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Chapter 4: Problem 84

Let \(x\) equal the number observed on the throw of a single balanced die. a. Find and graph the probability distribution for \(x\). b. What is the average or expected value of \(x ?\) c. What is the standard deviation of \(x ?\) d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the graph in part a. What proportion of all the measurements would fall into this range?

Short Answer

Expert verified
Answer: The probability distribution of rolling a balanced die is uniform, with a probability of \(\frac{1}{6}\) for each outcome (1, 2, 3, 4, 5, and 6). The interval \(\mu \pm 2\sigma\) is approximately from -0.24 to 7.24, which covers the entire range of possible outcomes. Therefore, the proportion of measurements that fall within the interval \(\mu \pm 2\sigma\) is 100%.

Step by step solution

01

Part a: Finding and graphing the probability distribution

A balanced die has six faces, each with an equal probability of showing up when the die is rolled. The possible outcomes are the numbers 1 through 6, and the probability of each is \(\frac{1}{6}\). Thus, the probability distribution for \(x\) is given by the following table: | x | 1 | 2 | 3 | 4 | 5 | 6 | |-----|-----|-----|-----|-----|-----|-----| | P(x)| 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | To graph this distribution, plot the probabilities on the vertical axis against the outcomes on the horizontal axis, and use bars to represent the probabilities for each outcome.
02

Part b: Finding the expected value or average of x

The expected value (average) of a discrete random variable is given by the formula: $$E(x) = \sum_{i} x_i P(x_i)$$ For our given probability distribution, the expected value is: $$E(x) = 1\cdot\frac{1}{6} + 2\cdot\frac{1}{6} + 3\cdot\frac{1}{6} + 4\cdot\frac{1}{6} + 5\cdot\frac{1}{6} + 6\cdot\frac{1}{6} = \frac{21}{6} = 3.5$$
03

Part c: Finding the standard deviation

To compute the standard deviation, we first need to calculate the variance of the distribution. The variance is given by: $$\sigma^2 = \sum_{i} (x_i - \mu)^2 P(x_i)$$ For our given probability distribution, the variance is: $$\sigma^2 = (1-3.5)^2\cdot\frac{1}{6} + (2-3.5)^2\cdot\frac{1}{6}+ (3-3.5)^2\cdot\frac{1}{6} + (4-3.5)^2\cdot\frac{1}{6} + (5-3.5)^2\cdot\frac{1}{6} + (6-3.5)^2\cdot\frac{1}{6} =\frac{17.5}{6}$$ Now we can find the standard deviation by taking the square root of the variance: $$\sigma = \sqrt{\frac{17.5}{6}} \approx 1.87$$
04

Part d: Locating the interval \(\mu \pm 2\sigma\) and calculating the proportion

The interval \(\mu \pm 2\sigma\) is found by adding and subtracting two times the standard deviation from the mean: $$\mu - 2\sigma= 3.5 - 2\cdot1.87 = 3.5 - 3.74 = -0.24$$ $$\mu + 2\sigma= 3.5 + 2\cdot1.87 = 3.5 + 3.74 = 7.24$$ Since the x-values are the integers from 1 to 6, the interval falls entirely within the range of possible outcomes. Therefore, every outcome falls into this range with a proportion of 100%. As a result, we can conclude that 100% of the measurements fall into the interval \(\mu \pm 2\sigma\).

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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 is a key concept in probability theory, which provides the average outcome or mean of a random variable over numerous trials. For a discrete random variable like the roll of a die, the expected value is calculated by multiplying each possible outcome by its probability and summing all these products. In this specific exercise of rolling a single die, the outcomes are 1 through 6, with equal probability for each outcome of \( \frac{1}{6} \).

The formula for expected value, \( E(x) = \sum_{i} x_i P(x_i) \), simplifies to:
  • \( E(x) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \)
  • Resulting in \( E(x) = 3.5 \). This means that over a large number of rolls, the average value is expected to converge to 3.5.
Standard Deviation
Standard deviation is a measure that describes how spread out the values of a random variable are from the mean. It helps us understand variability in the data. To find the standard deviation, we first need to calculate the variance, which is found by averaging the squared differences between each outcome and the expected value.

