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Combining random variables

Learning goals

  • Defines a new random variable as a sum, difference, or linear combination of existing variables.
  • Computes the mean of the combined variable using linearity of expectation.
  • Computes variance or standard deviation using the stated independence assumptions.
  • Distinguishes adding variances from adding standard deviations.
  • Interprets the combined variable in context.

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Combining random variables 1
Random variables \(X\) and \(Y\) have means \(E(X)=4 \qquad\text{and}\qquad E(Y)=7.\) Find \(E(X+Y)\). Is independence required for this calculation?

Hints

- Recall the rule for the expected value of a sum. - Separate the rule for means from the rule for variances.

Solution

1. Expected values add: \(E(X+Y)=E(X)+E(Y)=4+7=11.\) 2. Linearity of expectation does not require \(X\) and \(Y\) to be independent.

Answer

\(E(X+Y)=11\). Independence is not required.
Combining random variables 2
A random variable \(X\) has standard deviation \(5\). Define \(Y=X+12.\) Find the standard deviation of \(Y\). Explain why adding \(12\) does or does not change the spread.

Hints

- Imagine moving every value of a distribution the same distance to the right. - Ask whether the distances between values change after adding the same constant to all of them.

Solution

1. Adding a constant shifts every possible value by the same amount. 2. The distances between values, and therefore the spread, do not change. 3. Thus, \(\operatorname{SD}(Y)=\operatorname{SD}(X)=5.\)

Answer

\(\operatorname{SD}(Y)=5\). Adding a constant shifts the distribution but does not change its spread.
Combining random variables 3
Two fair coins are tossed. A head earns \(2\) points, and a tail earns \(0\) points. Let \(X\) and \(Y\) be the points from the first and second coins, and let \(S=X+Y\). List the possible values of \(S\). What does \(S\) describe?

Hints

- List the possible outcomes of the two coin tosses. - Turn each coin outcome into points before finding the total. - Different sequences of heads and tails can produce the same total.

Solution

1. Two tails give \(S=0+0=0\). 2. One head and one tail give \(S=2+0=2\), regardless of their order. 3. Two heads give \(S=2+2=4\). 4. Therefore, \(S\) is the total number of points earned from both tosses.

Answer

Possible values: \(0\), \(2\), and \(4\). \(S\) is the combined score from both coins.
Combining random variables 4
Let \(X\) be the number of books checked out of a library and \(Y\) the number returned during the same visit. Define \(D=X-Y\). If \(X=2\) and \(Y=5\), find \(D\) and explain what its sign means.

Hints

- Read the order of subtraction in the definition of \(D\). - Which action increases the number of books the visitor takes home? - Interpret the sign rather than only giving a number.

Solution

1. Subtract the number returned from the number checked out: \(D=2-5=-3\). 2. A negative result means more books were returned than checked out, by three books.

Answer

\(D=-3\). Three more books were returned than checked out.
Combining random variables 5
The average number of points from one game is \(E(X)=2.5\), and from another game it is \(E(Y)=1.5\). A student plays both and records \(T=X+Y\). Find \(E(T)\). Is independence needed to add these expected values?

Hints

- Think of the total score as adding two amounts each time the student plays. - How would their long-run average totals compare with the two separate averages? - Keep the rule for an average separate from the rule for variance.

Solution

1. The mean of a sum is the sum of the means, so \(E(T)=2.5+1.5=4\). 2. This rule holds even if the game scores are related; independence is not needed for adding expected values.

Answer

\(E(T)=4\) points. Independence is not needed to add expectations.
Combining random variables 6
A fair six-sided die is rolled. Set \(I=1\) if the die shows a six, and \(I=0\) otherwise. What values can \(I\) take, and what is \(P(I=1)\)?

Hints

- This variable records whether a particular event happens, not the face number itself. - Check what happens when the die lands on the selected face. - Use the equally likely die faces to find the requested probability.

Solution

1. By its definition, \(I\) equals \(1\) for a six and \(0\) for every other result. 2. A fair die has one face labeled six out of six equally likely faces, so \(P(I=1)=\frac{1}{6}\).

Answer

The possible values are \(0\) and \(1\); \(P(I=1)=\frac{1}{6}\).
Combining random variables 7
A digital scan has two independent stages. The capture time \(C\), in seconds, has mean \(4.2\) and standard deviation \(0.6\). The processing time \(P\), in seconds, has mean \(3.5\) and standard deviation \(0.8\). Let \(T=C+P\) be the total scan time. Find and interpret the mean and standard deviation of \(T\).

Hints

- Combine the centers of the two stages directly. - Use the stated relationship between the stages when combining their spreads. - Convert the combined variance back to the original time unit.

Solution

1. The mean of the sum is \(\mu_T=4.2+3.5=7.7\). 2. Independence allows the variances to add: \(\sigma_T^2=0.6^2+0.8^2=1\). 3. Therefore, \(\sigma_T=1.0\) second. 4. Total scan times average \(7.7\) seconds and typically differ from that mean by about \(1.0\) second.

Answer

\(\mu_T=7.7\,\text{seconds}\) and \(\sigma_T=1.0\,\text{second}\). Total scan times average \(7.7\) seconds and typically differ from that mean by about \(1.0\) second.
Combining random variables 8
A repair visit has a fixed charge of \(\$65\) plus \(\$42\) per hour. The random repair time \(H\), in hours, has mean \(2.8\) and standard deviation \(0.7\). Let \(C=65+42H\) be the total charge. Find and interpret the mean and standard deviation of \(C\).

Hints

- Express the total charge as a linear transformation of repair time. - Separate the role of the fixed charge from the hourly multiplier. - Interpret both parameters in dollars.

Solution

1. The mean is \(\mu_C=65+42(2.8)=182.6\). 2. The fixed charge does not affect spread, so \(\sigma_C=42(0.7)=29.4\). 3. Repair charges average \(\$182.60\), with a typical deviation of \(\$29.40\) from that average.

Answer

\(\mu_C=\$182.60\) and \(\sigma_C=\$29.40\). Repair charges average \(\$182.60\), with a typical deviation of \(\$29.40\) from that average.
Combining random variables 9
A performance measure \(X\) has mean \(40\) and standard deviation \(6\). A penalty score is defined by \(Y=100-1.5X\). a) Find the mean and standard deviation of \(Y\). b) Describe how the negative coefficient affects the ordering of individuals by score.

Hints

- Apply the full transformation to the center. - Standard deviation responds to the magnitude of a multiplier, not its sign. - Compare two hypothetical original scores to see what the negative coefficient does to their order.

Solution

1. The mean is \(\mu_Y=100-1.5(40)=40\). 2. The standard deviation is \(\sigma_Y=|-1.5|(6)=9\). 3. Because the coefficient is negative, a larger value of \(X\) produces a smaller value of \(Y\), so the ordering is reversed.

Answer

a) \(\mu_Y=40\) and \(\sigma_Y=9\). b) The negative coefficient reverses the ranking: higher \(X\)-values correspond to lower \(Y\)-values.
Combining random variables 10
A random variable \(X\) has mean \(65\) and standard deviation \(7\). Define \(Z=\frac{X-65}{7}\). Find the mean and standard deviation of \(Z\), and explain what the transformation accomplishes.

Hints

- Handle the subtraction and division as separate transformation steps. - Identify which step changes the center and which changes the scale. - Interpret the new unit in relation to the original spread.

Solution

1. The mean is \(\mu_Z=\frac{65-65}{7}=0\). 2. Subtracting \(65\) does not change spread, and dividing by \(7\) gives \(\sigma_Z=\frac{7}{7}=1\). 3. The transformation recenters the variable at \(0\) and rescales one standard deviation of \(X\) to one unit of \(Z\).

Answer

\(\mu_Z=0\) and \(\sigma_Z=1\). The transformation expresses values relative to the original mean in units of the original standard deviation.
Combining random variables 11
Daily sales over \(10\) independent days have the same distribution, with mean \(45\) units and standard deviation \(9\) units. Let \(S\) be the total sales and \(A\) be the average daily sales for the \(10\) days. Find the mean and standard deviation of both \(S\) and \(A\). Give exact standard deviations and decimal approximations to two decimal places.

