Combining Random Variables - AP Statistics

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?
Answer: $\mu_T=18$ and $\sigma_T=\sqrt{13}$ 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.
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?
Answer: $\mu_N=9.4$ and $\sigma_N=\sqrt{1.2^2+0.5^2}$ 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.
A delivery company models two independent random variables: $X$ , the time (in minutes) to drive to a customer with $\mu_X=18$ and $\sigma_X=5$ , and $Y$ , the time (in minutes) to return to the warehouse with $\mu_Y=16$ and $\sigma_Y=4$ . Define $R=X-Y$ (drive time minus return time). Which statement about $R$ is correct?
Answer: $\mu_R=2$ and $\sigma_R=\sqrt{5^2+4^2}$ This problem in AP Statistics tests combining independent random variables for R = X - Y. The mean is μ_R = 18 - 16 = 2. Independence gives Var_R = 5² + 4² = 25 + 16 = 41, so σ_R = √41 = √(5² + 4²). Distractor B uses the sum of means (34), not noting the difference. Choice D negates the mean, possibly flipping order. Mini-lesson: Subtract means for differences, but always add variances when independent, then sqrt; this ensures correct variability calculation.
Let $L$ be the number of pages Leah reads on Saturday and $K$ be the number of pages she reads on Sunday. Over many weekends, $\mu_L=40$ with $\sigma_L=9$ , and $\mu_K=55$ with $\sigma_K=12$ . Assume $L$ and $K$ are independent. Define $R=L+K$ , the total pages read over the weekend. Which statement about $R$ is correct?
Answer: $\mu_R=95$ and $\sigma_R=\sqrt{9^2+12^2}$ . This problem asks about the total pages read over a weekend. For R = L + K, the mean is μ_R = μ_L + μ_K = 40 + 55 = 95 pages. Since L and K are independent, their variances add: σ_R² = σ_L² + σ_K² = 9² + 12² = 81 + 144 = 225, giving σ_R = √225 = 15 = √(9² + 12²). Choice B correctly shows μ_R = 95 and σ_R = √(9² + 12²). Common mistakes include adding standard deviations directly (Choice C: 9 + 12 = 21) or subtracting them (Choice D: 12 - 9 = 3). The principle remains consistent: when adding independent random variables, add their means and add their variances.
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?
Answer: $\mu_A=80$ and $\sigma_A=\frac{\sqrt{61}}{2}$ 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.
A student records two independent random variables for commute time: $X$ , minutes on the bus with $\mu_X=25$ and $\sigma_X=4$ , and $Y$ , minutes walking with $\mu_Y=10$ and $\sigma_Y=3$ . Let $T=X+Y$ be total commute time. Which statement about $T$ is correct?
Answer: $\mu_T=35$ and $\sigma_T=\sqrt{4^2+3^2}$ This AP Statistics question involves combining independent random variables for T = X + Y. The mean is μ_T = 25 + 10 = 35. With independence, Var_T = 4² + 3² = 16 + 9 = 25, so σ_T = √25 = 5 = √(4² + 3²). Choice A uses the difference of means (15), a distractor for subtraction confusion. Choice D adds standard deviations directly, bypassing variances. Mini-lesson: Sum means for additions, add variances for independent variables, and sqrt for SD; this method accurately combines distributions.
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?
Answer: $\mu_Z=10$ and $\sigma_Z=\sqrt{4.0}$ 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.
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?
Answer: $\mu_H=1.5$ and $\sigma_H=\sqrt{1.1^2+1.4^2}$ . 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.
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?
Answer: $\mu_C=100$ and $\sigma_C=\sqrt{12^2+9^2}$ 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.
A bank models deposits. Let $X$ be the amount (in dollars) deposited on Monday and $Y$ be the amount (in dollars) deposited on Tuesday. Assume $X$ and $Y$ are independent, with $\mu_X=1200$ , $\sigma_X=300$ , $\mu_Y=900$ , and $\sigma_Y=200$ . Define $N=X-Y$ , the net difference (Monday minus Tuesday). Which statement about $N$ is correct?
Answer: $\mu_N=300$ and $\sigma_N=\sqrt{300^2+200^2}$ This question involves the net difference N = X - Y between Monday and Tuesday deposits. The mean is μ_N = μ_X - μ_Y = 1200 - 900 = 300 dollars. For independent variables, variances add even when subtracting: σ²_N = σ²_X + σ²_Y = 300² + 200² = 90000 + 40000 = 130000, so σ_N = √130000 = √(300² + 200²). Choice A incorrectly uses 100 for the standard deviation, confusing it with the difference of standard deviations. Choice E has the wrong sign for the mean. The fundamental principle remains: when combining independent random variables, variances always add, and the operation only affects how means combine.
A store monitors daily revenue from two independent departments. Let $X$ be the revenue (in hundreds of dollars) from electronics with $\mu_X=40$ and $\sigma_X=6$ . Let $Y$ be the revenue (in hundreds of dollars) from clothing with $\mu_Y=25$ and $\sigma_Y=8$ . Define $R=X+Y$ , the total revenue from both departments. Which statement about the combined variable is correct?
