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Chapter 3: Problem 29

The pmf for \(X=\) the number of major defects on a randomly selected appliance of a certain type is \begin{tabular}{l|ccccc} \(x\) & 0 & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.08\) & \(.15\) & \(.45\) & \(.27\) & \(.05\) \end{tabular} Compute the following: a. \(E(X)\) b. \(V(X)\) directly from the definition c. The standard deviation of \(X\) d. \(V(X)\) using the shortcut formula

Short Answer

Expert verified
a. 2.06, b. 0.9364, c. 0.9676, d. 0.9364

Step by step solution

01

Calculate E(X)

To find the expected value \(E(X)\), multiply each possible value of \(X\) by its probability and sum the results:\[ E(X) = \sum_{i=0}^{4} x_i \cdot p(x_i) = 0 \cdot 0.08 + 1 \cdot 0.15 + 2 \cdot 0.45 + 3 \cdot 0.27 + 4 \cdot 0.05 \]Calculate each term:- \(0 \cdot 0.08 = 0\)- \(1 \cdot 0.15 = 0.15\)- \(2 \cdot 0.45 = 0.90\)- \(3 \cdot 0.27 = 0.81\)- \(4 \cdot 0.05 = 0.20\)Sum them up to calculate \(E(X)\):\[ E(X) = 0 + 0.15 + 0.90 + 0.81 + 0.20 = 2.06 \]
02

Calculate V(X) from Definition

The variance \(V(X)\) is the expected value of \((X - E(X))^2\):\[ V(X) = \sum_{i=0}^{4} (x_i - 2.06)^2 \cdot p(x_i) \]Calculate each term:- \((0 - 2.06)^2 \cdot 0.08 = 4.2436 \cdot 0.08 = 0.3395\)- \((1 - 2.06)^2 \cdot 0.15 = 1.1236 \cdot 0.15 = 0.1685\)- \((2 - 2.06)^2 \cdot 0.45 = 0.0036 \cdot 0.45 = 0.0016\)- \((3 - 2.06)^2 \cdot 0.27 = 0.8836 \cdot 0.27 = 0.2386\)- \((4 - 2.06)^2 \cdot 0.05 = 3.7636 \cdot 0.05 = 0.1882\)Sum them to get \(V(X)\):\[ V(X) = 0.3395 + 0.1685 + 0.0016 + 0.2386 + 0.1882 = 0.9364 \]
03

Calculate Standard Deviation of X

The standard deviation \(\sigma_X\) is the square root of the variance:\[ \sigma_X = \sqrt{V(X)} = \sqrt{0.9364} \approx 0.9676 \]
04

Calculate V(X) Using the Shortcut Formula

The shortcut formula for variance is:\[ V(X) = E(X^2) - [E(X)]^2 \]First, calculate \(E(X^2)\):\[ E(X^2) = \sum_{i=0}^{4} x_i^2 \cdot p(x_i) = 0^2 \cdot 0.08 + 1^2 \cdot 0.15 + 2^2 \cdot 0.45 + 3^2 \cdot 0.27 + 4^2 \cdot 0.05 \]Calculate each term:- \(0^2 \cdot 0.08 = 0\)- \(1^2 \cdot 0.15 = 0.15\)- \(2^2 \cdot 0.45 = 1.80\)- \(3^2 \cdot 0.27 = 2.43\)- \(4^2 \cdot 0.05 = 0.80\)Sum these to find \(E(X^2)\):\[ E(X^2) = 0 + 0.15 + 1.80 + 2.43 + 0.80 = 5.18 \]Now use the shortcut formula:\[ V(X) = 5.18 - (2.06)^2 = 5.18 - 4.2436 = 0.9364 \]

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Key Concepts

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

Expected Value
When working with probability distributions, one key concept you will encounter is the expected value, often denoted as \(E(X)\). The expected value is a way to describe the center of a probability distribution and can be thought of as the "average" value of the random variable over many trials. To find the expected value, you multiply each possible value of your random variable \(X\) by its respective probability \(p(x)\), then sum these products together.
  • The formula for expected value is: \[ E(X) = \sum x_i \cdot p(x_i) \]
  • Example calculation: \(E(X) = 0 \cdot 0.08 + 1 \cdot 0.15 + 2 \cdot 0.45 + 3 \cdot 0.27 + 4 \cdot 0.05 = 2.06\)
In simpler terms, expected value indicates what you would expect on average if you were to repeat the experiment an infinite number of times under the same conditions.
Variance
The variance, denoted as \(V(X)\), measures the spread or dispersion of a probability distribution. It tells us how much the values of the random variable deviate from the expected value. A larger variance means that the data is more spread out.
  • Variance is calculated as \(V(X) = \sum (x_i - E(X))^2 \cdot p(x_i)\)
  • Variance from the example: \(V(X) = 0.3395 + 0.1685 + 0.0016 + 0.2386 + 0.1882 = 0.9364\)
A practical way to interpret variance is to think of it as the "average" squared deviation from the expected value. When you compute variance, less deviation means values are close to the mean, while larger deviations suggest the data is more variable.
Standard Deviation
Standard deviation, denoted by \(\sigma_X\), is a helpful statistic because it brings variance back into the realm of the original data units, making it easier to interpret by taking the square root of the variance. Essentially, standard deviation measures the average distance of each data point from the mean.
  • To calculate standard deviation, use: \( \sigma_X = \sqrt{V(X)} \)
  • Example calculation: \( \sigma_X = \sqrt{0.9364} \approx 0.9676\)
Standard deviation provides a clear measure of the spread of the data in the context of the original variable. A lower standard deviation indicates the data points are typically closer to the mean, while a higher value indicates more spread.
Probability Mass Function
In the context of discrete random variables like the number of defects, the probability mass function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. The PMF is vital as it summarizes how the probabilities are distributed over the different possibilities of the random variable.
  • The PMF for a discrete random variable \(X\) is given as \(p(x)\) for each \(x\).
  • For instance, in our example, some PMF values are \(p(0) = 0.08\), \(p(2) = 0.45\).
To understand a PMF, remember that:- The sum of all probabilities is equal to 1, because it accounts for all possible outcomes.- It helps in calculative methods, such as determining expected values and variances, by providing a foundational framework of probablility distributions.

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Most popular questions from this chapter

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is $$ \begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline p(y) & .60 & .25 & .10 & .05 \end{array} $$ a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

An educational consulting firm is trying to decide whether high school students who have never before used a handheld calculator can solve a certain type of problem more easily with a calculator that uses reverse Polish logic or one that does not use this logic. A sample of 25 students is selected and allowed to practice on both calculators. Then each student is asked to work one problem on the reverse Polish calculator and a similar problem on the other. Let \(p=P(S)\), where \(S\) indicates that a student worked the problem more quickly using reverse Polish logic than without, and let \(X=\) number of \(S\) 's. a. If \(p=.5\), what is \(P(7 \leq X \leq 18)\) ? b. If \(p=.8\), what is \(P(7 \leq X \leq 18)\) ? c. If the claim that \(p=.5\) is to be rejected when either \(X \leq\) 7 or \(X \geq 18\), what is the probability of rejecting the claim when it is actually correct? d. If the decision to reject the claim \(p=.5\) is made as in part (c), what is the probability that the claim is not rejected when \(p=.6\) ? When \(p=.8\) ? e. What decision rule would you choose for rejecting the claim \(p=.5\) if you wanted the probability in part (c) to be at most .01?

Write a general rule for \(E(X-c)\) where \(c\) is a constant. What happens when you let \(c=\mu\), the expected value of \(X\) ?

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?
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