The variance \( \sigma^2 \) is given by:
  • \( \sigma^2 = \sum_{i} (x_i - \mu)^2 P(x_i) \)
  • In our case, it becomes \( \sigma^2 = (1-3.5)^2 \cdot \frac{1}{6} + (2-3.5)^2 \cdot \frac{1}{6} + (3-3.5)^2 \cdot \frac{1}{6} + (4-3.5)^2 \cdot \frac{1}{6} + (5-3.5)^2 \cdot \frac{1}{6} + (6-3.5)^2 \cdot \frac{1}{6} = \frac{17.5}{6} \)
After finding variance, take the square root to get the standard deviation:
  • \( \sigma = \sqrt{\frac{17.5}{6}} \approx 1.87 \)
  • This value indicates the degree of variation or dispersion of die rolls around the mean of 3.5.
Variance
Variance is an essential statistic that indicates the degree to which each number in a set differs from the mean. It helps measure the randomness or reliability of outcomes in a probability distribution.

To compute variance for the experiment with a die:
  • First, subtract the expected value from each possible outcome.
  • Then, square these differences.
  • Finally, compute the average of these squared differences, weighted by the probability of each outcome.
  • Thus, \( \sigma^2 = (1-3.5)^2 \cdot \frac{1}{6} + (2-3.5)^2 \cdot \frac{1}{6} + \ldots + (6-3.5)^2 \cdot \frac{1}{6} = \frac{17.5}{6} \).
Variance shows how much the die roll outcomes differ from the mean and forms the foundation for computing standard deviation.
Discrete Random Variable
A discrete random variable is a type of random variable that takes on a finite or countable number of distinct values. In contrast to continuous random variables, which can take any value within a range, discrete random variables are associated with distinct, separate items.

In the exercise, the variable \( x \) representing the number rolled on a die is a perfect example of a discrete random variable. This is because:
  • The die can land on one of only six outcomes: 1, 2, 3, 4, 5, or 6.
  • Each outcome occurs with a probability of \( \frac{1}{6} \).
  • Its probability distribution is finite, listing all possible values and their associated probabilities.
Understanding discrete random variables is crucial as they form the backbone of many probability models and statistical analyses.

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Most popular questions from this chapter

Probability played a role in the rigging of the April \(24,1980,\) Pennsylvania state lottery. To determine each digit of the three-digit winning number, each of the numbers \(0,1,2, \ldots, 9\) is written on a Ping-Pong ball, the 10 balls are blown into a compartment, and the number selected for the digit is the one on the ball that floats to the top of the machine. To alter the odds, the conspirators injected a liquid into all balls used in the game except those numbered 4 and \(6,\) making it almost certain that the lighter balls would be selected and determine the digits in the winning number. They then proceeded to buy lottery tickets bearing the potential winning numbers. How many potential winning numbers were there (666 was the eventual winner)?

A retailer sells two styles of highpriced digital video recorders (DVR) that experience indicates are in equal demand. (Fifty percent of all potential customers prefer style \(1,\) and \(50 \%\) favor style \(2 .\) ) If the retailer stocks four of each, what is the probability that the first four customers seeking a DVR all purchase the same style?

Two cold tablets are unintentionally placed in a box containing two aspirin tablets. The four tablets are identical in appearance. One tablet is selected at random from the box and is swallowed by the first patient. A tablet is then selected at random from the three remaining tablets and is swallowed by the second patient. Define the following events as specific collections of simple events: a. The sample space \(S\) b. The event \(A\) that the first patient obtained a cold tablet c. The event \(B\) that exactly one of the two patients obtained a cold tablet d. The event \(C\) that neither patient obtained a cold tablet

The following information reflects the results of a survey reported by Mya Frazier in an \(A d\) Age Insights white paper. \({ }^{11}\) Working spouses were asked "Who is the household breadwinner?" Suppose that one person is selected at random from these 200 individuals. $$\begin{array}{lcccc} & \multicolumn{3}{c} {\text { Spouse or }} \\\& \text { You } & \text { Significant Other } & \text { About Equal } & \text { Totals } \\\\\hline \text { Men } & 64 & 16 & 20 & 100 \\\\\text { Women } & 32 & 45 & 23 & 100 \\\\\hline \text { Totals } & 96 & 61 & 43 & 200\end{array}$$ a. What is the probability that this person will identify his/herself as the household breadwinner? b. What is the probability that the person selected will be a man who indicates that he and his spouse/significant other are equal breadwinners? c. If the person selected indicates that the spouse or significant other is the breadwinner, what is the probability that the person is a man?

Suppose \(P(A)=.1\) and \(P(B)=.5\). a. If \(P(A \mid B)=.1,\) what is \(P(A \cap B) ?\) b. If \(P(A \mid B)=.1,\) are \(A\) and \(B\) independent? c. If \(P(A \cap B)=0,\) are \(A\) and \(B\) independent? d. If \(P(A \cup B)=.65,\) are \(A\) and \(B\) mutually exclusive?
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