# Transforming & Combining Random Variables ## Grade 9 Statistics & Probability Worksheet
Section A: Multiple Choice Questions
Q1: If X is a random variable with mean μX = 10 and standard deviation σX = 3, what is the mean of Y = 2X + 5? (a) 15 (b) 20 (c) 25 (d) 30
Solution:
Ans: (c) Explanation: When transforming a random variable with a linear transformation \(Y = aX + b\), the mean transforms as \(\mu_Y = a\mu_X + b\). Here, \(\mu_Y = 2(10) + 5 = 25\). The constant multiplies the mean, and then the added constant is included.
Q2: A random variable X has a standard deviation of 4. If Y = 3X, what is the standard deviation of Y? (a) 4 (b) 7 (c) 12 (d) 16
Solution:
Ans: (c) Explanation: When a random variable is multiplied by a constant \(a\), the standard deviation is multiplied by the absolute value of that constant: \(\sigma_Y = |a|\sigma_X\). Therefore, \(\sigma_Y = |3|(4) = 12\). Note that adding a constant does not affect standard deviation, but multiplying does.
Q3: If X and Y are independent random variables with variances Var(X) = 9 and Var(Y) = 16, what is Var(X + Y)? (a) 5 (b) 7 (c) 25 (d) 144
Solution:
Ans: (c) Explanation: For independent random variables, the variance of the sum equals the sum of the variances: \(\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) = 9 + 16 = 25\). Option (a) incorrectly subtracts, option (b) adds standard deviations instead of variances, and option (d) multiplies the variances.
Q4: Random variable X has mean 50 and standard deviation 5. What is the variance of W = X - 10? (a) 5 (b) 15 (c) 25 (d) 40
Solution:
Ans: (c) Explanation:Adding or subtracting a constant from a random variable does not change its variance or standard deviation. Since \(\text{Var}(X) = \sigma_X^2 = 5^2 = 25\), we have \(\text{Var}(W) = \text{Var}(X) = 25\). The transformation only shifts the distribution but does not affect spread.
Q5: If X and Y are independent with means μX = 8 and μY = 12, what is the mean of X - Y? (a) -4 (b) 4 (c) 20 (d) 96
Solution:
Ans: (a) Explanation: The mean of a difference of random variables is the difference of their means: \(\mu_{X-Y} = \mu_X - \mu_Y = 8 - 12 = -4\). This rule applies whether or not the variables are independent. Option (b) incorrectly takes the absolute value, option (c) adds the means, and option (d) multiplies them.
Q6: If X has Var(X) = 20 and Y has Var(Y) = 15, and X and Y are independent, what is Var(X - Y)? (a) 5 (b) 15 (c) 25 (d) 35
Solution:
Ans: (d) Explanation: For independent random variables, the variance of the difference equals the sum of the variances: \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) = 20 + 15 = 35\). Note that variances add for both sum and difference when variables are independent. Option (a) incorrectly subtracts the variances.
Q7: A random variable X has mean 100 and variance 36. What is the standard deviation of Z = -2X + 50? (a) 6 (b) 12 (c) 36 (d) 72
Solution:
Ans: (b) Explanation: The standard deviation of \(X\) is \(\sigma_X = \sqrt{36} = 6\). For the transformation \(Z = -2X + 50\), the standard deviation becomes \(\sigma_Z = |-2|\sigma_X = 2(6) = 12\). The negative sign becomes positive because we take the absolute value of the multiplier, and the added constant does not affect standard deviation.
Q8: Three independent random variables X, Y, and Z have variances 4, 9, and 5 respectively. What is Var(X + Y + Z)? (a) 6 (b) 18 (c) 36 (d) 180
Solution:
Ans: (b) Explanation: For independent random variables, the variance of the sum is the sum of the individual variances: \(\text{Var}(X + Y + Z) = \text{Var}(X) + \text{Var}(Y) + \text{Var}(Z) = 4 + 9 + 5 = 18\). Option (a) adds standard deviations instead, option (c) squares the sum, and option (d) multiplies the variances.
Section B: Fill in the Blanks
Q9: When a constant is added to a random variable, the mean increases by that constant but the __________ remains unchanged.
Solution:
Ans: variance (or standard deviation) Explanation:Adding a constant to a random variable shifts the entire distribution but does not change the spread of the distribution. Therefore, both variance and standard deviation remain the same.
Q10: If X and Y are independent random variables, then Var(X + Y) = Var(X) + __________.
Solution:
Ans: Var(Y) Explanation: For independent random variables, the variance of the sum equals the sum of the variances. This is a fundamental property used when combining independent random variables.
Q11: For a linear transformation Y = aX + b, the variance of Y is given by Var(Y) = __________ × Var(X).
Solution:
Ans: a² Explanation: When transforming a random variable with \(Y = aX + b\), the variance is multiplied by the square of the constant: \(\text{Var}(Y) = a^2\text{Var}(X)\). The constant \(b\) does not affect variance.
Q12: If two random variables X and Y are independent, then the variance of their difference Var(X - Y) = Var(X) + Var(Y), which means variances __________ for both sums and differences.
Solution:
Ans: add Explanation: A key property of independent random variables is that variances add for both the sum and the difference. This is because \(\text{Var}(X - Y) = \text{Var}(X + (-Y)) = \text{Var}(X) + \text{Var}(-Y) = \text{Var}(X) + \text{Var}(Y)\).
Q13: The expected value (mean) of the sum of two random variables equals the __________ of their individual expected values, regardless of whether they are independent.
Solution:
Ans: sum Explanation: The linearity of expectation states that \(E(X + Y) = E(X) + E(Y)\) for any random variables X and Y, whether or not they are independent. This is a fundamental property of expected value.
Q14: If the standard deviation of X is σX, then the standard deviation of 5X is __________.
