Free math worksheets

Build your own math worksheets from 30,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Combining random variables

Click problems to add them to your worksheet.

Difficulty: 1 – 5

55629312

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.

55629412

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.

54743012

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.

54743312

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.

54743612

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.

54743712

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.

54744912

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.

54745712

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.

54745812

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.

54755712

For samples of size \(16\), the sampling distribution of \(\bar X\) has mean \(12\) and standard deviation \(1.5\). Let \(S\) be the sample sum from the same samples. a) Find the mean and standard deviation of the sampling distribution of \(S\). b) If a particular sample has \(\bar x=14\), what is its sample sum?

Hints

- Write the exact relationship between the sum and mean for a fixed sample size. - Apply that transformation to both the center and spread. - Use the same relationship for the particular observed sample.

Solution

1. For every sample, \(S=16\bar X\). 2. Therefore, \(E(S)=16(12)=192\). 3. The standard deviation is \(16(1.5)=24\). 4. If \(\bar x=14\), then \(S=16(14)=224\).

Answer

a) Mean \(192\); standard deviation \(24\). b) The sample sum is \(224\).

53109312

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 \(36.29\).

53109412

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.

53109512

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.

53118412

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) \(E(X)=E(Y)=5\).

53206612

The spinner shown has \(6\) equal sections labeled \(0\), \(2\), and \(5\). The spinner is spun twice. The random variable \(X\) is the sum of the two 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) \(E(X)=3\)

53609012

The spinner shown has \(5\) equal sections. It is spun twice independently, and \(X\) is the product of the two numbers. a) Find the probability distribution of \(X\) and display it in a table. b) Find \(E(X)\). c) The section currently labeled \(5\) is replaced by a positive integer \(k\). Find \(k\) so that the expected product of two 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) \(k=10\).

54736112

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.

54743112

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.

54743212

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 \(\sigma_Y=6\).

54743412

Machine \(A\) produces parts with mean length \(100\,\text{mm}\) and standard deviation \(6\,\text{mm}\). Machine \(B\) produces parts with mean length \(98\,\text{mm}\) and standard deviation \(9\,\text{mm}\). Four independent parts from \(A\) and nine independent parts from \(B\) are selected, and the two samples are independent. Let \(D=\overline{X}_A-\overline{X}_B\). Find the mean and standard deviation of \(D\). Give the standard deviation exactly and to two decimal places.

Hints

- First determine the spread of each sample average. - Combine the two centers with the sign used in the definition of the difference. - Use independence when combining the two variance contributions.

Solution

1. For Machine \(A\), \(\mu_{\overline{X}_A}=100\) and \(\sigma_{\overline{X}_A}=6/\sqrt{4}=3\). For Machine \(B\), \(\mu_{\overline{X}_B}=98\) and \(\sigma_{\overline{X}_B}=9/\sqrt{9}=3\). 2. The mean difference is \(\mu_D=100-98=2\,\text{mm}\). 3. Independence gives \(\sigma_D=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}\,\text{mm}\approx4.24\,\text{mm}\).

Answer

\(\mu_D=2\,\text{mm}\) and \(\sigma_D=3\sqrt{2}\,\text{mm}\approx4.24\,\text{mm}\).

54743512

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 \(\sigma_Y=7.5\).

54743812

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.

54743912

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> \(P(S\ge3)=0.72\)

54744012

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.

54744112

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.

54744212

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.

54744312

A supplier considers two independent-packing plans. Plan A combines \(4\) cartons, each with mean weight \(12\,\text{kg}\) and standard deviation \(2\,\text{kg}\). Plan B combines \(2\) crates, each with 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?

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 \(3\sqrt{2}\approx4.24\,\text{kg}\). Plan B has the more variable total weight.

54744412

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 \(\sigma_R=\sqrt{73}\approx8.54\).

54744512

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> \(P(D>0)=0.55\)

54744712

Nine independent measurements have the same distribution. Their average \(\overline{X}\) has mean \(12\) and standard deviation \(1.5\). Find the mean and standard deviation of one measurement.

Hints

- Relate the center of an average to the center of its components. - Work backward from the way averaging changes spread. - Use the number of independent measurements in the scaling factor.

Solution

1. An average has the same mean as each identically distributed component, so \(\mu_X=12\). 2. For nine independent measurements, \(\sigma_{\overline{X}}=\frac{\sigma_X}{\sqrt{9}}\). 3. Therefore, \(1.5=\frac{\sigma_X}{3}\), so \(\sigma_X=4.5\).

Answer

One measurement has mean \(12\) and standard deviation \(4.5\).

54744812

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.

54745012

Independent instrument readings have standard deviation \(3.2\) units. A technician will average \(n\) readings. Find the smallest integer \(n\) for which the standard deviation of the average is at most \(0.8\) units.

Hints

- Express the spread of an average in terms of the number of independent readings. - Translate “at most” into an inequality. - Check that the final value is an integer count and satisfies the requirement.

