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Chapter 3: Problem 95

After shuffling a deck of 52 cards, a dealer deals out 5 . Let \(X=\) the number of suits represented in the five-card hand. a. Show that the pmf of \(X\) is $$ \begin{array}{c|cccc} x & 1 & 2 & 3 & 4 \\ \hline p(x) & .002 & .146 & .588 & .264 \end{array} $$ b. Compute \(\mu, \sigma^{2}\), and \(\sigma\).

Short Answer

Expert verified
\(\mu = 3.114\), \(\sigma^2 = 0.405311\), \(\sigma \approx 0.6366\).

Step by step solution

01

Understand the Random Variable Distribution

The problem gives us the probability mass function (pmf) of a random variable \(X\), where \(X\) represents the number of different suits in a five-card hand. The pmf is given as: \(\{ (1, 0.002), (2, 0.146), (3, 0.588), (4, 0.264) \}\). Each pair represents the value \(x\) and its corresponding probability \(p(x)\).
02

Calculate the Expected Value (\(\mu\))

The expected value \( \mu \) of \(X\) is calculated using the formula \( \mu = \sum_{x} x \cdot p(x) \). Substitute each \(x\) and \(p(x)\) into the formula:\[ \mu = 1 \cdot 0.002 + 2 \cdot 0.146 + 3 \cdot 0.588 + 4 \cdot 0.264 \]\[ \mu = 0.002 + 0.292 + 1.764 + 1.056 \]\[ \mu = 3.114 \]
03

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

The variance \( \sigma^2 \) is computed using the formula \( \sigma^2 = \sum_{x}(x - \mu)^2 \cdot p(x) \). First, find \((x - \mu)^2\) for each \(x\), then multiply by \(p(x)\) and sum up:\[ \sigma^2 = (1 - 3.114)^2 \cdot 0.002 + (2 - 3.114)^2 \cdot 0.146 + (3 - 3.114)^2 \cdot 0.588 + (4 - 3.114)^2 \cdot 0.264 \]\[ \sigma^2 = 4.467936 \cdot 0.002 + 1.241296 \cdot 0.146 + 0.013044 \cdot 0.588 + 0.785796 \cdot 0.264 \]\[ \sigma^2 = 0.008936 + 0.181234 + 0.007672 + 0.207469 \]\[ \sigma^2 = 0.405311 \]
04

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

The standard deviation \( \sigma \) is the square root of the variance \( \sigma^2 \). Thus,\[ \sigma = \sqrt{0.405311} \]\[ \sigma \approx 0.6366 \]
05

Conclusion

From the calculations, we find that the expected number of suits \( \mu \) is 3.114, the variance \( \sigma^2 \) is 0.405311, and the standard deviation \( \sigma \) is approximately 0.6366.