Hints

- Treat the total as a sum of independent daily variables. - Express the average as a scaled version of the total. - Apply the scaling rule to both center and spread.

Solution

1. For the total, \(\mu_S=10(45)=450,\qquad \sigma_S=9\sqrt{10}\approx28.46.\) 2. Since \(A=S/10\), \(\mu_A=45,\qquad \sigma_A=\frac{9}{\sqrt{10}}\approx2.85.\)

Answer

\(\mu_S=450\) units and \(\sigma_S=9\sqrt{10}\approx28.46\) units. \(\mu_A=45\) units and \(\sigma_A=\frac{9}{\sqrt{10}}\approx2.85\) units.
Combining random variables 12
A process measurement has standard deviation \(10\) units. A quality analyst averages \(25\) independent measurements and claims, “The standard deviation of the average is \(\frac{10}{25}=0.4\) unit.” Identify the error and find the correct standard deviation of the average.

Hints

- Express the average as a scaled sum. - Track variance through the sum before returning to standard deviation. - Compare how the number of independent measurements changes the spread.

Solution

1. The analyst divided the component standard deviation directly by the number of measurements. 2. For an average of \(25\) independent measurements, the standard deviation is divided by \(\sqrt{25}\), not by \(25\). 3. The correct value is \(\frac{10}{\sqrt{25}}=2\) units.

Answer

The analyst used \(25\) instead of \(\sqrt{25}\). The correct standard deviation is \(2\) units.
Combining random variables 13
A wildlife sensor's weekly number of false alarms, \(X\), has mean \(4\) and standard deviation \(1.2\). Two proposed performance scores are \(A=50+10X\) and \(B=80-6X\). Find the mean and standard deviation of each score. Which score has greater variability?

Hints

- Handle the center and spread of each scoring rule separately. - A fixed added amount affects only one of the two parameters. - Compare variability using nonnegative spread values.

Solution

1. For \(A=50+10X\), \(\mu_A=50+10(4)=90\) and \(\sigma_A=10(1.2)=12\). 2. For \(B=80-6X\), \(\mu_B=80-6(4)=56\) and \(\sigma_B=|-6|(1.2)=7.2\). 3. Since \(12>7.2\), score \(A\) has greater variability.

Answer

\(\mu_A=90\), \(\sigma_A=12\); \(\mu_B=56\), \(\sigma_B=7.2\). Score \(A\) has greater variability.
Combining random variables 14
Two independent random variables \(X\) and \(Y\) have standard deviations \(3\) and \(4\). Let \(S=X+Y\). Find the standard deviation of \(S\). Explain why it is not \(3+4\).

Hints

- Which quantity follows the addition rule when random variables are independent? - Convert each given standard deviation to that quantity first. - Make sure the final answer is a standard deviation, not a variance.

Solution

1. Independence lets the variances add: \(\operatorname{Var}(S)=3^2+4^2=25\). 2. Take the square root: \(SD(S)=\sqrt{25}=5\). 3. Standard deviations themselves cannot be added because variance, not standard deviation, is additive for independent variables.

Answer

\(SD(S)=5\). Add the variances and then take the square root, not the standard deviations.
Combining random variables 15
One fair coin is tossed. Set \(X=1\) for heads and \(X=0\) for tails. Consider two ways to define a second variable: - Case A: \(Y=X\). - Case B: \(Y=1-X\). Both versions of \(Y\) have probability \(0.5\) of being \(1\). Find \(P(X+Y=1)\) in each case and explain why the results differ.

Hints

- Check what \(X+Y\) becomes under each definition of \(Y\). - Consider heads and tails separately. - Having the same probabilities individually does not tell you how the two variables behave together.

Solution

1. In Case A, \(X+Y=2X\), which is either \(0\) or \(2\), so \(P(X+Y=1)=0\). 2. In Case B, \(X+Y=X+(1-X)=1\) on every toss, so \(P(X+Y=1)=1\). 3. The variables \(X\) and \(Y\) have the same separate distributions in both cases, but their dependence differs.

Answer

Case A: \(0\). Case B: \(1\). The relationship between \(X\) and \(Y\), not just their individual probabilities, determines the distribution of their sum.
Combining random variables 16
The bar charts give the distributions of independent random variables \(X\) and \(Y\). Let \(S=X+Y\). Construct the probability distribution of \(S\) in a table.
Figure for problem 563456

Hints

- Read possible values and their probabilities from each axis. - A value of the sum comes from one value of each variable. - Use independence to find the probability of each pair, then group pairs that have the same sum.

Solution

1. From the charts, \(P(X=0)=P(X=1)=0.5\), while \(P(Y=0)=0.25\) and \(P(Y=2)=0.75\). 2. By independence, the pairs \((X,Y)=(0,0),(0,2),(1,0),(1,2)\) have probabilities \(0.125,0.375,0.125,0.375\), respectively. 3. The corresponding sums are \(0,2,1,3\). Thus, \(P(S=0)=P(S=1)=0.125\) and \(P(S=2)=P(S=3)=0.375\).

Answer

<table> <tr><th>\(s\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(S=s)\)</th><td>\(0.125\)</td><td>\(0.125\)</td><td>\(0.375\)</td><td>\(0.375\)</td></tr> </table>
Combining random variables 17
A basketball team's total number of points in two parts of a practice drill is \(T=X+Y\). The average total is \(E(T)=10.5\) points, and the average from the first part is \(E(X)=4.25\) points. Find \(E(Y)\). Does your calculation require the parts to be independent?

Hints

- Write the relationship between the total score and its two parts. - Use that same relationship for the long-run averages. - The rule for means is different from the rule for variances.

Solution

1. Linearity of expectation gives \(E(T)=E(X)+E(Y)\). 2. Subtract the first average: \(E(Y)=10.5-4.25=6.25\) points. 3. Independence is not required for adding or subtracting expected values.

Answer

\(E(Y)=6.25\) points. Independence is not required.
Combining random variables 18
A student rides the bus to school on three days. A model gives probabilities \(0.8\), \(0.7\), and \(0.9\) that the bus arrives on time on Monday, Tuesday, and Wednesday. Let \(I_d=1\) when the bus arrives on time on day \(d\), and \(I_d=0\) otherwise. Find the expected total number \(I_M+I_T+I_W\) of on-time days. Must the three days be independent?

Hints

- What does each zero-one variable record? - Think about the average value of an indicator whose event occurs with a given probability. - Add averages for the count, even if the events may be related.

Solution

1. An indicator has expected value equal to the probability of the event it records. 2. Thus, \(E(I_M)=0.8\), \(E(I_T)=0.7\), and \(E(I_W)=0.9\). 3. The expected count is \(0.8+0.7+0.9=2.4\) days. Expectations add without requiring independence.

Answer

\(2.4\) on-time days. Independence is not required.
Combining random variables 19
A player rolls four independent fair six-sided dice. For each die, write \(I=1\) if it shows a six and \(I=0\) otherwise. Let \(T\) be the sum of these four indicators. a) Find \(E(T)\). b) Find the probability of at least one six. c) Explain why the answers to a) and b) are different.

Hints

- What does a single zero-one indicator average over many rolls? - Can you add the four expected values without listing all outcomes? - For the probability, consider the event that none of the dice shows a six.

Solution

1. Each indicator has expected value \(1\cdot\frac{1}{6}+0\cdot\frac{5}{6}=\frac{1}{6}\). 2. The expected total is \(E(T)=4\cdot\frac{1}{6}=\frac{2}{3}\). 3. The chance of at least one six is \(1-\left(\frac{5}{6}\right)^4=\frac{671}{1296}\approx0.5177\). 4. The expected count includes outcomes with more than one six, whereas the probability counts all outcomes with at least one six as one event.

