AP Statistics Quiz: Combining Random Variables

20 questions · exam conditions

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Combining Random VariablesQuestion 1 of 20

In a science lab, measurement error (in millimeters) comes from two independent sources. Let $X$ be the error from the scale calibration with $\mu_X=0.2$ and $\sigma_X=0.6$ . Let $Y$ be the error from the observer's reading with $\mu_Y=-0.1$ and $\sigma_Y=0.8$ . Define $E=X+Y$ , the total measurement error. Which statement about the combined variable is correct?

\mu_E=0.1

and

\sigma_E=1.4

\mu_E=0.3

and

\sigma_E=\sqrt{1.0}

\mu_E=0.1

and

\sigma_E=\sqrt{1.0}

\mu_E=-0.3

and

\sigma_E=\sqrt{1.0}

\mu_E=0.1

and

\sigma_E=0.2

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AP Statistics Quiz

AP Statistics Quiz: Combining Random Variables

Practice Combining Random Variables in AP Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Combining Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\mu_E=0.1$ and $\sigma_E=1.4$
$\mu_E=0.3$ and $\sigma_E=\sqrt{1.0}$
$\mu_E=0.1$ and $\sigma_E=\sqrt{1.0}$ (correct answer)
$\mu_E=-0.3$ and $\sigma_E=\sqrt{1.0}$
$\mu_E=0.1$ and $\sigma_E=0.2$

Explanation: This question tests combining independent random variables for total error E = X + Y in AP Statistics. The mean is μ_E = 0.2 + (-0.1) = 0.1 mm. The variance is σ_E² = (0.6)² + (0.8)² = 0.36 + 0.64 = 1.0, so σ_E = √1.0. Choice D distracts by using the wrong mean sign, perhaps subtracting incorrectly. In a mini-lesson, for sums of independent variables, add means directly (including negatives), and add variances; the standard deviation is the square root of that sum. This method works for errors or biases from independent sources.

Question 2

A phone's battery percentage decreases during two independent time periods. Let $D_1$ be the percent drop during the morning and $D_2$ be the percent drop during the afternoon. Assume $\mu_{D_1}=22$ , $\sigma_{D_1}=6$ , $\mu_{D_2}=28$ , and $\sigma_{D_2}=5$ . Let $D=D_1+D_2$ be the total percent drop. Which statement about $D$ is correct?

$\mu_D=50$ and $\sigma_D=\sqrt{6^2+5^2}$ (correct answer)
$\mu_D=6$ and $\sigma_D=\sqrt{6^2+5^2}$
$\mu_D=50$ and $\sigma_D=6+5$
$\mu_D=50$ and $\sigma_D=6-5$
$\mu_D=50$ and $\sigma_D=\sqrt{6^2-5^2}$

Explanation: This problem asks about the total battery percentage drop D = D₁ + D₂ over two time periods. The mean total drop is μD = μD₁ + μD₂ = 22 + 28 = 50 percent. For independent drops, the variances add: σ²D = σ²D₁ + σ²D₂ = 6² + 5² = 36 + 25 = 61, so σD = √61 = √(6² + 5²) ≈ 7.81. The key concept is that uncertainty compounds when we combine independent random variables—even though we might expect less variability in a total, the opposite is true. This explains why predicting total battery drain is harder than predicting drain in any single period.

Question 3

A factory tracks the number of defects from two independent machines each day. Let $X$ be the number of defects from Machine 1 with $\mu_X=4$ and $\sigma_X=1.2$ . Let $Y$ be the number of defects from Machine 2 with $\mu_Y=6$ and $\sigma_Y=1.6$ . Define $Z=X+Y$ , the total number of defects from both machines. Which statement about the combined variable is correct?

$\mu_Z=10$ and $\sigma_Z=2.8$
$\mu_Z=2$ and $\sigma_Z=0.4$
$\mu_Z=10$ and $\sigma_Z=\sqrt{4.0}$ (correct answer)
$\mu_Z=10$ and $\sigma_Z=\sqrt{2.8}$
$\mu_Z=10$ and $\sigma_Z=\sqrt{1.6}$

Explanation: This problem covers combining independent random variables for total defects Z = X + Y in AP Statistics. The mean is μ_Z = 4 + 6 = 10 defects. The variance is σ_Z² = (1.2)² + (1.6)² = 1.44 + 2.56 = 4.0, so σ_Z = √4.0 = 2. Choice A is a distractor adding standard deviations (1.2 + 1.6 = 2.8). For a mini-lesson, sums of independent counts have additive means and variances; take the square root after adding variances for the standard deviation. This applies to defect models assuming independence between machines.

