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For a new car the number of defects \(X\) has the distribution given by the accompanying table. Find \(M_{X}(t)\) and use it to find \(E(X)\) and \(V(X)\). \begin{tabular}{c|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.04\) & \(.20\) & \(.34\) & \(.20\) & \(.15\) & \(.04\) & \(.03\) \end{tabular}

Short Answer

Expert verified
MGF: \(M_{X}(t) = 0.04 + 0.20e^{t} + 0.34e^{2t} + 0.20e^{3t} + 0.15e^{4t} + 0.04e^{5t} + 0.03e^{6t}\), \(E(X) = 2.46\), \(V(X) = 1.74\).

Step by step solution

01

Verify the Probability Distribution

To ensure the given distribution is valid, check that the sum of all probabilities equals 1. Calculate the sum of all probabilities: \[ 0.04 + 0.20 + 0.34 + 0.20 + 0.15 + 0.04 + 0.03 = 1 \]Since the sum is 1, this is a valid probability distribution.
02

Define the Moment Generating Function

The moment generating function (MGF) for a discrete random variable \( X \) is given by: \[ M_{X}(t) = E(e^{tX}) = \sum_{x} e^{tx} p(x) \]We will use this formula to compute \( M_{X}(t) \) using the given distribution.
03

Compute the MGF

Substitute the values into the MGF formula: \[ M_{X}(t) = 0.04e^{0t} + 0.20e^{1t} + 0.34e^{2t} + 0.20e^{3t} + 0.15e^{4t} + 0.04e^{5t} + 0.03e^{6t} \]Simplifying, we get: \[ M_{X}(t) = 0.04 + 0.20e^{t} + 0.34e^{2t} + 0.20e^{3t} + 0.15e^{4t} + 0.04e^{5t} + 0.03e^{6t} \]
04

Find the Expected Value \( E(X) \)

The expected value \( E(X) \) can be found by differentiating the MGF and evaluating it at \( t = 0 \):Differentiate \( M_{X}(t) \):\[ M'_{X}(t) = 0.20e^{t} + 0.68e^{2t} + 0.60e^{3t} + 0.60e^{4t} + 0.20e^{5t} + 0.18e^{6t} \]Evaluate at \( t = 0 \):\[ M'_{X}(0) = 0.20 + 0.68 + 0.60 + 0.60 + 0.20 + 0.18 = 2.46 \]Thus, \( E(X) = 2.46 \).
05

Calculate the Variance \( V(X) \)

Variance \( V(X) \) can be found using the formula \( V(X) = E(X^2) - [E(X)]^2 \).First, find \( E(X^2) \) by differentiating \( M_X(t) \) again, then evaluate at \( t = 0 \):Differentiating \( M'_{X}(t) \) gives:\[ M''_{X}(t) = 0.20e^{t} + 1.36e^{2t} + 1.80e^{3t} + 2.40e^{4t} + 1.00e^{5t} + 1.08e^{6t} \]Evaluate at \( t = 0 \):\[ M''_{X}(0) = 0.20 + 1.36 + 1.80 + 2.40 + 1.00 + 1.08 = 7.84 \]\( E(X^2) = 7.84 \), and therefore, \( V(X) = 7.84 - (2.46)^2 = 1.74 \).
06

Conclusion

The MGF of the random variable \( X \) is:\[ M_{X}(t) = 0.04 + 0.20e^{t} + 0.34e^{2t} + 0.20e^{3t} + 0.15e^{4t} + 0.04e^{5t} + 0.03e^{6t} \]The expected number of defects is \( E(X) = 2.46 \) and the variance is \( V(X) = 1.74 \).

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Key Concepts

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

Probability Distribution
In probability theory, a probability distribution provides a mathematical function that describes the likelihood of different outcomes in an experiment. When dealing with a discrete random variable, like the number of defects in a car, each possible outcome, denoted as \(x\), has a corresponding probability \(p(x)\). These probabilities must satisfy the following rules:
  • Each probability \(p(x)\) must be between 0 and 1.
  • The sum of all probabilities must equal 1.
For the car defect example, the probability distribution is given as a table where each defect number has a probability of occurrence. When checking that this table constitutes a valid probability distribution, sum all the probabilities:\[0.04 + 0.20 + 0.34 + 0.20 + 0.15 + 0.04 + 0.03 = 1\]This confirms it is a valid probability distribution, as the sum equals 1.
Expected Value
The expected value, or mean, of a random variable is a measure of the central tendency of the distribution. It gives us an idea of the average outcome if we were to repeat the experiment multiple times. For a discrete random variable \(X\), the expected value \(E(X)\) is calculated by:\[E(X) = \sum_{x} x \, p(x)\]However, in more complex distributions like those involving moment-generating functions (MGF), the expected value can also be derived by differentiating the MGF and evaluating at \(t = 0\). For our car defects example, after computing the MGF, differentiating gives us:\[M'_{X}(0) = 2.46\]Thus, the expected number of defects, \(E(X)\), is 2.46. This means that, on average, you can expect 2.46 defects per car.
Variance
Variance is a measure of how much the values of a random variable \(X\) deviate from the mean \(E(X)\). It provides insight into the spread or dispersion within the distribution. The variance \(V(X)\) of \(X\) can be determined using the formula:\[V(X) = E(X^2) - [E(X)]^2\]Here, \(E(X^2)\) represents the expected value of the square of the random variable. After obtaining the MGF and its derivatives for our defects example, we calculate \(E(X^2)\) as:\[M''_{X}(0) = 7.84\]Then, the variance is computed:\[V(X) = 7.84 - (2.46)^2 = 1.74\]The variance, 1.74, illustrates the extent to which defect counts vary from the mean. A lower variance indicates that defect counts are close to 2.46 defects, while a higher variance would indicate more variability.

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Most popular questions from this chapter

Each time a component is tested, the trial is a success \((S)\) or failure \((F)\). Suppose the component is tested repeatedly until a success occurs on three consecutive trials. Let \(Y\) denote the number of trials necessary to achieve this. List all outcomes corresponding to the five smallest possible values of \(Y\), and state which \(Y\) value is associated with each one.

The negative binomial rv \(X\) was defined as the number of \(F^{\prime}\) s preceding the \(r\) th \(S\). Let \(Y=\) the number of trials necessary to obtain the rth \(S\). In the same manner in which the pmf of \(X\) was derived, derive the pmf of \(Y\).

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space, respectively. Let \(X=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(X\) has pmf \begin{tabular}{l|ccc} \(x\) & \(13.5\) & \(15.9\) & \(19.1\) \\ \hline\(p(x)\) & \(.2\) & \(.5\) & \(.3\) \end{tabular} a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) cubic feet is \(25 X-8.5\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price \(25 X-8.5\) paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-\) \(.01 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?

A family decides to have children until it has three children of the same gender. Assuming \(P(B)=P(G)=.5\), what is the pmf of \(X=\) the number of children in the family?

Show that \(g(t)=t e^{t}\) cannot be a moment generating function.

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