Variance of a Random Variable - College Statistics

1. A dataset consists of 100 observations of a random variable X, with a sample mean of 50 and a sample variance of 20. A new observation, $x_{101} = 200$, is added to the dataset. How will the new sample variance compare to the original variance?
   Answer: It will increase substantially. When analyzing how an extreme outlier affects sample variance, you need to understand that variance measures how spread out data points are from the mean. An outlier that's far from the existing data will dramatically increase this spread. Let's see what happens when we add $x_{101} = 200$. The original mean was 50, so this new point is 150 units away from the center of the data. When calculating variance, we square the deviations from the mean, so this large deviation gets amplified enormously. Even though the new mean will shift upward (from 50 to about 51.5), the point 200 will still be roughly 148.5 units from the new mean. That squared deviation of approximately 22,000 is massive compared to the original variance of 20. Option A incorrectly assumes larger sample sizes automatically reduce variance. While larger samples give more stable estimates, adding an extreme outlier has the opposite effect. Option B commits a serious error in thinking one observation out of 101 is negligible—this ignores how variance calculation works. A single extreme value can dominate the entire variance calculation because deviations are squared. Option C suggests we can't predict the direction, but we absolutely can: any point this far from the existing data will increase variance substantially. Remember this key principle: variance is extremely sensitive to outliers because it squares deviations. When you see an outlier that's many standard deviations away from the mean (here, 200 is over 30 standard deviations from 50), expect variance to increase dramatically, regardless of sample size.
2. A process produces ball bearings whose diameters, X, are independent and identically distributed random variables with variance $\sigma^2$. Let A be the diameter of a single ball bearing that is then doubled. Let B be the sum of the diameters of two independently drawn ball bearings. Which of the following correctly compares the variances of A and B?
   Answer: $\text{Var}(A) = 4\sigma^2$ and $\text{Var}(B) = 2\sigma^2$ Let X1 and X2 be two i.i.d. random variables representing the diameters. Variable A represents doubling a single measurement: $A = 2X_1$. Its variance is $\text{Var}(A) = \text{Var}(2X_1) = 2^2 \text{Var}(X_1) = 4\sigma^2$. Variable B represents summing two independent measurements: $B = X_1 + X_2$. Its variance is $\text{Var}(B) = \text{Var}(X_1 + X_2) = \text{Var}(X_1) + \text{Var}(X_2) = \sigma^2 + \sigma^2 = 2\sigma^2$. Thus, $\text{Var}(A) = 4\sigma^2$ and $\text{Var}(B) = 2\sigma^2$. The other options miscalculate or swap these two distinct concepts.
3. Two independent random variables, X and Y, have the same mean $\mu$. However, $\text{Var}(X) = 4$ and $\text{Var}(Y) = 16$. Which statement accurately describes the distributions of X and Y?
   Answer: A randomly selected value from the distribution of X is more likely to be close to $\mu$ than a randomly selected value from Y. Variance measures the dispersion of data points around the mean. A smaller variance implies that the data points tend to be closer to the mean. Since $\text{Var}(X) = 4$ is smaller than $\text{Var}(Y) = 16$, the distribution of X is more tightly clustered (more concentrated) around the common mean $\mu$. Therefore, a randomly chosen value of X is more likely to be near $\mu$ than a randomly chosen value of Y. Distractor A makes the opposite claim. Distractor C is incorrect because Chebyshev's inequality is stated in terms of standard deviations, not fixed units. An interval of 2 units is 1 standard deviation for X (SD=2) but only 0.5 standard deviations for Y (SD=4). Distractor D is a specific transformation; if Y=2X, Var(Y)=4Var(X)=16, but this requires E[X]=0 for E[Y]=E[X], and it is not a necessary condition.
4. For a random variable X, it is found that the expectation of its square, $E[X^2]$, is exactly equal to the square of its expectation, $(E[X])^2$. Which of the following is a necessary conclusion?