Question 4

A store tracks revenue from online sales $O$ and in-store sales $S$ (in hundreds of dollars) for a day. Assume $O$ and $S$ are independent with $\mu_O=18$ , $\sigma_O=4$ , $\mu_S=30$ , and $\sigma_S=6$ . Define $D=S-O$ , the difference (in-store minus online). Which statement about $D$ is correct?

$\mu_D=12$ and $\sigma_D=6-4$
$\mu_D=48$ and $\sigma_D=\sqrt{6^2+4^2}$
$\mu_D=12$ and $\sigma_D=\sqrt{6^2+4^2}$ (correct answer)
$\mu_D=-12$ and $\sigma_D=\sqrt{6^2+4^2}$
$\mu_D=12$ and $\sigma_D=6+4$

Explanation: This problem involves finding the mean and standard deviation when subtracting random variables. For D = S - O, the mean is μD = μS - μO = 30 - 18 = 12 hundred dollars (or $1,200). Even though we're subtracting, the variance formula for independent variables is σ²D = σ²S + σ²O, giving us σD = √(6² + 4²) = √52 ≈ 7.21. Many students incorrectly subtract standard deviations (6 - 4 = 2) or add them (6 + 4 = 10), but the correct approach always involves adding variances. This counterintuitive result shows that subtraction doesn't reduce variability when dealing with independent random variables.

Question 5

In a school cafeteria, let $X$ be the number of ounces of milk poured into a carton and $Y$ be the number of ounces spilled while carrying it. Assume $X$ and $Y$ are independent random variables with means $\mu_X=10$ and $\mu_Y=0.6$ , and standard deviations $\sigma_X=1.2$ and $\sigma_Y=0.5$ . Define a new random variable $N=X-Y$ , the number of ounces actually consumed. Which statement about $N$ is correct?

$\mu_N=10.6$ and $\sigma_N=1.7$
$\mu_N=9.4$ and $\sigma_N=0.7$
$\mu_N=9.4$ and $\sigma_N=\sqrt{1.2^2+0.5^2}$ (correct answer)
$\mu_N=10.6$ and $\sigma_N=\sqrt{1.2^2+0.5^2}$
$\mu_N=9.4$ and $\sigma_N=1.2-0.5$

Explanation: This problem tests your understanding of combining random variables through subtraction. When we define N = X - Y, we need to find the mean and standard deviation of this new variable. For the mean, we use the rule that μ(X-Y) = μX - μY = 10 - 0.6 = 9.4 ounces. For the standard deviation, since X and Y are independent, we use the rule that σ²(X-Y) = σ²X + σ²Y (note the plus sign even for subtraction!), so σ(X-Y) = √(1.2² + 0.5²). The key insight is that variances always add when combining independent random variables, regardless of whether we're adding or subtracting the variables themselves.

Question 6

A student's commute time is the sum of bus wait time $W$ and bus ride time $R$ (in minutes). Assume $W$ and $R$ are independent with $\mu_W=6$ , $\sigma_W=2$ , $\mu_R=18$ , and $\sigma_R=4$ . Let $T=W+R$ . Which statement about $T$ is correct?

$\mu_T=24$ and $\sigma_T=2+4$
$\mu_T=12$ and $\sigma_T=\sqrt{2^2+4^2}$
$\mu_T=24$ and $\sigma_T=\sqrt{2^2+4^2}$ (correct answer)
$\mu_T=24$ and $\sigma_T=4-2$
$\mu_T=24$ and $\sigma_T=\sqrt{4^2-2^2}$

Explanation: This question involves finding the total commute time T = W + R by combining two independent time components. The mean total time is μT = μW + μR = 6 + 18 = 24 minutes. Since wait time and ride time are independent, their variances add: σ²T = σ²W + σ²R = 2² + 4² = 4 + 16 = 20, giving us σT = √20 = √(2² + 4²) ≈ 4.47 minutes. Students often mistakenly add the standard deviations (2 + 4 = 6) or subtract them (4 - 2 = 2), but the correct formula always uses the square root of the sum of squares. This principle shows why total commute time has more variability than either component alone.