Answer

a) \(E(T)=\frac{2}{3}\) sixes. b) \(P(T\ge1)=\frac{671}{1296}\approx0.5177\). c) An expected count and the probability of at least one event are different quantities.
Combining random variables 20
A microscope slide is divided into \(100\) equal square cells. A total of \(120\) pollen grains are distributed independently and uniformly among the cells. For one selected cell, find the probability that it contains: a) no pollen grains; b) exactly one pollen grain; c) exactly two pollen grains. Round each probability to four decimal places. Also find the expected number of the \(100\) cells that contain exactly one pollen grain, to two decimal places.

Hints

- Model the number of grains in one fixed cell. - Use binomial point probabilities. - Apply linearity of expectation across the \(100\) cells.

Solution

1. For one selected cell, \(X\sim\operatorname{Bin}(120,0.01)\). 2. \(P(X=0)\approx0.2994\). 3. \(P(X=1)\approx0.3629\). 4. \(P(X=2)\approx0.2181\). 5. The expected number of cells containing exactly one grain is \(100P(X=1)\approx36.29\).

Answer

a) \(0.2994\). b) \(0.3629\). c) \(0.2181\). Expected exactly-one cells: approximately \(36.29\).
Combining random variables 21
For quality control, a glass panel is divided into \(80\) equal sections. The panel contains \(60\) tiny air bubbles that are distributed independently and uniformly across the panel. a) Find the probability that one specified section contains more than \(2\) air bubbles. Round to four decimal places. b) Find the expected number of the \(80\) sections that contain no air bubbles. Give an exact expression and a decimal approximation to two decimal places. c) Explain why the expected-value calculation in part b) does not require the events “section \(i\) is empty” to be independent across sections.

Hints

- First model the number of bubbles in one specified section. - For the expected number of empty sections, introduce one indicator for each section. - Add the indicator expectations rather than trying to model all empty-section events jointly.

Solution

1. For one specified section, each bubble lands there with probability \(1/80\), so its bubble count \(X\sim\operatorname{Bin}(60,1/80)\). 2. \(P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)]\approx0.0395\). 3. A specified section is empty with probability \(P(X=0)=(79/80)^{60}\). 4. Let \(I_j\) indicate that section \(j\) is empty, and let \(T=I_1+\cdots+I_{80}\). Then \(E(T)=\sum_{j=1}^{80}E(I_j)=80(79/80)^{60}\approx37.61\). 5. Linearity of expectation holds whether or not the indicators \(I_j\) are independent, so independence between empty-section events is not needed.

Answer

a) \(0.0395\). b) \(80(79/80)^{60}\approx37.61\) empty sections. c) Linearity of expectation does not require independence.
Combining random variables 22
A fair six-sided die is rolled independently \(6\) times. Focus on one specified number, such as \(6\). 1. Find the probability that the specified number does not appear. Give an exact expression and a decimal approximation to four decimal places. 2. Find the probability that it appears exactly once. Give an exact expression and a decimal approximation to four decimal places. 3. Find the expected number of the six possible die values that do not appear in the six rolls. Give an exact expression and a decimal approximation to two decimal places. Determine whether this is more than one-third of the possible values.

Hints

- Model appearances of one specified value with a binomial distribution. - For part 3, use one missing-value indicator for each die face. - Apply linearity of expectation.

Solution

1. \(P(X=0)=(5/6)^6\approx0.3349\). 2. \(P(X=1)=6(1/6)(5/6)^5=(5/6)^5\approx0.4019\). 3. The expected number missing is \(6\left(\frac56\right)^6\approx2.01.\) Since one-third of \(6\) is \(2\), the expected number missing is slightly greater than one-third.

Answer

1. \((5/6)^6\approx0.3349\). 2. \((5/6)^5\approx0.4019\). 3. \(6(5/6)^6\approx2.01\), slightly more than one-third of the six values.
Combining random variables 23
Compare two random experiments. Experiment A: Two fair four-sided dice labeled \(1\) through \(4\) are rolled. The random variable \(X\) is their sum. Experiment B: A fair coin labeled \(1\) and \(2\), and a fair six-sided die labeled \(1\) through \(6\), are tossed and rolled. The random variable \(Y\) is their sum. a) Find \(P(X=5)\) and \(P(Y=5)\). Which is greater? b) Determine whether \(P(X\ge7)\) and \(P(Y\ge7)\) are equal. c) Calculate and compare \(E(X)\) and \(E(Y)\).

Hints

- Count the total equally likely outcomes for each experiment separately. - List all ordered pairs producing the requested sums. - “At least \(7\)” includes every sum of \(7\) or more. - Use linearity of expectation for each sum.

Solution

1. Experiment A has \(16\) equally likely ordered outcomes. Four produce a sum of \(5\), so \(P(X=5)=\frac{4}{16}=0.25\). 2. Experiment B has \(12\) equally likely ordered outcomes. Two produce a sum of \(5\), so \(P(Y=5)=\frac{2}{12}=\frac{1}{6}\approx0.1667\). Experiment A has the greater probability. 3. In Experiment A, sums of at least \(7\) occur in \(3\) outcomes, so \(P(X\ge7)=\frac{3}{16}=0.1875\). In Experiment B, they occur in \(3\) outcomes, so \(P(Y\ge7)=\frac{3}{12}=0.25\). The probabilities are not equal. 4. By linearity of expectation, \(E(X)=2.5+2.5=5\) and \(E(Y)=1.5+3.5=5\).

Answer

a) \(P(X=5)=0.25\) and \(P(Y=5)=\frac{1}{6}\approx0.1667\); Experiment A is greater. b) No. \(P(X\ge7)=0.1875\) and \(P(Y\ge7)=0.25\). c) \(E(X)=E(Y)=5\).
Combining random variables 24
Spin the spinner twice independently. Let \(X\) be the sum of the results. a) Create a probability distribution table for \(X\). b) Calculate \(E(X)\).
Figure for problem 532066

Hints

- Determine the probability of each label on one spin. - List all possible sums from two spins. - Add the probabilities of all ordered pairs that produce the same sum. - Compute an expected value by multiplying each possible value by its probability. - You can also use the expected value of one spin and linearity of expectation.

Solution

1. On one spin, \(P(0)=\frac{3}{6}=\frac{1}{2}\), \(P(2)=\frac{2}{6}=\frac{1}{3}\), and \(P(5)=\frac{1}{6}\). 2. Combining the two independent spins gives \(P(X=0)=\frac{1}{4}\), \(P(X=2)=\frac{1}{3}\), \(P(X=4)=\frac{1}{9}\), \(P(X=5)=\frac{1}{6}\), \(P(X=7)=\frac{1}{9}\), and \(P(X=10)=\frac{1}{36}\). 3. The expected value is \(E(X)=0\left(\frac{1}{4}\right)+2\left(\frac{1}{3}\right)+4\left(\frac{1}{9}\right)+5\left(\frac{1}{6}\right)+7\left(\frac{1}{9}\right)+10\left(\frac{1}{36}\right)=3\). 4. Equivalently, one spin has expected value \(0\left(\frac{1}{2}\right)+2\left(\frac{1}{3}\right)+5\left(\frac{1}{6}\right)=1.5\), so two spins have expected sum \(2\cdot1.5=3\).

Answer

a) <table> <thead> <tr><th>\(k\)</th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th><th>\(5\)</th><th>\(7\)</th><th>\(10\)</th></tr> </thead> <tbody> <tr><th>\(P(X=k)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{36}\)</td></tr> </tbody> </table> b) \(E(X)=3\).
Combining random variables 25
Spin the spinner twice independently. Let \(X\) be the product of the two numbers. a) Find the probability distribution of \(X\) and display it in a table. b) Find \(E(X)\). c) Replace the largest number on the spinner with a positive integer \(k\). Find \(k\) so that the expected product of two independent spins is \(16\).
Figure for problem 536090

Hints

- Read the one-spin values from the displayed spinner. - List the \(25\) ordered pairs and group equal products. - For independent spins, the expected product equals the product of the one-spin expectations. - In part c, find the new one-spin mean first.