   Answer: X is not a random variable; it is a constant. This question tests your understanding of variance and what it means when a random variable has zero variance. The key insight comes from the variance formula: $\text{Var}(X) = E[X^2] - (E[X])^2$. Since you're told that $E[X^2] = (E[X])^2$, substituting into the variance formula gives us $\text{Var}(X) = E[X^2] - (E[X])^2 = 0$. When a random variable has zero variance, it means there's no variability in its values—the variable takes on the same value with probability 1. In other words, X is actually a constant, not truly random. Option D is correct because zero variance is the mathematical definition of a constant random variable (also called a degenerate random variable). Option A is wrong because X being constant doesn't require it to equal zero. For example, if X always equals 5, then $E[X] = 5$ and $E[X^2] = 25$, so $E[X^2] = (E[X])^2$ is satisfied. Option B is incorrect because a standard normal distribution has variance 1, not 0. Any continuous distribution with positive variance will have $E[X^2] > (E[X])^2$. Option C is wrong because symmetry has nothing to do with variance. Many asymmetric distributions exist where $E[X^2] \neq (E[X])^2$, and symmetry alone doesn't guarantee zero variance. Study tip: Remember that $\text{Var}(X) = E[X^2] - (E[X])^2$. Whenever you see these two expressions being equal, immediately think "zero variance equals constant value."
5. A random variable X, measured in kilograms, has a variance of 25 kg². A new variable Y is created by first converting X to grams (1 kg = 1000 g) and then adding an independent random measurement error E, which has a mean of 0 g and a variance of 5000 g². What is the variance of Y in g²?
   Answer: 25,005,000 First, convert the variance of X to units of g². Let $X_g$ be the variable X in grams. Since $X_g = 1000X$, the variance is $\text{Var}(X_g) = 1000^2 \text{Var}(X) = 1,000,000 \times 25 = 25,000,000 \text{ g}^2$. The new variable is $Y = X_g + E$. Since X and E are independent, the variance of their sum is the sum of their variances: $\text{Var}(Y) = \text{Var}(X_g) + \text{Var}(E) = 25,000,000 \text{ g}^2 + 5000 \text{ g}^2 = 25,005,000 \text{ g}^2$. Distractors A and B fail to square the conversion factor of 1000. Distractor C incorrectly squares the variance of X itself.
6. A biologist measures the weights of a species of bird in grams (g). The random variable W represents the weight of a randomly selected bird. The standard deviation of W is found to be 10 g. Which of the following statements correctly describes the variance of W?
   Answer: The variance is 100 g². Variance is the square of the standard deviation. If the standard deviation is 10 g, the variance is $(10 \text{ g})^2 = 100 \text{ g}^2$. The units of variance are always the square of the units of the random variable. Distractor A has the correct value but incorrect units. Distractor B has the correct units but did not square the value. Distractor D is incorrect because variance is a measure of spread and is not dependent on the value of the mean.
7. Consider a discrete random variable X with the probability distribution P(X = -1) = 0.5 and P(X = 1) = 0.5. A new random variable Y is created with the distribution P(Y = -10) = 0.5 and P(Y = 10) = 0.5. How does the variance of Y compare to the variance of X?
   Answer: $\text{Var}(Y) = 100 \times \text{Var}(X)$ When you encounter problems comparing variances of related random variables, focus on how transformations affect variance. The key insight here is recognizing that Y is essentially a scaled version of X. Let's calculate the variance of X first. With $P(X = -1) = 0.5$ and $P(X = 1) = 0.5$, the expected value is $E[X] = (-1)(0.5) + (1)(0.5) = 0$. The variance is $\text{Var}(X) = E[X^2] - (E[X])^2 = (1)(0.5) + (1)(0.5) - 0^2 = 1$. For Y, with $P(Y = -10) = 0.5$ and $P(Y = 10) = 0.5$, we get $E[Y] = (-10)(0.5) + (10)(0.5) = 0$. The variance is $\text{Var}(Y) = E[Y^2] - (E[Y])^2 = (100)(0.5) + (100)(0.5) - 0^2 = 100$. Notice that $Y = 10X$ (both variables take values that are 10 times the corresponding X values with identical probabilities). Using the variance transformation rule: when $Y = aX$, then $\text{Var}(Y) = a^2 \cdot \text{Var}(X)$. Here, $a = 10$, so $\text{Var}(Y) = 10^2 \cdot \text{Var}(X) = 100 \cdot \text{Var}(X)$. Answer A is correct. Answer B incorrectly uses the scaling factor 10 instead of its square. Answer C falls into the trap of thinking equal probabilities mean equal variances, ignoring the different outcome values. Answer D is wrong because the relationship follows directly from variance transformation properties. Study tip: Remember that variance scales by the square of the transformation factor. When comparing distributions, look for scaling relationships rather than just probability patterns.
8. Let X be the daily rainfall (in mm) and Y be the daily number of umbrellas sold at a local store. These two variables are positively correlated. Suppose $\text{Var}(X) = 25$ and $\text{Var}(Y) = 100$. Consider the random variable $Z = X + Y$. Which of the following must be true about the variance of Z?