Question 7

A delivery company models distance traveled on two independent road segments: $U$ miles on segment 1 and $V$ miles on segment 2. Suppose $\mu_U=35$ , $\sigma_U=5$ , $\mu_V=45$ , and $\sigma_V=7$ . Let $L=U+V$ be the total distance. Which statement about $L$ is correct?

$\mu_L=80$ and $\sigma_L=5+7$
$\mu_L=10$ and $\sigma_L=\sqrt{5^2+7^2}$
$\mu_L=80$ and $\sigma_L=\sqrt{5^2+7^2}$ (correct answer)
$\mu_L=80$ and $\sigma_L=7-5$
$\mu_L=80$ and $\sigma_L=\sqrt{7^2-5^2}$

Explanation: This question tests combining distances to find total distance L = U + V. The mean total distance is μL = μU + μV = 35 + 45 = 80 miles. Since the road segments are independent, their variances add: σ²L = σ²U + σ²V = 5² + 7² = 25 + 49 = 74, giving us σL = √74 = √(5² + 7²) ≈ 8.60 miles. A common mistake is to add standard deviations directly (5 + 7 = 12), but the correct formula uses the Pythagorean theorem structure. This demonstrates an important principle: when combining independent sources of variation, the total variability is less than the sum but more than the difference of the individual standard deviations.

Question 8

A store tracks customer spending. Let $X$ be dollars spent on groceries and $Y$ be dollars spent on household items during a visit. Assume $X$ and $Y$ are independent, with $\mu_X=65$ , $\sigma_X=12$ , $\mu_Y=35$ , and $\sigma_Y=9$ . Define $C=X+Y$ , the total dollars spent. Which statement about $C$ is correct?

$\mu_C=100$ and $\sigma_C=\sqrt{12^2+9^2}$ (correct answer)
$\mu_C=30$ and $\sigma_C=\sqrt{12^2+9^2}$
$\mu_C=100$ and $\sigma_C=21$
$\mu_C=100$ and $\sigma_C=3$ because spending categories offset
$\mu_C=65$ and $\sigma_C=12$ because groceries are the larger category

Explanation: This question tests combining independent spending amounts through addition. For the total C = X + Y, the mean is μ_C = μ_X + μ_Y = 65 + 35 = 100. To find the standard deviation, we add the variances: σ²_C = σ²_X + σ²_Y = 12² + 9² = 144 + 81 = 225, so σ_C = √225 = 15. Choice A correctly expresses this as √(12² + 9²), which equals 15. Choice C incorrectly adds the standard deviations (12 + 9 = 21), a common mistake. The key insight is that when independent random variables are combined, their uncertainties compound through variance addition, not standard deviation addition.

Question 9

A delivery company tracks travel times. Let $X$ be the minutes for Route 1 and $Y$ be the minutes for Route 2. Assume $X$ and $Y$ are independent, with $\mu_X=30$ , $\sigma_X=4$ , $\mu_Y=28$ , and $\sigma_Y=5$ . Define $S=X+Y$ , the total minutes to complete both routes in a day. Which statement about $S$ is correct?

$\mu_S=58$ and $\sigma_S=9$
$\mu_S=58$ and $\sigma_S=\sqrt{41}$ (correct answer)
$\mu_S=2$ and $\sigma_S=\sqrt{41}$
$\mu_S=58$ and $\sigma_S=1$ because independent times average out
$\mu_S=30$ and $\sigma_S=4$ since Route 1 is longer on average

Explanation: This question asks about the sum of two independent random variables representing delivery times. When adding independent variables, the means add: μ_S = μ_X + μ_Y = 30 + 28 = 58. For the standard deviation, we must add the variances first: σ²_S = σ²_X + σ²_Y = 4² + 5² = 16 + 25 = 41, so σ_S = √41. Choice A incorrectly adds the standard deviations directly (4 + 5 = 9), which is a common error. The fundamental principle is that uncertainty (variance) accumulates when we combine independent random variables, whether through addition or subtraction, so we always add variances before taking the square root.