Solution

1. Reading the spinner gives one-spin values \(1,2,3,4,5\), each with probability \(1/5\). There are \(25\) equally likely ordered pairs. 2. Grouping equal products gives the distribution in the answer. 3. From the distribution, \(E(X)=9\). Equivalently, independence gives \(E(X)=E(S_1)E(S_2)=3\cdot3=9\). 4. After replacing \(5\) by \(k\), the one-spin mean is \((10+k)/5\). 5. Require \(((10+k)/5)^2=16\). Since the mean is positive, \((10+k)/5=4\), so \(k=10\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(8\)</th><th>\(9\)</th><th>\(10\)</th><th>\(12\)</th><th>\(15\)</th><th>\(16\)</th><th>\(20\)</th><th>\(25\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac1{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac3{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac1{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac1{25}\)</td><td>\(\frac2{25}\)</td><td>\(\frac1{25}\)</td></tr> </table> b) \(E(X)=9\). c) \(k=10\).
Combining random variables 26
Four event indicators \(I_1,I_2,I_3,I_4\) have event probabilities \(0.20,0.35,0.50,\) and \(0.80\). Let \(N=I_1+I_2+I_3+I_4\) be the number of events that occur. Find \(E(N)\). Does this calculation require the events to be independent? Explain.

Hints

- Express the count as a sum of indicator variables. - Recall the expected value of an event indicator. - Separate the calculation of a mean from the calculation of a joint distribution.

Solution

1. Linearity of expectation gives \(E(N)=E(I_1)+E(I_2)+E(I_3)+E(I_4)\). 2. For an indicator, \(E(I_i)=P(I_i=1)\). 3. Thus \(E(N)=0.20+0.35+0.50+0.80=1.85\). 4. Linearity of expectation does not require independence; dependence affects the distribution of \(N\), but not this mean calculation.

Answer

\(E(N)=1.85\). Independence is not required.
Combining random variables 27
On a randomly selected workday, the number of incoming service requests \(X\) has mean \(52\) and standard deviation \(4\), while the number completed \(Y\) has mean \(47\) and standard deviation \(3\). Assume \(X\) and \(Y\) are independent. Let \(D=X-Y\) be the daily backlog increase. a) Find the mean and standard deviation of \(D\). b) Javier subtracts the standard deviations and reports \(1\). Explain the error.

Hints

- Treat the center and spread as separate calculations. - Consider what happens to a negative coefficient when it is squared. - Use the independence information when combining variability.

Solution

1. The mean is \(\mu_D=52-47=5\). 2. Independence allows the variances to add even for a difference: \(\sigma_D^2=4^2+(-1)^2(3^2)=25\). 3. Therefore, \(\sigma_D=5\). 4. Standard deviations are not subtracted because variability is combined through squared coefficients and variances.

Answer

a) \(\mu_D=5\,\text{requests}\) and \(\sigma_D=5\,\text{requests}\). b) The sign in a difference changes the mean contribution but disappears when the coefficient is squared for variance, so the variances add rather than the standard deviations subtracting.
Combining random variables 28
Independent random variables \(X\) and \(Y\) satisfy \(T=X+Y\). You know that \(\mu_T=50\), \(\sigma_T=6.5\), \(\mu_X=18\), and \(\sigma_X=2.5\). Find \(\mu_Y\) and \(\sigma_Y\).

Hints

- Work backward from the parameter rules for a sum. - Treat the center and variability equations separately. - Remember to solve for a nonnegative standard deviation.

Solution

1. From the mean of the sum, \(\mu_Y=50-18=32\). 2. Independence gives \(6.5^2=2.5^2+\sigma_Y^2\). 3. Therefore, \(\sigma_Y^2=42.25-6.25=36\), so \(\sigma_Y=6\).

Answer

\(\mu_Y=32\) and \(\sigma_Y=6\).
Combining random variables 29
A random variable \(X\) has mean \(8\) and standard deviation \(2.5\). A transformed score is \(Y=aX+7\). Choose \(a\) so that \(Y\) has mean \(31\), and then find the standard deviation of \(Y\).

Hints

- Use the target center to determine the unknown coefficient. - The added constant affects the center but not the spread. - After finding the coefficient, consider its magnitude when transforming standard deviation.

Solution

1. The mean condition gives \(8a+7=31\). 2. Solving gives \(a=3\). 3. The standard deviation is multiplied by \(|a|\), so \(\sigma_Y=3(2.5)=7.5\).

Answer

\(a=3\) and \(\sigma_Y=7.5\).
Combining random variables 30
Independent random variables \(X\) and \(Y\) have \(\mu_X=20\), \(\sigma_X=5\), \(\mu_Y=14\), and \(\sigma_Y=2\). Define \(S=X+Y\) and \(D=X-Y\). Find the mean and standard deviation of both \(S\) and \(D\), and explain why their standard deviations are equal. Give exact standard deviations and decimal approximations to two decimal places.

Hints

- Compute the two centers using their respective signs. - Compare how the coefficient of \(Y\) enters the variance calculation. - Focus on the coefficient after it is squared.

Solution

1. \(\mu_S=34\) and \(\mu_D=6\). 2. Independence gives \(\sigma_S^2=5^2+2^2=29\). 3. For \(D=X-Y\), the coefficient of \(Y\) is \(-1\), whose square is \(1\), so \(\sigma_D^2=29\). 4. Hence, \(\sigma_S=\sigma_D=\sqrt{29}\approx5.39.\)

Answer

\(\mu_S=34\), \(\sigma_S=\sqrt{29}\approx5.39\); \(\mu_D=6\), \(\sigma_D=\sqrt{29}\approx5.39\). The spreads are equal because the sign change does not change the squared variance contribution.
Combining random variables 31
Independent random variables \(X\) and \(Y\) have the distributions <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.70\)</td><td>\(0.30\)</td></tr> </table> <table> <tr><th>\(y\)</th><th>\(1\)</th><th>\(3\)</th></tr> <tr><th>\(P(Y=y)\)</th><td>\(0.40\)</td><td>\(0.60\)</td></tr> </table> Let \(S=X+Y\). Construct the probability distribution of \(S\), and find \(P(S\ge3)\).

Hints

- List every pair of possible component values and its resulting sum. - Use the relationship between the variables to obtain each pair probability. - Combine probabilities when different pairs produce the same sum.

Solution

1. Independence gives \(P(S=1)=P(X=0)P(Y=1)=0.70(0.40)=0.28\). 2. The value \(S=3\) can occur as \(0+3\) or \(2+1\), so \(P(S=3)=0.70(0.60)+0.30(0.40)=0.54\). 3. The value \(S=5\) occurs as \(2+3\), so \(P(S=5)=0.30(0.60)=0.18\). 4. Therefore, \(P(S\ge3)=0.54+0.18=0.72\).

Answer

<table> <tr><th>\(s\)</th><th>\(1\)</th><th>\(3\)</th><th>\(5\)</th></tr> <tr><th>\(P(S=s)\)</th><td>\(0.28\)</td><td>\(0.54\)</td><td>\(0.18\)</td></tr> </table> \(P(S\ge3)=0.72\)
Combining random variables 32
A project has a fixed total of \(30\) labor hours split between planning and production. Let \(X\) be the planning hours, with mean \(12\) and standard deviation \(3\). Let \(Y=30-X\) be the production hours. a) Find the mean and standard deviation of \(Y\). b) Find the mean and standard deviation of \(X+Y\). c) Explain why adding the variances of \(X\) and \(Y\) would be incorrect.

Hints

- Treat production hours as a transformation of planning hours. - Simplify the sum before applying any variance rule. - Consider how one component must change when the other component changes.

Solution

1. The transformation \(Y=30-X\) gives \(\mu_Y=30-12=18\) and \(\sigma_Y=|-1|(3)=3\). 2. Since \(X+Y=X+(30-X)=30\), the sum is constant. 3. Therefore, \(\mu_{X+Y}=30\) and \(\sigma_{X+Y}=0\). 4. The variables are perfectly dependent: an increase in planning hours forces an equal decrease in production hours, so their variations cancel.

