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The pmf for \(X=\) the number of major defects on a randomly selected appliance of a certain type is \begin{tabular}{c|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. \(E(X) = 2.06\); b. \(V(X) = 0.9364\); c. Standard deviation = \(0.9677\); d. \(V(X) = 0.9364\) again from the shortcut.

Step by step solution

01

Compute Expected Value E(X)

The expected value, or mean, of a discrete random variable \(X\) is calculated as \(E(X) = \sum x \, p(x)\). Here is the computation using the given probability mass function:\[E(X) = (0)(0.08) + (1)(0.15) + (2)(0.45) + (3)(0.27) + (4)(0.05)\] After simplifying, \[E(X) = 0 + 0.15 + 0.90 + 0.81 + 0.20 = 2.06\] Therefore, \(E(X) = 2.06\).
02

Compute Variance V(X) from Definition

The variance of \(X\) is calculated as \(V(X) = \sum (x - E(X))^2 \, p(x)\). Using \(E(X) = 2.06\), compute the variance: \[V(X) = (0 - 2.06)^2(0.08) + (1 - 2.06)^2(0.15) + (2 - 2.06)^2(0.45) + (3 - 2.06)^2(0.27) + (4 - 2.06)^2(0.05)\] Which gives: \[V(X) = (4.2436)(0.08) + (1.1236)(0.15) + (0.0036)(0.45) + (0.8836)(0.27) + (3.7636)(0.05)\] Simplifying,\[V(X) = 0.339488 + 0.16854 + 0.00162 + 0.238572 + 0.18818 = 0.9364\] Thus, \(V(X) = 0.9364\).
03

Compute Standard Deviation of X

The standard deviation is the square root of the variance. We calculated \(V(X) = 0.9364\). Therefore, the standard deviation \(\sigma\) is: \[\sigma = \sqrt{0.9364} = 0.9677\] Thus, the standard deviation of \(X\) is approximately \(0.9677\).
04

Compute Variance V(X) using Shortcut Formula

The shortcut formula for variance is \(V(X) = E(X^2) - [E(X)]^2\). First, calculate \(E(X^2)\): \[E(X^2) = \sum x^2 \, p(x) = (0^2)(0.08) + (1^2)(0.15) + (2^2)(0.45) + (3^2)(0.27) + (4^2)(0.05)\]\[= 0 + 0.15 + 1.8 + 2.43 + 0.8 = 5.18\] Now, compute the variance: \[V(X) = 5.18 - (2.06)^2 = 5.18 - 4.2436 = 0.9364\] Hence, using the shortcut formula, \(V(X) = 0.9364\), confirming the previous calculation.

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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 referred to as the mean, is a central concept in probability and statistics, particularly in dealing with discrete random variables. It's the average outcome that one would anticipate after many repetitions of an experiment. To compute the expected value for a discrete random variable, you use the formula: \(E(X) = \sum x \cdot p(x)\).
In this context, we are looking at the probability mass function (pmf), which assigns a probability to all possible outcomes of the random variable. Each outcome is multiplied by its respective probability, and these products are summed to find the expected value. This can be viewed as a weighted average, where each outcome contributes to the mean according to its probability.
For example, with our given probabilities for defects on an appliance, the expected value was calculated by considering the sum of each possible outcome extrapolated by its probability:
  • 0 defects (probability = 0.08): \(0 \cdot 0.08 = 0\)
  • 1 defect (probability = 0.15): \(1 \cdot 0.15 = 0.15\)
  • 2 defects (probability = 0.45): \(2 \cdot 0.45 = 0.9\)
  • 3 defects (probability = 0.27): \(3 \cdot 0.27 = 0.81\)
  • 4 defects (probability = 0.05): \(4 \cdot 0.05 = 0.2\)
The expected number of major defects is thus 2.06, signifying that, on average, there are just over two major defects on these appliances.
Variance
Variance is a measure of the spread or dispersion of a set of values around the mean. In the context of discrete random variables, it quantifies how much the values of the variable differ from the expected value.
To calculate variance, the formula \(V(X) = \sum (x - E(X))^2 \cdot p(x)\) is used. Here, each outcome's deviation from the mean (expected value) is squared, weighted by its probability, and summed to get the variance.
This squaring ensures that all deviations are treated as positive contributions, focusing on the magnitude of deviation rather than direction. For example, using our appliance defect data and previously calculated expected value of 2.06, every value shifts by subtracting the mean, square each of these differences, and then weigh them by their probability:
  • \((0 - 2.06)^2 \cdot 0.08 = 0.339488\)
  • \((1 - 2.06)^2 \cdot 0.15 = 0.16854\)
  • \((2 - 2.06)^2 \cdot 0.45 = 0.00162\)
  • \((3 - 2.06)^2 \cdot 0.27 = 0.238572\)
  • \((4 - 2.06)^2 \cdot 0.05 = 0.18818\)
When summed, these terms yield a variance of 0.9364. This value helps us understand that while the average number of defects is slightly over 2, the number can typically vary by a magnitude quantified by the variance.
Standard Deviation
The standard deviation provides a clearer measure of dispersion that is easier to interpret than variance as it is in the same unit as the random variable. It is simply the square root of the variance, calculated as \(\sigma = \sqrt{V(X)}\).
In our appliance defect example, the variance was found to be 0.9364. The standard deviation is thus \(\sqrt{0.9364} = 0.9677\). Our interpretation is that while the average number of defects is 2.06, the number of defects tends to deviate from this average by about 0.97 defects on average.
This makes standard deviation a vital statistic in understanding variability for not just academic purposes, but practical applications too, making it a standard measure for all types of data distributions.
Discrete Random Variables
A discrete random variable is one that has countable distinct outcomes. In contrast to continuous variables, discrete variables only take specific values. For instance, the number of major defects in an appliance represents a discrete random variable because you can count such defects (0, 1, 2, etc.) but they don't fall in between these values.
For discrete random variables, the probability mass function (pmf) is essential. It specifies the probability that The random variable takes on a particular value. This function helps in predicting the likelihood of different outcomes, determining expected values, and calculating variance and standard deviation.
Understanding discrete random variables is crucial for statistical analysis and expectations because many real-world phenomena are discrete at their core, such as counting occurrences or tallying items.Thus, conceptualizing events as discrete random variables helps us effectively model and solve various practical and theoretical problems.

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