Question 10

A commuter's total travel time is the sum of two independent parts. Let $X$ be the time (in minutes) spent walking to the bus stop with $\mu_X=7$ and $\sigma_X=2$ . Let $Y$ be the time (in minutes) spent riding the bus with $\mu_Y=23$ and $\sigma_Y=6$ . Define $T=X+Y$ , the total commute time. Which statement about the combined variable is correct?

$\mu_T=30$ and $\sigma_T=8$
$\mu_T=30$ and $\sigma_T=4$
$\mu_T=16$ and $\sigma_T=\sqrt{40}$
$\mu_T=30$ and $\sigma_T=\sqrt{40}$ (correct answer)
$\mu_T=30$ and $\sigma_T=\sqrt{16}$

Explanation: This question involves combining independent random variables for total commute time T = X + Y in AP Statistics. The mean is μ_T = 7 + 23 = 30 minutes. The variance is σ_T² = 2² + 6² = 4 + 36 = 40, so σ_T = √40. Choice B distracts by perhaps halving something incorrectly or using a wrong formula. In a mini-lesson, for independent times, add means for total expected time and add variances for total variability; the standard deviation is the square root of the variance sum. Simplifying √40 = 2√10 is optional but not required here.

Question 11

A store tracks two independent random variables: $X$ , the amount (in dollars) a customer spends on groceries, with $\mu_X=42$ and $\sigma_X=8$ , and $Y$ , the amount (in dollars) the same customer spends on household items, with $\mu_Y=18$ and $\sigma_Y=5$ . Define $D=X-Y$ (difference in spending). Which statement about $D$ is correct?

$\mu_D=24$ and $\sigma_D=\sqrt{8^2+5^2}$ (correct answer)
$\mu_D=60$ and $\sigma_D=\sqrt{8^2+5^2}$
$\mu_D=24$ and $\sigma_D=8-5$
$\mu_D=24$ and $\sigma_D=\sqrt{8^2-5^2}$
$\mu_D=-24$ and $\sigma_D=\sqrt{8^2+5^2}$

Explanation: This problem focuses on the AP Statistics topic of combining independent random variables for the difference D = X - Y. The mean of the difference is the difference of the means: μ_D = 42 - 18 = 24. Since X and Y are independent, the variance of D is still the sum of the variances: Var_D = 8² + 5² = 64 + 25 = 89, so σ_D = √89 = √(8² + 5²). Distractors like choice C subtract the standard deviations, which is a mistake because variances, not standard deviations, add when combining independent variables. Another distractor, choice E, uses a negative mean, confusing the order of subtraction. Mini-lesson: For independent X and Y, μ_{X-Y} = μ_X - μ_Y and Var_{X-Y} = Var_X + Var_Y; this addition of variances reflects that uncertainty accumulates whether adding or subtracting.

Question 12

A school compares quiz scores. Let $X$ be a student's score on Quiz 1 and $Y$ be the score on Quiz 2. Assume $X$ and $Y$ are independent, with $\mu_X=78$ , $\sigma_X=10$ , $\mu_Y=82$ , and $\sigma_Y=12$ . Define $A=(X+Y)/2$ , the average of the two quiz scores. Which statement about $A$ is correct?

$\mu_A=80$ and $\sigma_A=\sqrt{10^2+12^2}$
$\mu_A=160$ and $\sigma_A=11$
$\mu_A=80$ and $\sigma_A=11$ because standard deviations average: $(10+12)/2$
$\mu_A=80$ and $\sigma_A=\tfrac{1}{2}\sqrt{10^2+12^2}$ (correct answer)
$\mu_A=4$ and $\sigma_A=\tfrac{1}{2}\sqrt{10^2+12^2}$

Explanation: This question involves finding the mean and standard deviation of an average. For A = (X + Y)/2, the mean is μ_A = (μ_X + μ_Y)/2 = (78 + 82)/2 = 80. For the standard deviation, we use the property that Var(cX) = c²Var(X), so σ²_A = (1/2)²(σ²_X + σ²_Y) = (1/4)(10² + 12²) = (1/4)(100 + 144) = 61, giving σ_A = √61 = (1/2)√244 = (1/2)√(10² + 12²). Choice C incorrectly assumes standard deviations average like means. The key concept is that when we multiply a random variable by a constant, the variance is multiplied by the square of that constant.

