/*! 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 32 An appliance dealer sells three ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ \begin{array}{l|rrr} x & 13.5 & 15.9 & 19.1 \\ \hline p(x) & .2 & .5 & .3 \end{array} $$ 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?

Short Answer

Expert verified
a) \(E(X) = 16.46\), \(E(X^2) = 272.668\), \(V(X) = 2.6484\); b) \(E(25X-8.5) = 402\); c) \(V(25X-8.5) = 1655.25\); d) \(E(h(X)) = 13.73332\).

Step by step solution

01

Compute the Expected Value E(X)

The expected value is found by summing up the products of each value of \(X\) and its probability. The formula is \(E(X) = \sum_{i} x_i p(x_i)\).\[E(X) = (13.5 \times 0.2) + (15.9 \times 0.5) + (19.1 \times 0.3) = 16.46\]
02

Compute E(X²)

To find \(E(X^2)\), compute \(x_i^2 p(x_i)\) for each \(x_i\) and sum them.\[E(X^2) = (13.5^2 \times 0.2) + (15.9^2 \times 0.5) + (19.1^2 \times 0.3) = 272.668\]
03

Compute Variance V(X)

Variance \(V(X)\) is computed by the formula \(V(X) = E(X^2) - (E(X))^2\).\[V(X) = 272.668 - (16.46)^2 = 2.6484\]
04

Compute Expected Price Paid E(25X-8.5)

The expected price can be calculated using \(E(25X - 8.5) = 25E(X) - 8.5\).\[E(25X - 8.5) = 25 \times 16.46 - 8.5 = 402.\]
05

Compute Variance of Price V(25X-8.5)

Since variance is unaffected by linear constants (only by the coefficient attached to \(X\)), use \(V(25X - 8.5) = 25^2V(X)\).\[V(25X - 8.5) = 625 \times 2.6484 = 1655.25\]
06

Compute Expected Actual Capacity E(h(X))

For \(h(X) = X - 0.01X^2\), calculate \(E(h(X)) = E(X) - 0.01E(X^2)\).\[E(h(X)) = 16.46 - 0.01 \times 272.668 = 13.73332\]

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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 denoted as \(E(X)\), is a crucial concept in probability and statistics that represents the mean or average of a random variable. It provides a central tendency or a typical value the random variable takes. To compute the expected value of a discrete random variable like in our exercise, we sum up the products of each possible value of the random variable and their respective probabilities.
For example, if a random variable \(X\) represents the amount of storage space of a freezer purchased, with probabilities assigned to each possible storage size, the expected value is calculated using the formula:
  • \(E(X) = \sum_{i} x_i p(x_i)\)
The calculation involves:
  • For \(x = 13.5\), \(p(x) = 0.2\), multiplying gives \(13.5 \times 0.2\).
  • For \(x = 15.9\), \(p(x) = 0.5\), multiplying gives \(15.9 \times 0.5\).
  • For \(x = 19.1\), \(p(x) = 0.3\), multiplying gives \(19.1 \times 0.3\).
Add these values together, and you find the expected value, which, in this example, is \(16.46\) cubic feet. This value represents the average storage space that one might expect the next customer to purchase, considering the probabilities.
Variance
Variance, denoted as \(V(X)\), measures the spread or the dispersion of a set of values. For a random variable, it tells us how much the values deviate from the expected value or mean. Unlike the expected value, variance gives us an idea about the variability or uncertainty associated with the random variable.
The formula for variance is:
  • \(V(X) = E(X^2) - (E(X))^2\)
This equation shows that we need \(E(X^2)\), which is the expected value of the square of the random variable, and we've already computed \(E(X)\).
  • For \(x = 13.5\), calculate \((13.5^2 \times 0.2)\).
  • For \(x = 15.9\), calculate \((15.9^2 \times 0.5)\).
  • For \(x = 19.1\), calculate \((19.1^2 \times 0.3)\).
Summing these gives \(E(X^2) = 272.668\). Therefore, \(V(X) = 272.668 - (16.46)^2 = 2.6484\). This small variance indicates that the freezer sizes do not stray far from the mean.
Probability
Probability is a foundational aspect of statistics that quantifies how likely an event will occur. Each outcome or set of outcomes of a random variable has a designated probability. Probabilities must sum up to 1, ensuring that one of the possible outcomes will happen. In a Probability Mass Function (pmf), we assign probabilities to discrete random variables.
In our freezer example, the probabilities for each storage space indicate how likely each size is to be purchased:
  • A probability of 0.2 for 13.5 cubic feet.
  • A probability of 0.5 for 15.9 cubic feet.
  • A probability of 0.3 for 19.1 cubic feet.
These probabilities are crucial as they directly feed into calculations of expected value and variance, furnishing robust insights into the likelihood of various outcomes.
Random Variable
A random variable, as seen with \(X\) in our exercise, is a variable that can take on multiple values, each with an associated probability. It is a vital concept in probability theory, representing outcomes of a stochastic process.
There are two main types of random variables:
  • Discrete, which takes on distinct, separate values, like in our example with specific freezer sizes.

  • Continuous, which can take any value within a given range.
In our problem, \(X\) represents discrete storage capacities available for purchase, and each model's probability - given by the pmf - describes how likely it is for the storage space to be chosen. Through these values and their probabilities, we can perform various statistical analyses, such as calculating the expected value and variance, which offer insight into this random variable's behavior.

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Most popular questions from this chapter

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