   Answer: $\text{Var}(Z) > 125$ The formula for the variance of a sum is $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)$. Since X and Y are positively correlated, their covariance, $\text{Cov}(X,Y)$, is positive. Therefore, $\text{Var}(Z) = 25 + 100 + 2\text{Cov}(X,Y) = 125 + \text{a positive number}$, which means $\text{Var}(Z)$ must be greater than 125. Distractor A would be correct only if X and Y were independent. Distractor B would be correct if they were negatively correlated. Distractor D is incorrect as the means are not part of the variance of a sum calculation.
9. A random variable X has a mean $\mu = 10$ and variance $\sigma^2 = 4$. What is the value of $E[X^2]$, the expected value of the square of X?
   Answer: 104 When you encounter questions about the expected value of transformed random variables, you need to recall the fundamental relationship between variance, mean, and the second moment. This is a direct application of the variance formula. The key insight is using the variance formula: $\sigma^2 = E[X^2] - (E[X])^2$. Rearranging this equation gives us $E[X^2] = \sigma^2 + (E[X])^2$. With the given values $\mu = 10$ and $\sigma^2 = 4$, we can substitute: $E[X^2] = 4 + (10)^2 = 4 + 100 = 104$. Looking at the wrong answers: Choice B (96) represents the common error of subtracting the variance from the squared mean: $100 - 4 = 96$. This reflects confusion about whether to add or subtract in the variance formula. Choice C (16) appears to come from incorrectly using $(\mu + \sigma)^2 = (10 + 2)^2 = 144$ or some other computational error involving the standard deviation $\sigma = 2$. Choice D (100) is simply $\mu^2$, showing the mistake of ignoring the variance term entirely and thinking $E[X^2] = (E[X])^2$. The correct answer is A (104). Study tip: Always remember that $E[X^2] \neq (E[X])^2$ in general. The variance formula $E[X^2] = \sigma^2 + \mu^2$ is your bridge between these quantities. Memorize this rearranged form—it's frequently tested and essential for understanding the relationship between moments of a distribution.
10. Let X be a random variable representing the daily change in a stock's price, with $E[X] = 0.10$ and $\text{Var}(X) = 2.25$. An investor creates a portfolio Y that is a perfect short position on this stock, so $Y = -X$. Which of the following correctly describes the expected value and variance of Y?
    Answer: $E[Y] = -0.10$ and $\text{Var}(Y) = 2.25$ The expected value is a linear operator, so $E[Y] = E[-X] = -E[X] = -0.10$. For variance, $\text{Var}(Y) = \text{Var}(-X) = \text{Var}(-1 \cdot X) = (-1)^2 \text{Var}(X) = 1 \cdot \text{Var}(X) = 2.25$. Variance measures spread, which is non-negative; reflecting the distribution across the y-axis does not change its spread. Distractor A incorrectly states that variance can be negative. Distractor B has the wrong sign for the expected value. Distractor D incorrectly applies the transformation constant to the variance.
11. A quality control inspector takes a single measurement, X, of a product's length. The machine that produces the product has a known process variance of $\sigma^2 = 9 \text{ mm}^2$. The inspector's boss suggests that the mean of 9 independent measurements, $\bar{X}_9$, would be a better estimate of the true length. Which statement correctly compares the variance of the single measurement X to the variance of the sample mean $\bar{X}_9$?
    Answer: $\text{Var}(\bar{X}_9) = 1 \text{ mm}^2$, which is one-ninth of $\text{Var}(X)$. When you encounter questions about sample means versus individual measurements, you're dealing with the fundamental concept of how averaging reduces variability. The key insight is that sample means are less variable than individual observations. The variance of a sample mean follows the formula: $\text{Var}(\bar{X}_n) = \frac{\sigma^2}{n}$, where $\sigma^2$ is the population variance and $n$ is the sample size. For a single measurement, $\text{Var}(X) = \sigma^2 = 9 \text{ mm}^2$. For the mean of 9 measurements: $\text{Var}(\bar{X}_9) = \frac{9}{9} = 1 \text{ mm}^2$. This is exactly one-ninth of the original variance, confirming answer A. Looking at the incorrect options: B calculates $\frac{9}{3} = 3$, which incorrectly uses 3 instead of 9 in the denominator—this reflects confusion about the sample size. C incorrectly assumes equal variances because both estimators are unbiased, but bias and variance are completely different properties. While both $X$ and $\bar{X}_9$ are unbiased estimators of the true mean, their variances differ dramatically. D multiplies rather than divides by the sample size, getting $9 \times 9 = 81$—this reverses the fundamental relationship entirely. Study tip: Remember the formula $\text{Var}(\bar{X}_n) = \frac{\sigma^2}{n}$ and that averaging always reduces variance by a factor of $n$. Sample means become more precise (less variable) as sample size increases, which is why the inspector's boss correctly suggests using multiple measurements.