Question 13

Let $G$ be the amount (in grams) of protein in a randomly selected serving of Brand G yogurt and $H$ be the amount (in grams) of protein in a randomly selected serving of Brand H yogurt. Over many servings, $\mu_G=14$ with $\sigma_G=2.5$ , and $\mu_H=11$ with $\sigma_H=2.0$ . Assume $G$ and $H$ are independent. Define $W=G-H$ . Which statement about $W$ is correct?

$\mu_W=3$ and $\sigma_W=2.5-2.0=0.5$ .
$\mu_W=25$ and $\sigma_W=\sqrt{2.5^2+2.0^2}$ .
$\mu_W=-3$ and $\sigma_W=\sqrt{2.5^2+2.0^2}$ .
$\mu_W=3$ and $\sigma_W=2.5+2.0=4.5$ .
$\mu_W=3$ and $\sigma_W=\sqrt{2.5^2+2.0^2}$ . (correct answer)

Explanation: This problem involves the difference between protein amounts in two yogurt brands. For W = G - H, the mean is μ_W = μ_G - μ_H = 14 - 11 = 3. For independent random variables, variances add even when we subtract: σ_W² = σ_G² + σ_H² = 2.5² + 2.0² = 6.25 + 4 = 10.25, so σ_W = √10.25 = √(2.5² + 2.0²). Choice E correctly shows μ_W = 3 and σ_W = √(2.5² + 2.0²). Students often mistakenly subtract standard deviations (Choice A shows 2.5 - 2.0 = 0.5) or add them directly (Choice D shows 2.5 + 2.0 = 4.5). Remember: for independent variables, variances always add regardless of whether you're adding or subtracting the original variables.

Question 14

Let $F$ be the amount of time (in hours) a phone battery lasts after a full charge with Feature A turned on, and let $G$ be the amount of time (in hours) it lasts with Feature A turned off, for randomly selected days. Over many days, $\mu_F=9.5$ with $\sigma_F=1.1$ , and $\mu_G=11.0$ with $\sigma_G=1.4$ . Assume $F$ and $G$ are independent across days. Define $H=G-F$ , the increase in battery life when the feature is off. Which statement about $H$ is correct?

$\mu_H=1.5$ and $\sigma_H=1.4-1.1=0.3$ .
$\mu_H=-1.5$ and $\sigma_H=\sqrt{1.1^2+1.4^2}$ .
$\mu_H=20.5$ and $\sigma_H=\sqrt{1.1^2+1.4^2}$ .
$\mu_H=1.5$ and $\sigma_H=1.1+1.4=2.5$ .
$\mu_H=1.5$ and $\sigma_H=\sqrt{1.1^2+1.4^2}$ . (correct answer)

Explanation: This problem examines the increase in battery life when a feature is turned off. For H = G - F, the mean is μ_H = μ_G - μ_F = 11.0 - 9.5 = 1.5 hours (positive because battery lasts longer without the feature). Since F and G are independent across days, their variances add: σ_H² = σ_F² + σ_G² = 1.1² + 1.4² = 1.21 + 1.96 = 3.17, giving σ_H = √3.17 = √(1.1² + 1.4²). Choice E correctly shows μ_H = 1.5 and σ_H = √(1.1² + 1.4²). Common errors include subtracting standard deviations (Choice A: 1.4 - 1.1 = 0.3) or adding them directly (Choice D: 1.1 + 1.4 = 2.5). Remember: for independent variables, variances always add when combining, whether by addition or subtraction.

Question 15

A delivery company tracks daily packages. Let $X$ be the number of packages delivered by Driver 1 in a day and $Y$ be the number delivered by Driver 2 in a day. Over many days, $\mu_X=95$ with $\sigma_X=20$ , and $\mu_Y=105$ with $\sigma_Y=25$ . Assume $X$ and $Y$ are independent. Define $S=X+Y$ . Which statement about $S$ is correct?