Answer

a) \(\mu_Y=18\) hours and \(\sigma_Y=3\) hours. b) \(\mu_{X+Y}=30\) hours and \(\sigma_{X+Y}=0\) hours. c) The variables are not independent; they always move in opposite directions by equal amounts.
Combining random variables 33
Three independent course components have these parameters: assignments \(A\) have mean \(80\) and standard deviation \(6\), projects \(P\) have mean \(74\) and standard deviation \(8\), and the final exam \(F\) has mean \(90\) and standard deviation \(4\). The overall score is \(G=0.2A+0.3P+0.5F\). Find the mean and standard deviation of \(G\). Give the standard deviation exactly and to two decimal places.

Hints

- Match each coefficient to the corresponding component. - Combine means linearly. - Square the coefficients when combining independent variance contributions.

Solution

1. \(\mu_G=0.2(80)+0.3(74)+0.5(90)=83.2\). 2. Independence gives \(\sigma_G^2=(0.2)^2(6^2)+(0.3)^2(8^2)+(0.5)^2(4^2)=11.2.\) 3. Therefore, \(\sigma_G=\sqrt{11.2}=\frac{2\sqrt{70}}5\approx3.35.\)

Answer

\(\mu_G=83.2\) points and \(\sigma_G=\frac{2\sqrt{70}}5\approx3.35\) points.
Combining random variables 34
A system's total calibration error is the sum of \(n\) independent component errors. Each component error has mean \(0.3\) units and standard deviation \(1.2\) units. The total error has standard deviation \(4.8\) units. a) Find \(n\). b) Find the mean total error.

Hints

- Relate the total variability to the number of independent components. - Work with variance before solving for the count. - Once the count is known, combine the component centers.

Solution

1. Independence gives \(4.8^2=n(1.2^2)\). 2. Therefore, \(n=\left(\frac{4.8}{1.2}\right)^2=16\). 3. The total mean is \(16(0.3)=4.8\) units.

Answer

a) \(n=16\) b) The mean total error is \(4.8\) units.
Combining random variables 35
A supplier compares two packing plans. Plan A combines \(4\) cartons with independent weights. Each carton has mean weight \(12\,\text{kg}\) and standard deviation \(2\,\text{kg}\). Plan B combines \(2\) crates with independent weights. Each crate has mean weight \(24\,\text{kg}\) and standard deviation \(3\,\text{kg}\). For each plan, find the mean and standard deviation of the total weight. Which plan has the more variable total weight? Give exact standard deviations and round any decimal approximations to two decimal places.

Hints

- Treat each plan as a sum of independent, identically distributed components. - Compare the centers and spreads separately. - Use standard deviation, not the number of containers, to decide which total is more variable.

Solution

1. Plan A has mean \(4(12)=48\,\text{kg}\) and standard deviation \(2\sqrt{4}=4\,\text{kg}\). 2. Plan B has mean \(2(24)=48\,\text{kg}\) and standard deviation \(3\sqrt{2}\approx4.24\,\text{kg}\). 3. The means are equal, but Plan B has the larger standard deviation, so its total weight is more variable.

Answer

Plan A: mean \(48\,\text{kg}\), standard deviation \(4\,\text{kg}\). Plan B: mean \(48\,\text{kg}\), standard deviation \(3\sqrt{2}\,\text{kg}\approx4.24\,\text{kg}\). Plan B has the more variable total weight.
Combining random variables 36
Independent random variables \(X\) and \(Y\) have \(\mu_X=30\), \(\sigma_X=4\), \(\mu_Y=18\), and \(\sigma_Y=6\). Define \(R=2X-0.5Y+6\). Find the mean and standard deviation of \(R\). Give the standard deviation exactly and to two decimal places.

Hints

- Include the constant only in the center calculation. - Track each coefficient through the mean and variance rules. - Use independence before adding variance contributions.

Solution

1. \(\mu_R=2(30)-0.5(18)+6=57\). 2. Independence gives \(\sigma_R^2=2^2(4^2)+(-0.5)^2(6^2)=73.\) 3. Thus, \(\sigma_R=\sqrt{73}\approx8.54.\)

Answer

\(\mu_R=57\) and \(\sigma_R=\sqrt{73}\approx8.54\).
Combining random variables 37
The joint probability distribution of \(X\) and \(Y\) is shown. <table> <tr><th></th><th>\(Y=0\)</th><th>\(Y=2\)</th><th>\(Y=5\)</th></tr> <tr><th>\(X=1\)</th><td>\(0.10\)</td><td>\(0.20\)</td><td>\(0.15\)</td></tr> <tr><th>\(X=4\)</th><td>\(0.25\)</td><td>\(0.20\)</td><td>\(0.10\)</td></tr> </table> Let \(D=X-Y\). Construct the probability distribution of \(D\), and find \(P(D>0)\).

Hints

- Evaluate the difference for every joint outcome in the table. - Combine probabilities from cells that produce the same difference. - Select only positive values of the combined variable for the final event.

Solution

1. Map each joint outcome to \(X-Y\): the six cells give \(1,-1,-4,4,2,-1\). 2. The value \(-1\) occurs in two cells, so its probability is \(0.20+0.10=0.30\). 3. The full distribution is \(P(D=-4)=0.15\), \(P(D=-1)=0.30\), \(P(D=1)=0.10\), \(P(D=2)=0.20\), and \(P(D=4)=0.25\). 4. Therefore, \(P(D>0)=0.10+0.20+0.25=0.55\).

Answer

<table> <tr><th>\(d\)</th><th>\(-4\)</th><th>\(-1\)</th><th>\(1\)</th><th>\(2\)</th><th>\(4\)</th></tr> <tr><th>\(P(D=d)\)</th><td>\(0.15\)</td><td>\(0.30\)</td><td>\(0.10\)</td><td>\(0.20\)</td><td>\(0.25\)</td></tr> </table> \(P(D>0)=0.55\)
Combining random variables 38
A random variable \(X\) has mean \(5\) and standard deviation \(2\). Define \(Y=2X+1\) and \(T=X+Y\). a) Find the mean and standard deviation of \(Y\). b) Find the mean and standard deviation of \(T\). c) Explain why the variances of \(X\) and \(Y\) should not be added.

Hints

- First transform the original variable to obtain the parameters of the second variable. - Substitute the definition of the second variable into the total and simplify. - Check whether the two variables can vary independently of one another.

Solution

1. For \(Y=2X+1\), \(\mu_Y=2(5)+1=11\) and \(\sigma_Y=2(2)=4\). 2. Since \(T=X+(2X+1)=3X+1\), \(\mu_T=3(5)+1=16\) and \(\sigma_T=3(2)=6\). 3. The variables \(X\) and \(Y\) are perfectly dependent because \(Y\) is determined by \(X\), so the independent-variance rule does not apply.

Answer

a) \(\mu_Y=11\) and \(\sigma_Y=4\). b) \(\mu_T=16\) and \(\sigma_T=6\). c) \(Y\) is determined by \(X\), so the two variables are not independent.
Combining random variables 39
Independent standardized scores \(Z_1\) and \(Z_2\) each have mean \(0\) and standard deviation \(1\). Define \(C=0.6Z_1+0.8Z_2\). Find the mean and standard deviation of \(C\). Explain why the standard deviation is \(1\) even though the coefficients do not add to \(1\).

Hints

- Combine the centers separately from the spreads. - Use the relationship between the two standardized scores. - Compare the sum of the coefficient squares with the sum of the coefficients.

Solution

1. The mean is \(\mu_C=0.6(0)+0.8(0)=0\). 2. Independence gives \(\sigma_C^2=(0.6)^2(1^2)+(0.8)^2(1^2)=0.36+0.64=1\). 3. Therefore, \(\sigma_C=1\). 4. Variability depends on the squares of the coefficients, and \((0.6)^2+(0.8)^2=1\); the ordinary coefficient sum is not the relevant condition.

