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Chapter 6: Problem 5

In a certain manufacturing process, the (Fahrenheit) temperature never varies by more than \(2^{\circ}\) from \(62^{\circ}\). The temperature is, in fact, a random variable \(F\) with distribution $$P_{F}=\left(\begin{array}{ccccc}60 & 61 & 62 & 63 & 64 \\\1 / 10 & 2 / 10 & 4 / 10 & 2 / 10 & 1 / 10\end{array}\right)$$ (a) Find \(E(F)\) and \(V(F)\). (b) Define \(T=F-62 .\) Find \(E(T)\) and \(V(T),\) and compare these answers with those in part (a). (c) It is decided to report the temperature readings on a Celsius scale, that is, \(C=(5 / 9)(F-32) .\) What is the expected value and variance for the readings now?

Short Answer

Expert verified
(a) \(E(F) = 62\), \(V(F) = 0.8\). (b) \(E(T) = 0\), \(V(T) = 0.8\), same variance. (c) \(E(C) \approx 16.67\), \(V(C) \approx 0.247\).

Step by step solution

01

Calculate Expectation of F

The expectation or expected value of a random variable is obtained by multiplying each possible outcome by its probability and summing these products. For our variable \( F \),\[ E(F) = \sum f_i P(F=f_i) = 60 \cdot \frac{1}{10} + 61 \cdot \frac{2}{10} + 62 \cdot \frac{4}{10} + 63 \cdot \frac{2}{10} + 64 \cdot \frac{1}{10}. \]Computing this gives:\[ E(F) = 6 + 12.2 + 24.8 + 12.6 + 6.4 = 62. \]
02

Calculate Variance of F

Variance measures how much the values of a random variable differ from the mean. We use the formula: \[ V(F) = \sum (f_i - E(F))^2 P(F=f_i). \]Substitute known values: \[ V(F) = (60 - 62)^2 \cdot \frac{1}{10} + (61 - 62)^2 \cdot \frac{2}{10} + (62 - 62)^2 \cdot \frac{4}{10} + (63 - 62)^2 \cdot \frac{2}{10} + (64 - 62)^2 \cdot \frac{1}{10}. \]This simplifies to: \[ V(F) = 4 \cdot \frac{1}{10} + 1 \cdot \frac{2}{10} + 0 \cdot \frac{4}{10} + 1 \cdot \frac{2}{10} + 4 \cdot \frac{1}{10} = 0.8. \]
03

Define T and Calculate E(T)

Since \( T = F - 62 \), each value of \( T \) is centered around zero:- \( F = 60 \) yields \( T = -2 \),- \( F = 61 \) yields \( T = -1 \),- \( F = 62 \) yields \( T = 0 \),- \( F = 63 \) yields \( T = 1 \),- \( F = 64 \) yields \( T = 2 \).The probabilities remain unchanged, but the outcomes are shifted. Let's find the expectation: \[ E(T)= (-2) \cdot \frac{1}{10} + (-1) \cdot \frac{2}{10} + 0 \cdot \frac{4}{10} + 1 \cdot \frac{2}{10} + 2 \cdot \frac{1}{10}. \]Calculations show: \[ E(T) = -0.2 - 0.2 + 0 + 0.2 + 0.2 = 0. \]
04

Calculate V(T)

The variance of \( T \) is calculated similarly as for \( F \):\[ V(T) = (-2-0)^2 \cdot \frac{1}{10} + (-1-0)^2 \cdot \frac{2}{10} + (0-0)^2 \cdot \frac{4}{10} + (1-0)^2 \cdot \frac{2}{10} + (2-0)^2 \cdot \frac{1}{10}. \]This simplifies to:\[ V(T) = 4 \cdot \frac{1}{10} + 1 \cdot \frac{2}{10} + 0 \cdot \frac{4}{10} + 1 \cdot \frac{2}{10} + 4 \cdot \frac{1}{10} = 0.8. \]
05

Convert to Celsius and Find E(C)

Convert to Celsius using \( C = \frac{5}{9}(F-32) \). To find \( E(C) \), use linearity:\[ E(C) = \frac{5}{9}(E(F) - 32). \]Substitute \( E(F)=62 \):\[E(C) = \frac{5}{9}(62 - 32) = \frac{5}{9} \times 30 = \frac{150}{9} \approx 16.67.\]
06

Find V(C)

Since variance is not affected by shifts (just scaling), but affected by square of the scaling factor, use:\[ V(C) = \left(\frac{5}{9}\right)^2 V(F). \]Substitute \( V(F)=0.8 \):\[ V(C) = \left(\frac{5}{9}\right)^2 \cdot 0.8 \approx 0.247. \]

