IB Mathematics: Applications and Interpretation Quiz: Discrete Random Variables
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Discrete Random VariablesQuestion 1 of 20

A player pays $5 to play a game with a spinner. The spinner has three sectors: Red (P=0.5P=0.5), Blue (P=0.3P=0.3), and Green (P=0.2P=0.2). Landing on Red wins $2, and landing on Blue wins $5. What must be the prize for landing on Green for the game to be considered fair?

$10.00
$12.50
$25.00
$37.50
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Discrete Random Variables

Practice Discrete Random Variables in IB Mathematics: Applications and Interpretation 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 Discrete Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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

A player pays $5 to play a game with a spinner. The spinner has three sectors: Red (P=0.5P=0.5), Blue (P=0.3P=0.3), and Green (P=0.2P=0.2). Landing on Red wins $2, and landing on Blue wins $5. What must be the prize for landing on Green for the game to be considered fair?

  1. $10.00
  2. $12.50 (correct answer)
  3. $25.00
  4. $37.50
Explanation: A fair game means the expected profit is zero, or the expected payout is equal to the cost of playing. The cost is $5. Let GG be the prize for landing on Green. The expected payout is E(Payout)=(2)(0.5)+(5)(0.3)+(G)(0.2)E(\text{Payout}) = (2)(0.5) + (5)(0.3) + (G)(0.2). For a fair game, E(Payout)=5E(\text{Payout}) = 5. 1.0+1.5+0.2G=51.0 + 1.5 + 0.2G = 5 2.5+0.2G=52.5 + 0.2G = 5 0.2G=2.50.2G = 2.5 G=2.50.2=12.5G = \frac{2.5}{0.2} = 12.5 The prize for Green must be 12.50.DistractorC(12.50. Distractor C (25.00) incorrectly assumes the expected payout from Green alone must equal the cost (0.2G=50.2G=5). Distractor D ($37.50) results from setting the expected profit to $5 instead of $0.

Question 2

A lottery ticket has the following probability distribution for the prize, XX. P(X=0)=0.6P(X=0)=0.6, P(X=10)=0.3P(X=10)=0.3, and P(X=k)=0.1P(X=k)=0.1. The expected prize value of the ticket is $8. Find the value of the prize kk.

  1. 30
  2. 50 (correct answer)
  3. 80
  4. 110
Explanation: The expected value is given by the formula E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x). We are given E(X)=8E(X) = 8. E(X)=(0)(0.6)+(10)(0.3)+(k)(0.1)E(X) = (0)(0.6) + (10)(0.3) + (k)(0.1) 8=0+3+0.1k8 = 0 + 3 + 0.1k 8=3+0.1k8 = 3 + 0.1k Subtract 3 from both sides: 5=0.1k5 = 0.1k Divide by 0.1: k=50.1=50k = \frac{5}{0.1} = 50 Distractor C (80) results from ignoring the $10 prize (8=0.1k8 = 0.1k). Distractor D (110) results from an algebraic error (8+3=0.1k8+3=0.1k).

Question 3

A company models its daily sales of a specific product, XX, as a discrete random variable. After analyzing historical data, they calculate the expected value to be E(X)=8.2E(X) = 8.2. Which statement is the best interpretation of this expected value?

  1. The company will sell exactly 8.2 products on any given day.
  2. The company is guaranteed to sell more than 8 products on most days.
  3. The most likely number of products sold on any given day is 8.
  4. Over a long period, the average number of products sold per day will be close to 8.2. (correct answer)
Explanation: When you encounter expected value questions, you're dealing with the fundamental concept of what happens "on average" over many repetitions of a random process. The expected value E(X)=8.2E(X) = 8.2 represents the theoretical mean of the distribution - the center point around which actual outcomes will cluster over time. Option D correctly captures this interpretation. Expected value tells you the long-run average: if this company tracked daily sales over hundreds or thousands of days, the mean of all those daily totals would approach 8.2. This is the law of large numbers in action. Option A reflects a common misconception about expected value. Since XX is discrete (you can't sell 8.2 products), the expected value doesn't have to be a possible outcome. It's a theoretical average, not a prediction for any single day. Option B incorrectly assumes the expected value guarantees outcomes above 8 on most days. Expected value tells you nothing about the probability distribution around that mean - the company could sell 0 products on many days and 20+ on others, still averaging 8.2. Option C confuses expected value with mode (the most frequent outcome). The expected value is the mean of the distribution, not necessarily the most likely single outcome. In fact, 8.2 isn't even a possible outcome for this discrete variable. Remember: expected value always describes long-term average behavior, never single-event predictions. When you see expected value questions on the IB exam, focus on the "over many trials" interpretation rather than what happens on any individual trial.

Question 4

A biased coin is tossed twice. The probability of getting a Head (H) is 0.6. Let XX be the number of Heads obtained. Find the variance of XX.