Solution:
Ans: 5σX Explanation: When a random variable is multiplied by a constant, the standard deviation is multiplied by the absolute value of that constant. Therefore, \(\sigma_{5X} = |5|\sigma_X = 5\sigma_X\).
Section C: Word Problems
Q15: A coffee shop sells cups of coffee with a mean volume of 350 mL and a standard deviation of 8 mL. If each cup is measured and then 50 mL of cream is added, what are the new mean and standard deviation of the total volume?
Solution:
Ans: Step 1: Let X represent the volume of coffee. We have \(\mu_X = 350\) mL and \(\sigma_X = 8\) mL. Step 2: The new volume is \(Y = X + 50\). Step 3: Using the transformation rules: \(\mu_Y = \mu_X + 50 = 350 + 50 = 400\) mL. Step 4: Adding a constant does not change standard deviation: \(\sigma_Y = \sigma_X = 8\) mL. Final Answer: Mean = 400 mL, Standard Deviation = 8 mL
Q16: The time spent on homework for math has a mean of 45 minutes with a standard deviation of 10 minutes. The time spent on science homework has a mean of 30 minutes with a standard deviation of 8 minutes. Assuming the times are independent, find the mean and standard deviation of the total homework time.
Solution:
Ans: Step 1: Let M = math homework time and S = science homework time. Given: \(\mu_M = 45\) min, \(\sigma_M = 10\) min, \(\mu_S = 30\) min, \(\sigma_S = 8\) min. Step 2: Total time \(T = M + S\). Step 3: Mean of total: \(\mu_T = \mu_M + \mu_S = 45 + 30 = 75\) minutes. Step 4: Since M and S are independent: \(\text{Var}(T) = \text{Var}(M) + \text{Var}(S) = 10^2 + 8^2 = 100 + 64 = 164\). Step 5: Standard deviation: \(\sigma_T = \sqrt{164} \approx 12.81\) minutes. Final Answer: Mean = 75 minutes, Standard Deviation ≈ 12.81 minutes
Q17: A store marks up the wholesale price of an item by tripling it and then adding $15. If the wholesale price has a mean of $40 and a standard deviation of $5, what are the mean and standard deviation of the retail price?
Solution:
Ans: Step 1: Let W = wholesale price. Given: \(\mu_W = 40\) dollars, \(\sigma_W = 5\) dollars. Step 2: Retail price \(R = 3W + 15\). Step 3: Mean of retail price: \(\mu_R = 3\mu_W + 15 = 3(40) + 15 = 120 + 15 = 135\) dollars. Step 4: Standard deviation: \(\sigma_R = |3|\sigma_W = 3(5) = 15\) dollars. Note: The added constant (15) does not affect standard deviation. Final Answer: Mean = $135, Standard Deviation = $15
Q18: Two independent random variables X and Y have the following properties: X has mean 20 and variance 25, while Y has mean 15 and variance 16. Calculate the mean and variance of the random variable D = X - Y.
Solution:
Ans: Step 1: Given: \(\mu_X = 20\), \(\text{Var}(X) = 25\), \(\mu_Y = 15\), \(\text{Var}(Y) = 16\). Step 2: For \(D = X - Y\), the mean is: \(\mu_D = \mu_X - \mu_Y = 20 - 15 = 5\). Step 3: Since X and Y are independent, the variance of the difference is the sum of variances: \(\text{Var}(D) = \text{Var}(X) + \text{Var}(Y) = 25 + 16 = 41\). Final Answer: Mean = 5, Variance = 41
Q19: A factory produces bolts with lengths that have a mean of 50 mm and a standard deviation of 0.5 mm. Quality control requires that each bolt length be measured and then converted to centimeters by dividing by 10. What are the mean and standard deviation of the bolt lengths in centimeters?
Solution:
Ans: Step 1: Let L = bolt length in mm. Given: \(\mu_L = 50\) mm, \(\sigma_L = 0.5\) mm. Step 2: Length in cm: \(C = \frac{L}{10} = 0.1L\). Step 3: Mean in cm: \(\mu_C = 0.1\mu_L = 0.1(50) = 5\) cm. Step 4: Standard deviation in cm: \(\sigma_C = |0.1|\sigma_L = 0.1(0.5) = 0.05\) cm. Final Answer: Mean = 5 cm, Standard Deviation = 0.05 cm
Q20: A student takes three independent quizzes. Quiz 1 has a mean score of 80 with variance 9, Quiz 2 has a mean of 75 with variance 16, and Quiz 3 has a mean of 85 with variance 4. If the total score T is the sum of all three quiz scores, find the mean and standard deviation of T.
Solution:
Ans: Step 1: Let \(Q_1, Q_2, Q_3\) represent the three quiz scores. Given: \(\mu_{Q_1} = 80\), \(\text{Var}(Q_1) = 9\); \(\mu_{Q_2} = 75\), \(\text{Var}(Q_2) = 16\); \(\mu_{Q_3} = 85\), \(\text{Var}(Q_3) = 4\). Step 2: Total score: \(T = Q_1 + Q_2 + Q_3\). Step 3: Mean of total: \(\mu_T = \mu_{Q_1} + \mu_{Q_2} + \mu_{Q_3} = 80 + 75 + 85 = 240\). Step 4: Since quizzes are independent: \(\text{Var}(T) = \text{Var}(Q_1) + \text{Var}(Q_2) + \text{Var}(Q_3) = 9 + 16 + 4 = 29\). Step 5: Standard deviation: \(\sigma_T = \sqrt{29} \approx 5.39\). Final Answer: Mean = 240, Standard Deviation ≈ 5.39
The document Worksheet (with Solutions): Transforming & Combining Random Variables is a part of the Grade 9 Course Statistics & Probability.
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