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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, denoted as \( \mu \), is a measure of the center or average of a probability distribution. It's a fundamental concept used to predict the long-run average outcome of a random event. In the context of our exercise, the random variable \( X \) represents the number of different suits in a five-card hand. The expected value is calculated using the probability mass function (pmf), where each outcome \( x \) is multiplied by its corresponding probability \( p(x) \), and the sum of these products gives us \( \mu \).
  • This process involves multiplying each value of \( x \) by its probability: \( 1 \cdot 0.002 + 2 \cdot 0.146 + 3 \cdot 0.588 + 4 \cdot 0.264 \).
  • The result comes out to 3.114, indicating that, on average, we expect 3.114 suits to be represented in a five-card hand.
This value gives us a theoretical measure of the number of suits we might expect from a random dealer hand.
Variance
Variance, denoted as \( \sigma^2 \), shows how much the values of the random variable deviate from the expected value, effectively capturing the variability or spread in the distribution. Calculating variance involves taking the squared difference between each possible value \( x \) and the expected value \( \mu \), then weighting it by the probability \( p(x) \), and summing all these values. This statistical measure gives us insight into the consistency or unpredictability of a random variable.
  • For this exercise, \( (x - 3.114)^2 \) is computed for each \( x \) in our data, multiplying by its probability and summing: \( 0.405311 \).
  • The variance of 0.405311 indicates that there is a mild spread around our expected value of 3.114 suits.
In simple terms, variance helps us understand the risk or uncertainty inherent in selecting a random five-card hand from the deck.
Standard Deviation
The standard deviation, denoted \( \sigma \), is another measure of variability and represents the average distance between each data point and the mean. It is the square root of variance, making it a more intuitive measure as it is in the same units as the data itself. For our card hand scenario, the standard deviation indicates the degree of fluctuation we might expect in the number of suits in a hand.
  • Taking the square root of our variance (0.405311), we find \( \sigma \approx 0.6366 \).
  • This standard deviation tells us that while 3.114 suits is the average, the actual value may often vary by about 0.64 suits.
This value provides an easily interpretable sense of how much change or variation we can expect when considering different hands dealt from the deck.
Random Variable
A random variable is a variable that assigns numerical values to different outcomes of a random phenomenon. In this exercise, \( X \), the random variable, represents the number of different suits in a five-card hand. Random variables are absolutely crucial in probability and statistics as they encapsulate uncertain outcomes in quantitative terms.
  • Discrete random variables, like \( X \), have a countable number of possible outcomes. Here, \( X \) can be 1, 2, 3, or 4 suits as per the pmf provided.
  • Understanding \( X \) helps us frame the probability problem and calculate the meaningful statistics, like expected value and variance, to make predictions.
Thus, random variables form the basis for all statistical inference and probability calculations, essential for making sense of uncertainty and randomness in various scenarios.

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Most popular questions from this chapter

Twenty pairs of individuals playing in a bridge tournament have been seeded \(1, \ldots, 20\). In the first part of the tournament, the 20 are randomly divided into 10 east-west pairs and 10 north-south pairs. a. What is the probability that \(x\) of the top 10 pairs end up playing east- west? b. What is the probability that all of the top five pairs end up playing the same direction? c. If there are \(2 n\) pairs, what is the pmf of \(X=\) the number among the top \(n\) pairs who end up playing east-west? What are \(E(X)\) and \(V(X)\) ?

A mail-order computer business has six telephone lines. Let \(X\) denote the number of lines in use at a specified time. Suppose the pmf of \(X\) is as given in the accompanying table. \begin{tabular}{c|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} Calculate the probability of each of the following events. a. \(\\{\) at most three lines are in use \(\\}\) b. \\{fewer than three lines are in use \(\\}\) c. \\{at least three lines are in use \(\\}\) d. \\{between two and five lines, inclusive, are in use \\} e. \\{between two and four lines, inclusive, are not in use \(\\}\) f. \\{at least four lines are not in use \(\\}\)

Grasshoppers are distributed at random in a large field according to a Poisson distribution with parameter \(\alpha=2\) per square yard. How large should the radius \(R\) of a circular sampling region be taken so that the probability of finding at least one in the region equals \(.99\) ?

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number \(X\) has a Poisson distribution with parameter \(\lambda=.2\). (Suggested in "Average Sample Number for Semi-Curtailed Sampling Using the Poisson Distribution," J. Quality Technology, 1983: 126-129.) a. What is the probability that a disk has exactly one missing pulse? b. What is the probability that a disk has at least two missing pulses? c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Consider a deck consisting of seven cards, marked \(1,2, \ldots\), 7. Three of these cards are selected at random. Define an rv \(W\) by \(W=\) the sum of the resulting numbers, and compute the pmf of \(W\). Then compute \(\mu\) and \(\sigma^{2}\). [Hint: Consider outcomes as unordered, so that \((1,3,7)\) and \((3,1,7)\) are not different outcomes. Then there are 35 outcomes, and they can be listed. (This type of rv actually arises in connection with a hypothesis test called Wilcoxon's rank-sum test, in which there is an \(x\) sample and a \(y\) sample and \(W\) is the sum of the ranks of the \(x\) 's in the combined sample.)]
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