Answer

\(\mu_C=0\) and \(\sigma_C=1\). The squared coefficients, not the coefficients themselves, determine the variance contribution.
Combining random variables 40
Independent random variables \(X\) and \(Y\) have \(\mu_X=4\), \(\sigma_X=1.5\), \(\mu_Y=7\), and \(\sigma_Y=2\). A new variable is \(W=aX+Y\). Find the value of \(a\) that makes \(\mu_W=-1\), and then find \(\sigma_W\).

Hints

- Use the target center to determine the unknown coefficient first. - Treat a negative coefficient carefully when finding spread. - Use the stated relationship between the component variables.

Solution

1. The mean condition is \(4a+7=-1\). 2. Solving gives \(a=-2\). 3. Independence gives \(\sigma_W=\sqrt{(-2)^2(1.5)^2+2^2}=\sqrt{13}\approx3.61\).

Answer

\(a=-2\) and \(\sigma_W=\sqrt{13}\approx3.61\).
Combining random variables 41
On a certain morning, the Celsius temperatures at two independent monitoring sites are random variables \(C_1\) and \(C_2\). Their means are \(18^\circ\text{C}\) and \(12^\circ\text{C}\), and their standard deviations are \(2^\circ\text{C}\) and \(1.5^\circ\text{C}\), respectively. Fahrenheit temperature is given by \(F=1.8C+32\). Find the mean and standard deviation of the Fahrenheit difference \(D=F_1-F_2\).

Hints

- Write both converted temperatures before taking their difference. - Look for terms that cancel when the two expressions are subtracted. - Apply the independence information only when combining variability.

Solution

1. Subtracting the two conversions gives \(D=(1.8C_1+32)-(1.8C_2+32)=1.8(C_1-C_2)\). 2. The mean is \(\mu_D=1.8(18-12)=10.8^\circ\text{F}\). 3. Independence gives \(\sigma_D=1.8\sqrt{2^2+1.5^2}=1.8(2.5)=4.5^\circ\text{F}\).

Answer

The mean difference is \(10.8^\circ\text{F}\), and the standard deviation is \(4.5^\circ\text{F}\).
Combining random variables 42
Independent random variables \(X\) and \(Y\) satisfy \(\sigma_X=4\). The variable \(R=2X+3Y\) has standard deviation \(13\). Find \(\sigma_Y\).

Hints

- Translate the spread of the combination into an equation. - Account for the coefficients before isolating the unknown spread. - The final standard deviation must be nonnegative.

Solution

1. Independence gives \(13^2=2^2(4^2)+3^2\sigma_Y^2\). 2. Thus, \(169=64+9\sigma_Y^2\), so \(\sigma_Y^2=\frac{105}{9}=\frac{35}{3}\). 3. Therefore, \(\sigma_Y=\sqrt{\frac{35}{3}}\approx3.42\).

Answer

\(\sigma_Y=\sqrt{\frac{35}{3}}\approx3.42\).
Combining random variables 43
Two independent signal measurements, \(X\) and \(Y\), have means \(8\) and \(5\) volts and standard deviations \(1.5\) and \(2\) volts. Both readings are affected by the same random calibration offset \(C\), whose distribution is unknown. The displayed readings are \(A=X+C\) and \(B=Y+C\). Find the mean and standard deviation of the displayed difference \(D=A-B\). Explain why no information about \(C\) is needed.

Hints

- Substitute the displayed-reading definitions into the difference. - Simplify before trying to combine any parameters. - Use independence only for variables that remain after simplification.

Solution

1. The common offset cancels: \(D=(X+C)-(Y+C)=X-Y\). 2. The mean is \(\mu_D=8-5=3\) volts. 3. Since \(X\) and \(Y\) are independent, \(\sigma_D=\sqrt{1.5^2+2^2}=2.5\) volts. 4. The random variable \(C\) does not appear in the simplified difference.

Answer

\(\mu_D=3\) volts and \(\sigma_D=2.5\) volts. The shared calibration offset cancels exactly.
Combining random variables 44
A study measures the same participants before and after a program. Let \(\bar X_{\text{before}}\) and \(\bar X_{\text{after}}\) have variances \(25\) and \(16\), with covariance \(18\). The observed means are \(68\) and \(72\). a) Give the observed change, defined as after minus before. b) Find the variance and standard deviation of \(\bar X_{\text{after}}-\bar X_{\text{before}}.\) Give the standard deviation exactly and to three decimal places. c) Compare the variance with the value obtained by incorrectly treating the two means as independent. Explain the effect of the positive covariance.

Hints

- Preserve the requested subtraction order. - Use the variance formula for a difference of dependent random variables. - Track the sign of the covariance term. - Compare the dependent and independent calculations.

Solution

1. The observed change is \(72-68=4.\) 2. For a difference of dependent random variables, \(\operatorname{Var}(\bar X_{\text{after}}-\bar X_{\text{before}}) =16+25-2(18)=5.\) 3. Therefore, \(\operatorname{SD}(\bar X_{\text{after}}-\bar X_{\text{before}}) =\sqrt5\approx2.236.\) 4. Incorrectly treating the means as independent would omit the covariance term and give \(16+25=41.\) Positive covariance reduces the variance of a difference, so the independence calculation greatly overstates the spread.

Answer

a) \(4\). b) Variance \(5\); standard deviation \(\sqrt5\approx2.236\). c) The incorrect independence calculation gives variance \(41\). Positive covariance reduces the variance of the difference.
Combining random variables 45
A fair spinner has \(32\) equal sections labeled \(1\) through \(32\). It is spun independently \(100\) times. a) Let \(X\) be the number of times the spinner lands on \(13\). Find \(P(X=0)\), \(P(X=1)\), \(P(X=2)\), and \(P(X>2)\), each to four decimal places. b) Find the expected number of different sections that are landed on at least once. Give an exact expression and a decimal approximation to two decimal places.

Hints

- Model the number of hits on one specified section with a binomial distribution. - Use a complement for more than \(2\). - For part b), use one indicator for each section and linearity of expectation.

Solution

1. \(X\sim\operatorname{Bin}(100,\frac1{32})\). 2. \(P(X=0)\approx0.0418\), \(P(X=1)\approx0.1348\), and \(P(X=2)\approx0.2153\). 3. \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.6081\). 4. For any particular section, the probability of being hit at least once is \(1-\left(\frac{31}{32}\right)^{100}.\) 5. By linearity of expectation, the expected number of distinct sections reached is \(32\left[1-\left(\frac{31}{32}\right)^{100}\right]\approx30.66.\)

Answer

a) \(0.0418,\ 0.1348,\ 0.2153,\ 0.6081\), respectively. b) \(32[1-(31/32)^{100}]\approx30.66\) sections.
Combining random variables 46
A company produces a batch of \(1000\) microchips. During production, \(50\) tiny surface defects occur and are distributed independently and uniformly across the chips. a) Find the probability that a specified chip has no surface defects. Round to four decimal places. b) Find the expected number of chips in the batch that have exactly \(2\) surface defects. Give a decimal approximation to two decimal places. c) For one specified chip, explain what counts as a “success” and what represents the number of trials.

Hints

- Treat each defect placement as one trial. - Use a binomial model for one specified chip. - Multiply the one-chip probability by \(1000\) to obtain the expected number of chips.

Solution

1. For one specified chip, \(X\sim\operatorname{Bin}(50,0.001)\). 2. \(P(X=0)=(0.999)^{50}\approx0.9512\). 3. \(P(X=2)=\binom{50}{2}(0.001)^2(0.999)^{48}\approx0.0011676\). 4. By linearity of expectation, the expected number of chips with exactly \(2\) defects is \(1000P(X=2)\approx1.17.\) 5. A success is one defect landing on the specified chip; the \(50\) defects are the trials.

Answer

a) \(0.9512\). b) Approximately \(1.17\) chips. c) Success = one defect lands on the specified chip; \(n=50\) trials = the \(50\) defects.
Combining random variables 47
In a biological study, \(200\) equal square sample areas are examined for bacterial colonies. A total of \(80\) colonies are distributed independently and uniformly across the full area. a) Find the probability that a specified sample area contains at least \(3\) colonies. Round to four decimal places. b) Find the expected number of the \(200\) sample areas that contain no colonies. Give a decimal approximation to two decimal places. c) Briefly explain why a binomial model is appropriate for one specified area.