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Key Concepts

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

Random Variables
Random variables are fundamental concepts in probability and statistics. They represent outcomes of a random phenomenon. In our exercise, the temperature in a manufacturing process is modeled as a random variable, denoted by \(F\). The temperature can take specific values (60, 61, 62, 63, 64 degrees Fahrenheit), each with a certain probability.
Understanding random variables helps predict outcomes and understand the distribution of events in a controlled yet unpredictable environment. The probabilities associated with each temperature value sum to 1, representing a complete probability distribution. This feature is crucial for accurately calculating further statistical properties, such as the expected value and variance.
Expected Value
The expected value, denoted as \(E(X)\), is a measure of the central tendency of a random variable. It is also known as the mean or average. To find the expected value, multiply each possible outcome by its probability and sum all these products.
For the temperature variable \(F\), the expected value is calculated as follows:
  • Multiply each temperature by its respective probability: \(60 \times \frac{1}{10}, 61 \times \frac{2}{10}, 62 \times \frac{4}{10}, 63 \times \frac{2}{10}, 64 \times \frac{1}{10}\).
  • Sum these results: \(6 + 12.2 + 24.8 + 12.6 + 6.4\).
This gives an expected value of \(62\) degrees Fahrenheit, indicating that 62 is the "average" or central temperature around which other values are distributed.
Variance
The variance \(V(X)\) of a random variable provides a quantitative measure of how much the values differ from the mean. It indicates the spread of the data.
To calculate variance, use the formula: \[ V(F) = \sum (f_i - E(F))^2 P(F=f_i) \]For our temperature variable, the calculations involve:
  • Subtracting the expected value from each temperature value (e.g., \(60 - 62, 61 - 62\)), squaring it (to eliminate negative values), and multiplying by the probability of that temperature.
  • Summing these products: \((60 - 62)^2 \cdot \frac{1}{10}, (61 - 62)^2 \cdot \frac{2}{10}\), etc.
The variance of \(0.8\) tells us about the degree of temperature fluctuation around the expected value, with larger values indicating greater spread.
Temperature Conversion
Converting temperatures from Fahrenheit to Celsius is a common task in many scientific fields. The formula used is:\[ C = \frac{5}{9}(F - 32) \]This conversion changes the scale and zero point of temperature, which, in turn, affects statistical measures like expected value and variance.
For our variable \(F\) and its conversion to \(C\), the expected value in Celsius, \(E(C)\), is calculated as:\[ E(C) = \frac{5}{9}(E(F) - 32) \]Substituting the expected Fahrenheit temperature gives \(16.67\) degrees Celsius. Variance in Celsius, \(V(C)\), adjusts by the square of the conversion factor, as variance is scale-dependent:\[ V(C) = \left(\frac{5}{9}\right)^2 V(F) \]This calculation yields a variance of approximately \(0.247\), reflecting the reduced spread after scaling.

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Most popular questions from this chapter

Recall that in the martingale doubling system (see Exercise 1.1.10), the player doubles his bet each time he loses. Suppose that you are playing roulette in a fair casino where there are no 0 's, and you bet on red each time. You then win with probability \(1 / 2\) each time. Assume that you enter the casino with 100 dollars, start with a 1 -dollar bet and employ the martingale system. You stop as soon as you have won one bet, or in the unlikely event that black turns up six times in a row so that you are down 63 dollars and cannot make the required 64 -dollar bet. Find your expected winnings under this system of play.

In a second version of roulette in Las Vegas, a player bets on red or black. Half of the numbers from 1 to 36 are red, and half are black. If a player bets a dollar on black, and if the ball stops on a black number, he gets his dollar back and another dollar. If the ball stops on a red number or on 0 or 00 he loses his dollar. Find the expected winnings for this bet.

Let \(X\) be a random variable with range [-1,1] and \(f_{X}\) its density function. Find \(\mu(X)\) and \(\sigma^{2}(X)\) if, for \(|x|>1, f_{X}(x)=0,\) and for \(|x|<1,\) (a) \(f_{X}(x)=(3 / 4)\left(1-x^{2}\right)\). (b) \(f_{X}(x)=(\pi / 4) \cos (\pi x / 2)\). (c) \(f_{X}(x)=(x+1) / 2\). (d) \(f_{X}(x)=(3 / 8)(x+1)^{2}\).

If \(X\) and \(Y\) are any two random variables, then the covariance of \(X\) and \(Y\) is defined by \(\operatorname{Cov}(X, Y)=E((X-E(X))(Y-E(Y))) .\) Note that \(\operatorname{Cov}(X, X)=\) \(V(X) .\) Show that, if \(X\) and \(Y\) are independent, then \(\operatorname{Cov}(X, Y)=0 ;\) and show, by an example, that we can have \(\operatorname{Cov}(X, Y)=0\) and \(X\) and \(Y\) not independent.

A coin is tossed until the first time a head turns up. If this occurs on the \(n\) th toss and \(n\) is odd you win \(2^{n} / n,\) but if \(n\) is even then you lose \(2^{n} / n\). Then if your expected winnings exist they are given by the convergent series $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ called the alternating harmonic series. It is tempting to say that this should be the expected value of the experiment. Show that if we were to do this, the expected value of an experiment would depend upon the order in which the outcomes are listed.
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