12. Let X and Y be independent random variables representing the scores of two different students on a standardized test. Both scores are drawn from a distribution with a mean of 500 and a variance of 100. What is the variance of the difference in their scores, $D = X - Y$?
    Answer: 200 For independent random variables X and Y, the variance of their difference is the sum of their variances: $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$. This is because the variability from both sources adds up, regardless of whether we are summing or differencing the variables themselves. Here, $\text{Var}(D) = 100 + 100 = 200$. Distractor A is a common mistake, subtracting the variances. Distractor B incorrectly assumes the variance of the difference is equal to the original variance. Distractor D calculates the standard deviation of the sum/difference ($\sqrt{200} \approx 141.4$) which is not the variance.
13. Let X be a random variable with mean $\mu$ and variance $\sigma^2$. Let $\bar{X}_n$ denote the sample mean of a random sample of size n. A researcher computes $\bar{X}_{100}$ from a sample of size 100. If the researcher wishes to obtain a new sample mean, $\bar{X}_k$, with a variance that is half the variance of $\bar{X}_{100}$, what should be the new sample size k?
    Answer: 200 The variance of the sample mean is given by $\text{Var}(\bar{X}_n) = \sigma^2/n$. The variance of the original sample mean is $\text{Var}(\bar{X}_{100}) = \sigma^2/100$. We want the new variance, $\text{Var}(\bar{X}_k)$, to be half of this: $\text{Var}(\bar{X}_k) = \frac{1}{2} \text{Var}(\bar{X}_{100})$. So, $\sigma^2/k = \frac{1}{2} (\sigma^2/100)$, which simplifies to $1/k = 1/200$. Therefore, $k = 200$. To halve the variance of the sample mean, the sample size must be doubled. Distractor A would double the variance. Distractor D would quarter the variance.
14. Let X be a Bernoulli random variable representing the success (X=1) or failure (X=0) of a single trial, with $P(X=1) = p$. The variance of a Bernoulli random variable is given by $\text{Var}(X) = p(1-p)$. For what value of p is the uncertainty of the outcome, as measured by variance, maximized?
    Answer: A value of p equal to 0.5. The function $f(p) = p(1-p) = p - p^2$ describes a downward-opening parabola for $p \in [0,1]$. Its maximum occurs at the vertex. The vertex of a parabola $ax^2+bx+c$ is at $x = -b/2a$. For $-p^2+p$, the vertex is at $p = -1/(2(-1)) = 0.5$. Conceptually, the outcome of the trial is most uncertain when success and failure are equally likely, which corresponds to $p=0.5$. At the extremes $p=0$ or $p=1$, the outcome is certain and the variance is 0, representing minimum uncertainty.
15. The temperature, C, in a controlled environment is a random variable measured in degrees Celsius with a variance of 4 (°C)². The temperature is converted to degrees Fahrenheit, F, using the formula $F = \frac{9}{5}C + 32$. What is the variance of the temperature in degrees Fahrenheit, Var(F)?
    Answer: 12.96 (°F)² The conversion is a linear transformation $F = aC + b$ with $a = 9/5$ and $b = 32$. The variance of F is $\text{Var}(F) = a^2 \text{Var}(C)$. So, $\text{Var}(F) = (\frac{9}{5})^2 \times \text{Var}(C) = (1.8)^2 \times 4 = 3.24 \times 4 = 12.96$. The units of variance are the square of the original units, so the variance is in (°F)². Distractor A incorrectly multiplies by $a$ instead of $a^2$. Distractors C and D incorrectly include the constant shift $b$ in the variance calculation.
16. A financial analyst is comparing two investment portfolios, A and B. Portfolio A has a return with a standard deviation of $100. Portfolio B has a return with a variance of 900 dollars². Which statement provides the most accurate conceptual comparison of their risk?
    Answer: Portfolio A is riskier because its variance is $100^2 = 10,000$, which is greater than Portfolio B's variance of 900. To compare the risk (spread) of the two portfolios, we need to use the same measure for both. The variance of Portfolio A is $(SD_A)^2 = (\$100)^2 = 10,000 \text{ dollars}^2$. The variance of Portfolio B is given as $900 \text{ dollars}^2$. Since $10,000 > 900$, Portfolio A has a much larger variance and is therefore considered riskier or more volatile. Distractor A makes an invalid comparison between variance and standard deviation. Distractor C misinterprets the magnitude of the difference in risk. Distractor D is incorrect as the measures are directly convertible.