Hints

- Use a complement for “at least \(3\).” - Find the probability that one specified area is empty. - Use linearity of expectation across the \(200\) areas.

Solution

1. For one specified area, \(X\sim\operatorname{Bin}(80,0.005)\). 2. Using the complement, \(P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)\approx0.0077.\) 3. \(P(X=0)=(0.995)^{80}\). 4. By linearity of expectation, the expected number of empty areas is \(200(0.995)^{80}\approx133.93.\) 5. Each colony placement is an independent trial with constant probability \(1/200\) of landing in the specified area.

Answer

a) \(P(X\ge3)\approx0.0077\). b) Approximately \(133.93\) sample areas. c) The \(80\) placements are independent trials with the same probability \(1/200\) for the specified area.
Combining random variables 48
A fair \(20\)-sided die numbered \(1\) through \(20\) is rolled independently. a) What is the probability that \(20\) does not appear in \(15\) rolls? Give an exact expression and a decimal approximation to four decimal places. b) After \(n\) rolls, exactly \(4\) of the \(20\) possible numbers have not appeared. Estimate \(n\) by setting the observed number of missing values equal to its expected value. Report the real-valued estimate to two decimal places and then give the nearest whole-number estimate.

Hints

- First find the probability that one specified number is still missing. - Use linearity of expectation across all \(20\) possible numbers. - Use logarithms to solve the equation with \(n\) in the exponent.

Solution

1. The probability of avoiding \(20\) on one roll is \(19/20\), so \(P(\text{no }20\text{ in }15)=\left(\frac{19}{20}\right)^{15}\approx0.4633.\) 2. A specified number is missing after \(n\) rolls with probability \((19/20)^n\). 3. Hence the expected number missing is \(20(19/20)^n\). 4. Setting this equal to \(4\) gives \(n=\frac{\ln(0.2)}{\ln(0.95)}\approx31.38.\) The nearest whole-number estimate is \(31\) rolls.

Answer

a) \((19/20)^{15}\approx0.4633\). b) \(n\approx31.38\), so about \(31\) rolls.
Combining random variables 49
A bakery mixes \(450\) raisins into dough for a batch of \(150\) raisin rolls. Assume the raisins are distributed independently and uniformly among the rolls. a) Find the probability that a specified roll contains exactly \(3\) raisins. Round to four decimal places. b) Find the expected number of rolls in the batch that contain more than \(4\) raisins. Give a decimal approximation to two decimal places. c) What is the minimum total number of raisins needed so that the probability a specified roll contains no raisins is at most \(5\%\)?

Hints

- Use a binomial model for one specified roll. - Multiply the one-roll tail probability by \(150\) for the expected count. - Use logarithms for the inverse zero-raisin condition.

Solution

1. For a specified roll, \(X\sim\operatorname{Bin}(450,\frac1{150})\). 2. \(P(X=3)\approx0.2248\). 3. \(P(X>4)\approx0.1842\). 4. Thus, the expected number of rolls with more than \(4\) raisins is \(150P(X>4)\approx27.63\). 5. With \(n\) raisins, require \((149/150)^n\le0.05\), giving \(n\ge447.86\ldots\). The minimum integer is \(448\).

Answer

a) \(0.2248\). b) Approximately \(27.63\) rolls. c) \(448\) raisins.
Combining random variables 50
A quality-control study examines \(200\) computer monitors. A total of \(n\) pixel defects are assumed to be distributed independently and uniformly among the monitors. a) Suppose the average number of pixel defects is \(1.5\) per monitor. Find the probability that a specified monitor has no pixel defects, to four decimal places, and find the expected number of the \(200\) monitors that have at least one defect, to two decimal places. b) In another batch of \(200\) monitors, exactly \(90\) monitors have no pixel defects. Use proportion matching: equate the observed empty-monitor proportion \(90/200\) to the model probability that a specified monitor is empty, then estimate the total number \(n\) of pixel defects. Give the real-valued estimate to two decimal places, then the nearest whole-number estimate.

Hints

- For a specified monitor, first find the probability that one defect goes somewhere else. - Use an indicator for whether each monitor has at least one defect when finding the expected affected count. - In part b), follow the stated proportion-matching rule before solving the exponential equation. - Use logarithms to solve for the unknown total number of defects.

Solution

1. An average of \(1.5\) defects per monitor corresponds to \(300\) defects. 2. For one specified monitor, the probability that one defect does not land there is \(199/200\). Independence of defect locations gives \(P(X=0)=\left(\frac{199}{200}\right)^{300}\approx0.2223\). 3. The expected number of monitors with at least one defect is \(200\left[1-\left(\frac{199}{200}\right)^{300}\right]\approx155.54\). 4. For part b), the requested proportion-matching estimate sets the observed empty proportion equal to the model empty probability: \(\left(\frac{199}{200}\right)^n=90/200=0.45\). 5. Taking logarithms gives \(n=\frac{\ln(0.45)}{\ln(199/200)}\approx159.30\). 6. The corresponding nearest whole-number estimate is \(159\) defects.

Answer

a) \(P(X=0)\approx0.2223\); expected affected monitors \(\approx155.54\). b) Proportion matching gives \(n\approx159.30\), or about \(159\) defects.
Combining random variables 51
A regional hospital has \(500\) employees. For a simplified model, assume their birthdays are independently and uniformly distributed across the \(365\) days of a year. a) Find the probability that no employee has a birthday on one specified day. Round to four decimal places. b) Find the expected number of days in the year on which no employee has a birthday. Round to two decimal places. c) Find the expected number of days on which more than one employee has a birthday. Round to two decimal places.

Hints

- Start with the chance of a birthday falling on one specified day. - Extend a single-day probability across the year using the sum of expected counts. - For more than one birthday, consider the complement of zero or exactly one.

Solution

1. For one specified day, the birthday count is binomial with \(n=500\) and \(p=\frac{1}{365}\). 2. The probability of no birthdays that day is \(P(X=0)=\left(\frac{364}{365}\right)^{500}\approx0.2537\). 3. The expected number of days with no birthdays is \(365P(X=0)\approx92.59\). 4. The chance of exactly one birthday on a specified day is \(P(X=1)=500\cdot\frac{1}{365}\left(\frac{364}{365}\right)^{499}\approx0.3484\). 5. Hence \(P(X>1)=1-P(X=0)-P(X=1)\approx0.3979\). 6. The expected number of days with more than one birthday is \(365P(X>1)\approx145.23\).

Answer

a) \(0.2537\). b) Approximately \(92.59\) days. c) Approximately \(145.23\) days.
Combining random variables 52
A random-number generator creates a list of \(100\) values. Each value is selected independently and uniformly from the integers \(1\) through \(100\). a) Find the expected number of values in \(\{1,\ldots,100\}\) that do not appear in the list. Give an exact expression and a decimal approximation to two decimal places. b) Find the expected number of values that appear exactly twice in the list. Give an exact expression and a decimal approximation to two decimal places.

Hints

- Focus first on one fixed integer. - Use an indicator for whether that integer has the requested frequency. - Apply linearity of expectation across the \(100\) possible integers.

Solution

1. For one specified integer, \(X\sim\operatorname{Bin}(100,0.01)\). 2. The probability it is missing is \((0.99)^{100}\), so the expected number missing is \(100(0.99)^{100}\approx36.60.\) 3. The probability it appears exactly twice is \(\binom{100}{2}(0.01)^2(0.99)^{98}.\) 4. Therefore, the expected number appearing exactly twice is \(100\binom{100}{2}(0.01)^2(0.99)^{98}\approx18.49.\)

Answer

a) \(100(0.99)^{100}\approx36.60\). b) \(100\binom{100}{2}(0.01)^2(0.99)^{98}\approx18.49\).
Combining random variables 53
A digital music player has a playlist of \(50\) different songs. In shuffle mode, each new song is selected independently and uniformly from all \(50\) songs, so immediate repeats are possible. A total of \(50\) songs are played. a) What is the probability that one specified favorite song is never played? Round to four decimal places. b) What is the probability that the favorite song is played at least twice? Round to four decimal places. c) Find the expected number of the \(50\) songs that are never played. Give a decimal approximation to two decimal places.

