Combining Random Variables
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AP Statistics › Combining Random Variables
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 $\sigma_E=1.4$
$\mu_E=0.1$ and $\sigma_E=\sqrt{1.0}$
$\mu_E=0.3$ and $\sigma_E=\sqrt{1.0}$
$\mu_E=-0.3$ and $\sigma_E=\sqrt{1.0}$
$\mu_E=0.1$ and $\sigma_E=0.2$
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.
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 $\sigma_D=6-5$
$\mu_D=50$ and $\sigma_D=\sqrt{6^2-5^2}$
$\mu_D=50$ and $\sigma_D=\sqrt{6^2+5^2}$
$\mu_D=6$ and $\sigma_D=\sqrt{6^2+5^2}$
$\mu_D=50$ and $\sigma_D=6+5$
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.
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 $\sigma_Z=\sqrt{1.6}$
$\mu_Z=2$ and $\sigma_Z=0.4$
$\mu_Z=10$ and $\sigma_Z=2.8$
$\mu_Z=10$ and $\sigma_Z=\sqrt{4.0}$
$\mu_Z=10$ and $\sigma_Z=\sqrt{2.8}$
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.
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 $\sigma_D=6+4$
$\mu_D=12$ and $\sigma_D=6-4$
$\mu_D=12$ and $\sigma_D=\sqrt{6^2+4^2}$
$\mu_D=48$ and $\sigma_D=\sqrt{6^2+4^2}$
$\mu_D=-12$ and $\sigma_D=\sqrt{6^2+4^2}$
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.
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?
$\mu_N=9.4$ and $\sigma_N=0.7$
$\mu_N=10.6$ and $\sigma_N=\sqrt{1.2^2+0.5^2}$
$\mu_N=9.4$ and $\sigma_N=1.2-0.5$
$\mu_N=9.4$ and $\sigma_N=\sqrt{1.2^2+0.5^2}$
$\mu_N=10.6$ and $\sigma_N=1.7$
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.
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 $\sigma_T=\sqrt{4^2-2^2}$
$\mu_T=12$ and $\sigma_T=\sqrt{2^2+4^2}$
$\mu_T=24$ and $\sigma_T=4-2$
$\mu_T=24$ and $\sigma_T=2+4$
$\mu_T=24$ and $\sigma_T=\sqrt{2^2+4^2}$
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.
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 $\sigma_L=\sqrt{5^2+7^2}$
$\mu_L=80$ and $\sigma_L=5+7$
$\mu_L=80$ and $\sigma_L=7-5$
$\mu_L=80$ and $\sigma_L=\sqrt{7^2-5^2}$
$\mu_L=10$ and $\sigma_L=\sqrt{5^2+7^2}$
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.
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 $\sigma_C=\sqrt{12^2+9^2}$
$\mu_C=30$ and $\sigma_C=\sqrt{12^2+9^2}$
$\mu_C=65$ and $\sigma_C=12$ because groceries are the larger category
$\mu_C=100$ and $\sigma_C=3$ because spending categories offset
$\mu_C=100$ and $\sigma_C=21$
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.
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 $\sigma_S=9$
$\mu_S=2$ and $\sigma_S=\sqrt{41}$
$\mu_S=30$ and $\sigma_S=4$ since Route 1 is longer on average
$\mu_S=58$ and $\sigma_S=1$ because independent times average out
$\mu_S=58$ and $\sigma_S=\sqrt{41}$
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.
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 $\sigma_T=\sqrt{16}$
$\mu_T=30$ and $\sigma_T=8$
$\mu_T=30$ and $\sigma_T=\sqrt{40}$
$\mu_T=30$ and $\sigma_T=4$
$\mu_T=16$ and $\sigma_T=\sqrt{40}$
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.