17. Let X be the number rolled on a standard six-sided die. Let Y be the number on the opposite face of the die. For a standard die, the sum of opposite faces is always 7, so $Y = 7 - X$. What is the variance of the sum of the two faces, $\text{Var}(X+Y)$?
    Answer: 0 The sum of the two faces is a new random variable, $S = X + Y$. Since the problem states that $Y = 7 - X$, we can substitute this into the sum: $S = X + (7 - X) = 7$. The sum is always the constant value 7. The variance of any constant is zero. Therefore, $\text{Var}(S) = \text{Var}(7) = 0$. Distractor B, $35/6$, would be the correct answer if X and Y were independent, as it equals $2 \times \text{Var}(X)$. Distractor C is the variance of a single roll, $\text{Var}(X)$. Distractor D is incorrect because the dependence is perfectly defined, which allows the variance to be determined.
18. An investor builds a portfolio consisting of two stocks, Stock A and Stock B. Let $R_A$ and $R_B$ be the random returns of these stocks, with known positive variances. If the investor wants to create a portfolio $P = 0.5 R_A + 0.5 R_B$ with the minimum possible variance, what should be true about the relationship between the returns of the two stocks?
    Answer: They should be perfectly negatively correlated ($\rho = -1$). When you encounter portfolio optimization problems, you're dealing with the fundamental trade-off between risk and diversification. The key insight is understanding how correlation affects the variance of a combined portfolio. For a portfolio $P = w_A R_A + w_B R_B$, the variance formula is: $\text{Var}(P) = w_A^2 \text{Var}(R_A) + w_B^2 \text{Var}(R_B) + 2w_A w_B \text{Cov}(R_A, R_B)$ Since covariance equals $\rho \sigma_A \sigma_B$, this becomes: $\text{Var}(P) = w_A^2 \sigma_A^2 + w_B^2 \sigma_B^2 + 2w_A w_B \rho \sigma_A \sigma_B$ With equal weights (0.5 each), the correlation coefficient $\rho$ directly affects the covariance term. To minimize portfolio variance, you want the covariance term to be as negative as possible, which occurs when $\rho = -1$. Perfect negative correlation means when one stock goes up, the other goes down by a proportional amount, creating maximum diversification benefit. Option A ($\rho = +1$) maximizes portfolio variance because the stocks move together perfectly, eliminating diversification benefits. Option B ($\rho = 0$) provides some risk reduction but not the minimum possible variance. Option C incorrectly ignores the covariance term entirely—correlation absolutely affects portfolio variance through the cross-product term. Remember this key principle: negative correlation is your friend in portfolio construction. When stocks move in opposite directions, they hedge each other's risk, reducing overall portfolio volatility. Look for the most negative correlation to minimize risk in equal-weighted portfolios.
19. A company's daily profit, X, in dollars, is a random variable with a variance of $\sigma^2$. To account for a new flat tax of $100 and a 10% corporate surcharge on the remaining profit, the net daily profit, Y, is calculated as $Y = 0.90(X - 100)$. What is the variance of the net daily profit, Y?
    Answer: $0.81 \sigma^2$ The net profit is $Y = 0.90(X - 100) = 0.90X - 90$. This is a linear transformation of X of the form $aX + b$, with $a = 0.90$ and $b = -90$. The variance of a linear transformation is given by $\text{Var}(aX + b) = a^2 \text{Var}(X)$. Therefore, $\text{Var}(Y) = (0.90)^2 \text{Var}(X) = 0.81\sigma^2$. The constant shift of -90 does not affect the variance. Distractor A incorrectly uses $a$ instead of $a^2$. Distractors C and D incorrectly attempt to incorporate the shift term into the variance calculation.
20. The two histograms shown display the distributions of daily sales for two different products, Product A and Product B, over a period of 100 days. Both distributions have approximately the same mean, near 50 units. Based on the histograms, which statement about the variances is most likely true?
    Answer: The variance of sales for Product A is greater than the variance of sales for Product B. Variance is a measure of the spread or dispersion of a distribution. Visually comparing the two histograms, the distribution of sales for Product A is much more spread out than the distribution for Product B. The values for Product A range widely, while the values for Product B are more tightly clustered around the mean. Therefore, the variance of sales for Product A is greater. Distractor B incorrectly assesses the visual spread. Distractor C ignores the clear difference in spread. Distractor D is incorrect because spread is precisely what a histogram helps to visualize.

Start this quiz