/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A chemical supply company curren... [FREE SOLUTION] | 91Ó°ÊÓ

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A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a chemical, which it sells to customers in 5-lb containers. Let \(X=\) the number of containers ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

Short Answer

Expert verified
\(E(X) = 2.3\), \(V(X) = 0.81\). Expected pounds left: 88.5, variance: 20.25.

Step by step solution

01

Understand the Problem

We have a random variable \(X\) representing the number of 5-lb containers ordered by a customer, with a given probability mass function (pmf). We aim to find the expected value \(E(X)\) and the variance \(V(X)\), and we interpret the number of pounds left over after an order as a linear function of \(X\).
02

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

The expected value \(E(X)\) of a discrete random variable is calculated using \(E(X) = \sum x \, p(x)\). For our problem: \[E(X) = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3.\]
03

Compute the Variance \(V(X)\)

The variance \(V(X)\) is computed by \(V(X) = E(X^2) - [E(X)]^2\), where \(E(X^2) = \sum x^2 \, p(x)\). Calculate:\[E(X^2) = 1^2(0.2) + 2^2(0.4) + 3^2(0.3) + 4^2(0.1) = 0.2 + 1.6 + 2.7 + 1.6 = 6.1.\]Next, compute \(V(X)\):\[V(X) = 6.1 - (2.3)^2 = 6.1 - 5.29 = 0.81.\]
04

Express Pounds Left as a Linear Function of \(X\)

After an order, the pounds left, \(L\), can be expressed as \(100 - 5X\). This is because each container is 5 lb. The number of pounds left is thus a linear function of \(X\).
05

Compute Expected Pounds Left

Using the linearity of expectation, the expected number of pounds left is \(E(L) = E(100 - 5X) = 100 - 5E(X)\). We calculated \(E(X) = 2.3\), so:\[E(L) = 100 - 5 \times 2.3 = 88.5.\]
06

Compute Variance of Pounds Left

For a linear transformation \(aX + b\), \(V(aX + b) = a^2V(X)\). Thus, \(V(L) = 5^2V(X) = 25 \times 0.81 = 20.25\).

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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 symbolized as \(E(X)\), is a fundamental concept in probability and statistics. It is essentially the average or mean value that you would expect to see if you were to repeat an experiment or process a large number of times.
The formula for calculating the expected value of a discrete random variable \(X\) is given by:
  • \(E(X) = \sum x \, p(x)\)
Here, \(x\) represents each possible value that the random variable can take, \(p(x)\) is the probability of \(x\) occurring, and the sum is over all possible values of \(x\).
In our example, where \(X\) is the number of 5-lb containers ordered, the expected value is calculated by summing the products of each \(x\) and its corresponding probability \(p(x)\):
  • \(E(X) = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 2.3\)
This tells us that, on average, a customer would order 2.3 containers.
Variance
Variance, denoted as \(V(X)\), measures the spread or variability of the random variable \(X\) around its expected value \(E(X)\). It's useful for understanding how much the values of \(X\) deviate from the expected value on average.
The formula for variance is:
  • \(V(X) = E(X^2) - [E(X)]^2\)
To calculate it, you first need \(E(X^2)\), the expected value of the square of \(X\), which is:
  • \(E(X^2) = \sum x^2 \, p(x)\)
  • \( = 1^2(0.2) + 2^2(0.4) + 3^2(0.3) + 4^2(0.1) = 6.1\)
Using \(E(X^2)\) and \(E(X)\), we find the variance:
  • \(V(X) = 6.1 - 2.3^2 = 0.81\)
This means that there is some variability in the number of containers that customers order, but not a lot.
Probability Mass Function
The probability mass function (pmf) of a discrete random variable \(X\) provides the probabilities of each possible value \(X\) can take. It is a way to describe the distribution of a discrete random variable.
For a pmf, the following properties hold:
  • \(0 \leq p(x) \leq 1\) for all \(x\)
  • The sum of all probabilities must equal 1: \(\sum p(x) = 1\)
In the context of the chemical supply company's exercise, the pmf is:
  • \(p(1) = 0.2\)
  • \(p(2) = 0.4\)
  • \(p(3) = 0.3\)
  • \(p(4) = 0.1\)
This pmf shows the likelihood of a customer ordering 1, 2, 3, or 4 containers. Each probability indicates how often each order size is expected to occur in the long run. Understanding the pmf is key for deriving the expected value and variance, as it tells us the different values \(X\) can assume and with what frequency.

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Most popular questions from this chapter

A student who is trying to write a paper for a course has a choice of two topics, \(A\) and \(B\). If topic \(A\) is chosen, the student will order two books through interlibrary loan, whereas if topic \(\mathrm{B}\) is chosen, the student will order four books. The student believes that a good paper necessitates receiving and using at least half the books ordered for either topic chosen. If the probability that a book ordered through interlibrary loan actually arrives in time is \(.9\) and books arrive independently of one another, which topic should the student choose to maximize the probability of writing a good paper? What if the arrival probability is only .5 instead of .9?

A manufacturer of flashlight batteries wishes to control the quality of its product by rejecting any lot in which the proportion of batteries having unacceptable voltage appears to be too high. To this end, out of each large lot ( 10,000 batteries), 25 will be selected and tested. If at least 5 of these generate an unacceptable voltage, the entire lot will be rejected. What is the probability that a lot will be rejected if a. Five percent of the batteries in the lot have unacceptable voltages? b. Ten percent of the batteries in the lot have unacceptable voltages? c. Twenty percent of the batteries in the lot have unacceptable voltages? d. What would happen to the probabilities in parts (a)-(c) if the critical rejection number were increased from 5 to 6 ?

Suppose small aircraft arrive at an airport according to a Poisson process with rate \(\alpha=8 / \mathrm{h}\), so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\hat{\lambda}=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1 -h period? At least 6 ? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2 \frac{1}{2} h\) period? That at most 10 arrive during this period?

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]

Individual \(\mathrm{A}\) has a red die and \(\mathrm{B}\) has a green die (both fair). If they each roll until they obtain five "doubles" (1-1, ..., 6-6), what is the pmf of \(X=\) the total number of times a die is rolled? What are \(E(X)\) and \(V(X)\) ?

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