Hints

- Model the number of appearances of one specified song. - Use the complement of zero or one appearance for part b). - Use one never-played indicator for each song in part c).

Solution

1. For one specified song, \(X\sim\operatorname{Bin}(50,0.02)\). 2. \(P(X=0)=0.98^{50}\approx0.3642\). 3. \(P(X=1)=50(0.02)(0.98)^{49}=0.98^{49}\). 4. Therefore, \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642\). 5. By linearity of expectation, the expected number never played is \(50(0.98)^{50}\approx18.21\).

Answer

a) \(0.3642\). b) \(0.2642\). c) Approximately \(18.21\) songs.
Combining random variables 54
Events \(A\) and \(B\) have probabilities \(0.30\) and \(0.50\). Let \(N=I_A+I_B\) be the number of the two events that occur. a) Find \(\operatorname{Var}(N)\) if \(A\) and \(B\) are independent. b) Find \(\operatorname{Var}(N)\) if \(A\) and \(B\) are mutually exclusive. c) Explain why \(E(N)\) is the same in both cases but the variances differ.

Hints

- Express the event count as a sum of indicator variables. - Independence removes covariance; mutual exclusivity makes it negative. - Compare which moments depend only on marginal probabilities.

Solution

1. Always, \(E(N)=P(A)+P(B)=0.80\). 2. Under independence, the indicators have zero covariance, so \(\operatorname{Var}(N)=0.30\cdot0.70+0.50\cdot0.50=0.21+0.25=0.46\). 3. Under mutual exclusivity, \(E(I_AI_B)=0\), so \(\operatorname{Cov}(I_A,I_B)=0-0.30\cdot0.50=-0.15\). 4. Thus \(\operatorname{Var}(N)=0.46+2\cdot(-0.15)=0.16\). 5. Expectation adds without independence, but variance includes covariance and therefore depends on the relationship.

Answer

a) Independent: \(\operatorname{Var}(N)=0.46\). b) Mutually exclusive: \(\operatorname{Var}(N)=0.16\). c) The mean uses only the marginals; the variance also uses covariance.
Combining random variables 55
A random variable \(X\) has mean \(10\) and standard deviation \(4\). A transformed variable has the form \(Y=aX+b\). Find all pairs \((a,b)\) for which \(Y\) has mean \(25\) and standard deviation \(6\).

Hints

- Use the target spread to determine the possible magnitudes of the multiplier. - Remember that two multiplier signs can produce the same standard deviation. - Use the target mean separately for each possible multiplier.

Solution

1. The standard deviation condition gives \(|a|(4)=6\), so \(a=1.5\) or \(a=-1.5\). 2. If \(a=1.5\), the mean condition \(1.5(10)+b=25\) gives \(b=10\). 3. If \(a=-1.5\), the mean condition \(-1.5(10)+b=25\) gives \(b=40\). 4. Both transformations have the required center and spread.

Answer

\((a,b)=(1.5,10)\) or \((a,b)=(-1.5,40)\).
Combining random variables 56
Random variables \(X\) and \(Y\) each have mean \(10\) and standard deviation \(2\). Let \(S=X+Y\). Compare these three dependence structures: a) \(X\) and \(Y\) are independent. b) \(Y=X\). c) \(Y=20-X\). For each case, find the mean and standard deviation of \(S\), then rank the cases from least to greatest variability.

Hints

- The center of a sum does not require independence. - Rewrite the sum algebraically in the cases where one variable is defined from the other. - Compare how much the two variables reinforce or cancel one another.

Solution

1. In every case, \(\mu_S=10+10=20\). 2. In a), independence gives \(\sigma_S=\sqrt{2^2+2^2}=\sqrt{8}\approx2.83\). 3. In b), \(S=2X\), so \(\sigma_S=2(2)=4\). 4. In c), \(S=X+(20-X)=20\), so \(\sigma_S=0\). 5. The variability ranking is c), a), b).

Answer

a) \(\mu_S=20\), \(\sigma_S=\sqrt{8}\approx2.83\). b) \(\mu_S=20\), \(\sigma_S=4\). c) \(\mu_S=20\), \(\sigma_S=0\). Least to greatest variability: c), a), b).
Combining random variables 57
Independent random variables \(X\) and \(Y\) satisfy \(P(X=0)=0.60\), \(P(X=1)=0.40\), and \(Y\) can equal \(1\) or \(4\). Let \(q=P(Y=4)\), and define \(T=3X+Y\). It is known that \(P(T=4)=0.46\). a) Find \(q\). b) Construct the probability distribution of \(T\).

Hints

- Identify every pair of component values that can produce the stated combined value. - Express the known probability in terms of the unknown component probability. - After recovering the component distribution, group all pairs by their transformed totals.

Solution

1. The value \(T=4\) occurs from \((X,Y)=(0,4)\) or \((1,1)\). 2. Independence gives \(P(T=4)=0.60q+0.40(1-q)=0.40+0.20q\). 3. Solving \(0.40+0.20q=0.46\) gives \(q=0.30\). 4. Then \(P(T=1)=0.60(0.70)=0.42\), \(P(T=4)=0.46\), and \(P(T=7)=0.40(0.30)=0.12\).

Answer

a) \(q=0.30\). b) <table> <tr><th>\(t\)</th><th>\(1\)</th><th>\(4\)</th><th>\(7\)</th></tr> <tr><th>\(P(T=t)\)</th><td>\(0.42\)</td><td>\(0.46\)</td><td>\(0.12\)</td></tr> </table>
Combining random variables 58
Independent random variables \(X\) and \(Y\) each have mean \(50\) and standard deviation \(10\). A weighted score is \(W=aX+(1-a)Y\), where \(0\le a\le1\). Find all values of \(a\) for which \(W\) has standard deviation \(10\sqrt{0.58}\).

Hints

- Express the spread of the weighted score in terms of the unknown weight. - Use the fact that the two weights add to one. - Check every algebraic solution against the allowed interval.

Solution

1. Independence gives \(\sigma_W^2=a^2(10^2)+(1-a)^2(10^2)\). 2. The target variance condition becomes \(a^2+(1-a)^2=0.58\). 3. Simplifying gives \(2a^2-2a+0.42=0\). 4. The solutions are \(a=0.3\) and \(a=0.7\), and both lie in \([0,1]\).

Answer

\(a=0.3\) or \(a=0.7\).
Combining random variables 59
An energy trader will buy \(10\) contracts. Each contract tied to supplier A has the same random daily settlement \(X\), and each contract tied to supplier B has the same random daily settlement \(Y\). The variables \(X\) and \(Y\) are independent, both have mean \(100\), and their standard deviations are \(4\) and \(2\), respectively. If \(a\) contracts are tied to A and \(10-a\) are tied to B, the total settlement is \(T=aX+(10-a)Y\), where \(a\) is an integer from \(0\) to \(10\). Choose \(a\) to minimize the standard deviation of \(T\), and give that minimum standard deviation.

Hints

- Write the total variability as a function of the allocation. - Rewrite the resulting quadratic so its minimum is visible. - Confirm that the minimizing value is an allowed whole-number allocation.

Solution

1. The variance is \(\sigma_T^2=16a^2+4(10-a)^2\). 2. Expanding gives \(\sigma_T^2=20a^2-80a+400=20(a-2)^2+320\). 3. The minimum occurs at the allowed integer \(a=2\). 4. The minimum standard deviation is \(\sqrt{320}=8\sqrt{5}\approx17.89\).

Answer

Choose \(a=2\) contracts tied to A and \(8\) tied to B. The minimum standard deviation is \(8\sqrt{5}\approx17.89\).

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