$\mu_S=200$ and $\sigma_S=\sqrt{20^2+25^2}$ . (correct answer)
$\mu_S=10$ and $\sigma_S=\sqrt{20^2+25^2}$ .
$\mu_S=200$ and $\sigma_S=20+25$ .
$\mu_S=200$ and $\sigma_S=25-20$ .
$\mu_S=200$ and $\sigma_S=\sqrt{25^2-20^2}$ .

Explanation: This question asks about the sum of two independent random variables representing package deliveries. For S = X + Y, the mean is μ_S = μ_X + μ_Y = 95 + 105 = 200. Since X and Y are independent, their variances add: σ_S² = σ_X² + σ_Y² = 20² + 25² = 400 + 625 = 1025, giving σ_S = √1025 = √(20² + 25²). Choice A correctly identifies both μ_S = 200 and σ_S = √(20² + 25²). Common errors include adding standard deviations directly (Choice C shows 20 + 25) or subtracting them (Choice D shows 25 - 20). When combining independent random variables by addition, always add means and add variances (not standard deviations).

Question 16

A student buys two items. Let $X$ be the cost of the first item and $Y$ the cost of the second item (in dollars). Assume $X$ and $Y$ are independent with $\mu_X=14$ , $\sigma_X=3$ , $\mu_Y=9$ , and $\sigma_Y=2$ . Let $S=X-Y$ , the difference in costs. Which statement about $S$ is correct?

$\mu_S=5$ and $\sigma_S=3-2$
$\mu_S=-5$ and $\sigma_S=\sqrt{3^2+2^2}$
$\mu_S=23$ and $\sigma_S=\sqrt{3^2+2^2}$
$\mu_S=5$ and $\sigma_S=\sqrt{3^2+2^2}$ (correct answer)
$\mu_S=5$ and $\sigma_S=3+2$

Explanation: This problem asks for the difference in costs S = X - Y between two items. The mean difference is μS = μX - μY = 14 - 9 = 5 dollars. For independent purchases, the variance of the difference equals the sum of the variances: σ²S = σ²X + σ²Y = 3² + 2² = 9 + 4 = 13, so σS = √13 = √(3² + 2²) ≈ 3.61 dollars. Many students incorrectly compute 3 - 2 = 1 or 3 + 2 = 5 for the standard deviation, but the correct approach always involves adding variances, even when subtracting the original variables. This principle explains why the uncertainty in price differences can be substantial even when individual prices are relatively predictable.

Question 17

In a factory, let $X$ be the number of seconds it takes Machine A to seal a package and $Y$ be the number of seconds it takes Machine B to seal a package. Assume $X$ and $Y$ are independent, with $\mu_X=12$ , $\sigma_X=3$ , $\mu_Y=10$ , and $\sigma_Y=2$ . Define the combined variable $T=X+Y$ , the total time to seal two packages (one on each machine). Which statement about $T$ is correct?

$\mu_T=22$ and $\sigma_T=5$
$\mu_T=2$ and $\sigma_T=\sqrt{13}$
$\mu_T=22$ and $\sigma_T=\sqrt{13}$ (correct answer)
$\mu_T=22$ and $\sigma_T=1$ because variability cancels when adding
$\mu_T=12$ and $\sigma_T=3$ because $X$ dominates the sum

Explanation: This question tests understanding of combining independent random variables through addition. When we add two independent random variables, their means add directly: μ_T = μ_X + μ_Y = 12 + 10 = 22. However, variances (not standard deviations) add when variables are independent: σ²_T = σ²_X + σ²_Y = 3² + 2² = 9 + 4 = 13, so σ_T = √13. Choice A incorrectly adds the standard deviations (3 + 2 = 5) instead of adding variances. The key concept is that when combining independent random variables through addition or subtraction, we always add the variances, then take the square root to find the new standard deviation.

Question 18

A student takes two independent quizzes. Let $X$ be the score on Quiz 1 with $\mu_X=78$ and $\sigma_X=6$ . Let $Y$ be the score on Quiz 2 with $\mu_Y=82$ and $\sigma_Y=5$ . Define $A=\frac{X+Y}{2}$ , the average of the two quiz scores. Which statement about the combined variable is correct?

