribution where the outcomes are the ranks of job candidates, each rank from 1 to \(n\) has an equal probability of being selected.

The formula for calculating the expected value over a discrete uniform distribution is \(E(X) = \frac{n + 1}{2}\). This formula reflects the arithmetic mean of the integers from 1 to \(n\). For example, for 5 candidates ranked from 1 to 5, the expected rank of a randomly selected candidate would be \(\frac{5 + 1}{2} = 3\). This shows that on average, the middle rank is expected.

Understanding the expected value helps us predict what rank is likely to be chosen most frequently if we were to repeatedly select candidates at random.
Probability Mass Function
The probability mass function (pmf) is a key concept that defines the probability of a discrete random variable taking on each of its possible values. In the case of the discrete uniform distribution described here, each potential rank for a job candidate is equally probable. This is reflected in the pmf given by \(p(x) = \frac{1}{n}\) for each rank \(x = 1, 2, ..., n\). The pmf informs us about how the probabilities are distributed among different possible outcomes.

For a set of \(n\) job candidates, each rank is not more or less likely than any other, meaning every possible rank value is treated uniformly. This characteristic sets the discrete uniform distribution apart from other types of distributions where probability values differ.

If a candidate were to be chosen at random from a pool of 3 candidates, each rank (1, 2, or 3) would have a probability of \(\frac{1}{3}\). Such equal probability assures that there is no favoritism or bias toward any rank in the selection.
Variance
Variance, denoted as \(V(X)\), provides insight into the variability or spread of a set of values around their mean. Within the framework of the discrete uniform distribution in this scenario, variance quantifies the degree of dispersion in candidate ranks.

The formula for computing the variance of a discrete uniform distribution is given by \(V(X) = \frac{(n^2 - 1)}{12}\). This formula evaluates how far ranks deviate around the expected mean \(E(X)\). The variance takes into account all possible ranks, assessing the overall consistency from the central rank.

A practical way to understand variance is by considering it as a measure of risk or unpredictability. For instance, in a scenario with 4 candidates, plugging into the formula yields \(V(X) = \frac{(16 - 1)}{12} = \frac{15}{12} = 1.25\). This result tells us the degree to which candidates' ranks fluctuate around the expected rank.

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Most popular questions from this chapter

Each of 12 refrigerators has been retumed to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let \(X\) be the number among the first 6 examined that have a defective compressor. Compute the following: a. \(P(X=5)\) b. \(P(X \leq 4)\) c. The probability that \(X\) exceeds its mean yalue by more than 1 standard deviation. d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If \(X\) is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) \(P(X \leq 5)\) than to use the hypergeometric pmf.

A trial has just resulted in a hung jury because eight members of the jury were in favor of a guilty verdict and the other four were for acquittal. If the jurors leave the jury room in random onder and each of the first four leaving the room is accosted by a reporter in quest of an interview, what is the pmf of \(X=\) the number of jurors favoring acquittal among those interviewed? How many of those favoring acquittal do you expect to be interviewed?

Suppose that the number of plants of a particular type found in a rectangular region (called a quadrat by ecologists) in a certain geographic area is an rv \(X\) with pmf $$ p(x)=\left\\{\begin{array}{cc} c / x^{3} & x=1,2,3, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ Is \(E(X)\) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed).

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a chemical, which it sells to customers in 5-lb containers. Let \(X=\) the number of containers ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

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