Answer: $\mu_R=65$ and $\sigma_R=\sqrt{100}$ This problem involves combining independent random variables for total revenue R = X + Y in AP Statistics. The mean is μ_R = 40 + 25 = 65 hundred dollars. The variance is σ_R² = 6² + 8² = 36 + 64 = 100, so σ_R = √100 = 10. Choice A is a distractor that adds the standard deviations (6 + 8 = 14). For a mini-lesson, when adding independent X and Y, μ_{X+Y} = μ_X + μ_Y and σ_{X+Y}² = σ_X² + σ_Y²; here, √100 simplifies to 10, matching the correct statement. Independence is key to avoiding covariance adjustments.
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?
Answer: $\mu_D=3$ and $\sigma_D=\sqrt{41}$ 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.
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?
Answer: $\mu_A=80$ and $\sigma_A=\tfrac{1}{2}\sqrt{10^2+12^2}$ 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.
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?
Answer: $\mu_S=200$ and $\sigma_S=\sqrt{20^2+25^2}$ . 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).
Let $M$ be the number of minutes Jordan spends on math homework on a randomly selected weekday and $R$ be the number of minutes Jordan spends on reading on that weekday. Over many weekdays, $\mu_M=50$ with $\sigma_M=10$ , and $\mu_R=30$ with $\sigma_R=8$ . Assume $M$ and $R$ are independent. Define $D=M-R$ . Which statement about $D$ is correct?
Answer: $\mu_D=20$ and $\sigma_D=\sqrt{10^2+8^2}$ . This problem involves finding the mean and standard deviation of a difference of independent random variables. For the difference D = M - R, the mean is μ_D = μ_M - μ_R = 50 - 30 = 20. For independent variables, variances always add when combining, even for subtraction: σ_D² = σ_M² + σ_R² = 10² + 8² = 100 + 64 = 164, so σ_D = √164 = √(10² + 8²). Choice C correctly shows μ_D = 20 and σ_D = √(10² + 8²). A common error is thinking standard deviations subtract when variables are subtracted (Choice A shows 10 - 8 = 2). The key principle: variances always add for independent variables, whether you're computing X + Y or X - Y.
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?
Answer: $\mu_T=24$ and $\sigma_T=\sqrt{2^2+4^2}$ 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.
A factory records the weight of cereal in a box as $W$ ounces and the weight of the empty box as $B$ ounces. Assume $W$ and $B$ are independent with $\mu_W=18.2$ , $\sigma_W=0.4$ , $\mu_B=1.1$ , and $\sigma_B=0.1$ . Let $F=W+B$ be the total shipping weight. Which statement about $F$ is correct?
Answer: $\mu_F=19.3$ and $\sigma_F=\sqrt{0.4^2+0.1^2}$ This problem asks about adding two independent weights to find total shipping weight F = W + B. The mean of the sum is straightforward: μF = μW + μB = 18.2 + 1.1 = 19.3 ounces. For the standard deviation, we apply the independence rule that variances add: σ²F = σ²W + σ²B = (0.4)² + (0.1)² = 0.16 + 0.01 = 0.17, so σF = √0.17 = √(0.4² + 0.1²). A common error is to add or subtract the standard deviations directly, but the correct approach always involves the Pythagorean theorem-like formula for combining independent variables. This shows how even small sources of variation contribute to overall uncertainty.
A student's daily screen time is modeled with two independent random variables: $X$ , minutes on social media with $\mu_X=80$ and $\sigma_X=20$ , and $Y$ , minutes watching videos with $\mu_Y=50$ and $\sigma_Y=15$ . Define $T=X+Y$ as total screen time. Which statement about $T$ is correct?
Answer: $\mu_T=130$ and $\sigma_T=\sqrt{20^2+15^2}$ This question from AP Statistics examines combining independent random variables for T = X + Y. The mean is μ_T = 80 + 50 = 130. Independence allows Var_T = 20² + 15² = 400 + 225 = 625, so σ_T = √625 = 25 = √(20² + 15²). Choice A subtracts standard deviations, a distractor confusing means with variability. Choice C adds them directly, skipping the variance step. Mini-lesson: Always sum means for additions and add variances for independent variables, then sqrt for SD; this prevents underestimating variability in combined distributions.
A runner logs times. Let $X$ be the time (in minutes) for the first half of a route and $Y$ be the time (in minutes) for the second half. Assume $X$ and $Y$ are independent, with $\mu_X=24$ , $\sigma_X=2.5$ , $\mu_Y=26$ , and $\sigma_Y=3.5$ . Define $R=Y-X$ , the second-half time minus first-half time. Which statement about $R$ is correct?
Answer: $\mu_R=2$ and $\sigma_R=\sqrt{(3.5)^2+(2.5)^2}$ This problem asks about the difference in running times R = Y - X. The mean is μ_R = μ_Y - μ_X = 26 - 24 = 2, representing that the second half takes 2 minutes longer on average. For independent variables, variances add even when subtracting: σ²_R = σ²_Y + σ²_X = (3.5)² + (2.5)² = 12.25 + 6.25 = 18.5, so σ_R = √18.5. Choice E correctly shows this as √((3.5)² + (2.5)²). Choice D makes the error of adding standard deviations directly (3.5 + 2.5 = 6). Remember that when combining independent random variables, we always add variances regardless of whether we're adding or subtracting the variables themselves.
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?
Answer: $\mu_D=24$ and $\sigma_D=\sqrt{8^2+5^2}$ 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.

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