$\mu_A=160$ and $\sigma_A=11$
$\mu_A=80$ and $\sigma_A=\sqrt{61}$
$\mu_A=80$ and $\sigma_A=\frac{11}{2}$
$\mu_A=80$ and $\sigma_A=\frac{\sqrt{61}}{2}$ (correct answer)
$\mu_A=80$ and $\sigma_A=\frac{\sqrt{11}}{2}$

Explanation: This question examines combining independent random variables for the average A = (X + Y)/2 in AP Statistics. The mean is μ_A = (78 + 82)/2 = 80. The variance is σ_A² = (6² + 5²)/4 = (36 + 25)/4 = 61/4, so σ_A = √(61/4) = √61 / 2. Choice B distracts by not dividing the standard deviation correctly, using √61 without the /2. In a mini-lesson, for linear combinations like (1/2)X + (1/2)Y, the mean scales accordingly, and variance is (1/2)²σ_X² + (1/2)²σ_Y²; independence ensures variances add without correlation terms. This is useful for averages of independent scores.

Question 19

In a school cafeteria, let $X$ be the number of minutes a randomly selected student waits in the lunch line on Monday, with mean $\mu_X=8$ and standard deviation $\sigma_X=2$ . Let $Y$ be the number of minutes a randomly selected student waits in the lunch line on Tuesday, with mean $\mu_Y=10$ and standard deviation $\sigma_Y=3$ . Assume $X$ and $Y$ are independent. Define the combined variable $T=X+Y$ as the total waiting time over the two days. Which statement about the combined variable is correct?

$\mu_T=2$ and $\sigma_T=5$
$\mu_T=18$ and $\sigma_T=1$
$\mu_T=18$ and $\sigma_T=5$
$\mu_T=18$ and $\sigma_T=\sqrt{13}$ (correct answer)
$\mu_T=18$ and $\sigma_T=\sqrt{1}$

Explanation: This question assesses the AP Statistics skill of combining independent random variables, specifically for the sum T = X + Y. The mean of T is the sum of the individual means: μ_T = 8 + 10 = 18. The variance of T is the sum of the variances since X and Y are independent: σ_T² = 2² + 3² = 4 + 9 = 13, so σ_T = √13. A common distractor is choice C, which incorrectly adds the standard deviations (2 + 3 = 5) instead of combining variances. In a mini-lesson on combining random variables, remember that for independent X and Y, the mean of aX + bY is aμ_X + bμ_Y, and the variance is a²σ_X² + b²σ_Y²; here, a = b = 1 for the sum. This rule holds regardless of the distributions, as long as independence is assumed.

Question 20

A delivery company tracks package weights. Let $X$ be the weight (in pounds) of a randomly selected package from Warehouse A with $\mu_X=12$ and $\sigma_X=4$ . Let $Y$ be the weight (in pounds) of a randomly selected package from Warehouse B with $\mu_Y=9$ and $\sigma_Y=5$ . Assume $X$ and $Y$ are independent. Define $D=X-Y$ , the difference in weights (A minus B). Which statement about the combined variable is correct?

$\mu_D=3$ and $\sigma_D=1$
$\mu_D=21$ and $\sigma_D=9$
$\mu_D=3$ and $\sigma_D=9$
$\mu_D=3$ and $\sigma_D=\sqrt{41}$ (correct answer)
$\mu_D=-3$ and $\sigma_D=\sqrt{41}$

Explanation: This problem focuses on the AP Statistics topic of combining independent random variables for the difference D = X - Y. The mean is μ_D = μ_X - μ_Y = 12 - 9 = 3. The variance is σ_D² = σ_X² + σ_Y² = 16 + 25 = 41, since variances add for differences of independent variables, yielding σ_D = √41. Choice E is a distractor that flips the sign of the mean to -3 while correctly computing the standard deviation. For a mini-lesson, note that subtracting Y is like adding (-1)Y, so the mean becomes μ_X - μ_Y, but the variance contribution is (-1)²σ_Y² = σ_Y², added to σ_X². Independence ensures no covariance term is needed.