Solution

1. The standard deviation of the average is \(\frac{3.2}{\sqrt{n}}\). 2. Require \(\frac{3.2}{\sqrt{n}}\le0.8\), which gives \(\sqrt{n}\ge4\). 3. Therefore, \(n\ge16\), so the smallest integer is \(16\).

Answer

\(n=16\) readings.

54745112

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.

54745312

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 \(\sigma_W=\sqrt{13}\approx3.61\).

54745512

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 \(4.5^\circ\text{F}\).

54745612

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

\(\sigma_Y=\sqrt{\frac{35}{3}}\approx3.42\).

54746112

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.

54761612

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.

54848812

A company knows that \(16\%\) of the parts produced by a certain machine need rework. A quality engineer will inspect a random sample of \(250\) parts and treat the sampled outcomes as independent. Let \(X\) be the number that need rework and let \(\hat p=X/250\). a) Find the mean and standard deviation of \(X\). Give the standard deviation exactly and to four decimal places. b) Find the mean and standard deviation of \(\hat p\). Give the standard deviation exactly and to five decimal places. c) Explain the numerical relationship between the two means and the two standard deviations.

Hints

- Keep track of whether the random variable is a count or a proportion. - Use the rule for multiplying a random variable by a constant. - Check that each standard deviation has the same units as its random variable.

Solution

1. \(E(X)=250(0.16)=40\), and \(\sigma_X=\sqrt{250(0.16)(0.84)}=\sqrt{33.6}\approx5.7966.\) 2. Since \(\hat p=X/250\), \(E(\hat p)=\frac{40}{250}=0.16,\) and \(\sigma_{\hat p}=\frac{\sqrt{33.6}}{250} =\sqrt{\frac{0.16(0.84)}{250}} \approx0.02319.\) 3. Dividing \(X\) by \(250\) divides both its mean and standard deviation by \(250\).

Answer

a) Mean \(40\); standard deviation \(\sqrt{33.6}\approx5.7966\) parts. b) Mean \(0.16\); standard deviation \(\frac{\sqrt{33.6}}{250}\approx0.02319\). c) Both center and spread are divided by \(250\) because \(\hat p=X/250\).

54850112

A random sample of \(100\) independent observations from a population with proportion \(p=0.36\) is divided into two groups of \(50\). Let \(\hat p_1\) and \(\hat p_2\) be the sample proportions in the two groups, and let \(A=\frac{\hat p_1+\hat p_2}{2}.\) a) Find the standard deviation of each half-sample proportion. Give an exact expression and a decimal approximation to five decimal places. b) Find the standard deviation of \(A\), assuming the two groups are independent. c) Compare it with the standard deviation of the proportion from the full sample of \(100\).

Hints

- Compare the sample sizes used by the half-sample and full-sample proportions. - Use independence when averaging the two half-sample proportions. - Check whether averaging equal-size group proportions reproduces the overall proportion.

Solution

1. Each half-sample proportion has standard deviation \(\sqrt{\frac{0.36(0.64)}{50}} =\sqrt{0.004608} \approx0.06788.\) 2. Since the two half-sample proportions are independent, \(\sigma_A=\frac{1}{2}\sqrt{\sigma_{\hat p_1}^2+\sigma_{\hat p_2}^2} =0.048.\) 3. The full-sample proportion has standard deviation \(\sqrt{\frac{0.36(0.64)}{100}}=0.048.\) 4. The average of two equal-size half-sample proportions is exactly the full-sample proportion, so their standard deviations agree.

Answer

a) \(\sqrt{0.004608}\approx0.06788\) for each half-sample proportion. b) \(0.048\). c) The full-sample proportion also has standard deviation \(0.048\).

53107512

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.

53107712

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.

53107812

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.

53108112

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.

53108712

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.

53108812

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.

53109712

A company has \(500\) employees. Assume birthdays are distributed independently and uniformly 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. Give a decimal approximation to two decimal places. c) Find the expected number of days on which more than one employee has a birthday. Give a decimal approximation to two decimal places.

Hints

- Start with one specified day. - Use linearity of expectation to extend a one-day probability to all \(365\) days. - “More than one” is the complement of zero or exactly one.

Solution

1. For one specified day, \(X\sim\operatorname{Bin}(500,\frac1{365})\). 2. \(P(X=0)=(364/365)^{500}\approx0.2537\). 3. The expected number of empty-birthday days is \(365P(X=0)\approx92.59\). 4. \(P(X=1)=500(1/365)(364/365)^{499}\approx0.3484\). 5. Thus, \(P(X>1)=1-P(X=0)-P(X=1)\approx0.3979\). 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.

53109812

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) \(100\binom{100}{2}(0.01)^2(0.99)^{98}\approx18.49\).

53119612

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.

54736312

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.

54744612

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 \((a,b)=(-1.5,40)\).

54745212

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).

54745412

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>

54745912

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 \(a=0.7\).

54746012

A company 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 \(8\sqrt{5}\approx17.89\).

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.