  1. 0.48 (correct answer)
  2. 0.50
  3. 1.20
  4. 1.92
Explanation: First, establish the probability distribution for XX, the number of heads. P(H)=0.6P(H) = 0.6, P(T)=0.4P(T) = 0.4. P(X=0)=P(TT)=(0.4)(0.4)=0.16P(X=0) = P(TT) = (0.4)(0.4) = 0.16 P(X=1)=P(HT or TH)=(0.6)(0.4)+(0.4)(0.6)=0.24+0.24=0.48P(X=1) = P(HT \text{ or } TH) = (0.6)(0.4) + (0.4)(0.6) = 0.24 + 0.24 = 0.48 P(X=2)=P(HH)=(0.6)(0.6)=0.36P(X=2) = P(HH) = (0.6)(0.6) = 0.36 Next, calculate E(X)E(X) and E(X2)E(X^2). E(X)=0(0.16)+1(0.48)+2(0.36)=0+0.48+0.72=1.2E(X) = 0(0.16) + 1(0.48) + 2(0.36) = 0 + 0.48 + 0.72 = 1.2 E(X2)=02(0.16)+12(0.48)+22(0.36)=0+0.48+1.44=1.92E(X^2) = 0^2(0.16) + 1^2(0.48) + 2^2(0.36) = 0 + 0.48 + 1.44 = 1.92 Finally, calculate the variance: Var(X)=E(X2)−[E(X)]2=1.92−(1.2)2=1.92−1.44=0.48Var(X) = E(X^2) - [E(X)]^2 = 1.92 - (1.2)^2 = 1.92 - 1.44 = 0.48. Alternatively, using the binomial distribution formula Var(X)=np(1−p)Var(X) = np(1-p) for n=2,p=0.6n=2, p=0.6, Var(X)=2(0.6)(0.4)=0.48Var(X) = 2(0.6)(0.4) = 0.48. Distractor C is E(X)E(X), and D is E(X2)E(X^2). Distractor B is the variance for a fair coin.

Question 5

A baker makes a special type of cake. The number of cakes sold per day, XX, follows the probability distribution: P(X=0)=0.1P(X=0)=0.1, P(X=1)=0.2P(X=1)=0.2, P(X=2)=0.4P(X=2)=0.4, P(X=3)=0.3P(X=3)=0.3. Each cake costs $5 to make and is sold for $15. The baker always makes 3 cakes each day, and any unsold cakes are discarded at a total loss.

What is the baker's expected daily profit?

  1. $12.00
  2. $13.50 (correct answer)
  3. $15.00
  4. $16.50
Explanation: First, determine the profit for each outcome of XX (number of cakes sold), given that 3 cakes are always made. The total cost is 3 \times \5 = $15$. The profit is Revenue - Cost.
  • If X=0X=0, Revenue = $0. Profit = $0 - 15=−15 = -15.
  • If X=1X=1, Revenue = $15. Profit = $15 - $15 = $0.
  • If X=2X=2, Revenue = $30. Profit = $30 - $15 = $15.
  • If X=3X=3, Revenue = $45. Profit = $45 - $15 = $30. Now, calculate the expected profit: E(Profit)=(−15)(0.1)+(0)(0.2)+(15)(0.4)+(30)(0.3)E(\text{Profit}) = (-15)(0.1) + (0)(0.2) + (15)(0.4) + (30)(0.3) E(Profit)=−1.5+0+6.0+9.0=13.5E(\text{Profit}) = -1.5 + 0 + 6.0 + 9.0 = 13.5 The expected daily profit is 13.50.DistractorA(13.50. Distractor A (12.00) results from calculation errors. Distractor C (15.00)assumesnolossesfromunsoldcakes.DistractorD(15.00) assumes no losses from unsold cakes. Distractor D (16.50) overestimates the expected sales.

Question 6

The number of typos, XX, on a randomly chosen page of a book is a discrete random variable with E(X)=2.5E(X) = 2.5. An editor charges a fee based on the square of the number of typos, given by the formula C=4X2C = 4X^2. Which of the following statements must be true about the expected fee, E(C)E(C)?

  1. E(C)=25E(C) = 25
  2. The relationship cannot be determined without the full probability distribution.
  3. E(C)<25E(C) < 25
  4. E(C)>25E(C) > 25 (correct answer)
Explanation: When you encounter questions about the expected value of transformed random variables, the key concept is understanding how expectation behaves under non-linear transformations. This tests whether you know that E[g(X)]≠g(E[X])E[g(X)] \neq g(E[X]) when gg is non-linear. Given that E(X)=2.5E(X) = 2.5 and C=4X2C = 4X^2, we need E(C)=E(4X2)=4E(X2)E(C) = E(4X^2) = 4E(X^2). The crucial insight is that E(X2)≠[E(X)]2E(X^2) \neq [E(X)]^2 for any random variable with non-zero variance. Since XX represents typos on a page (a count that naturally varies), it must have positive variance. Using the variance formula: Var(X)=E(X2)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2, we can rearrange to get E(X2)=Var(X)+[E(X)]2E(X^2) = \text{Var}(X) + [E(X)]^2. Since Var(X)>0\text{Var}(X) > 0, we have E(X2)>[E(X)]2=(2.5)2=6.25E(X^2) > [E(X)]^2 = (2.5)^2 = 6.25. Therefore, E(C)=4E(X2)>4(6.25)=25E(C) = 4E(X^2) > 4(6.25) = 25. Option A incorrectly assumes E(C)=4[E(X)]2=4(2.5)2=25E(C) = 4[E(X)]^2 = 4(2.5)^2 = 25, falling into the common trap of treating expectation as if it's linear for all functions. Option B suggests we need the full distribution, but we only need to know that variance is positive. Option C incorrectly claims the expected fee is less than 25, contradicting our analysis. Remember: for any non-linear transformation of a random variable, you cannot simply plug the expected value into the function. Jensen's inequality tells us that for convex functions like x2x^2, the expectation of the transformed variable exceeds the transformation of the expectation.

Question 7

In a game, a player's score XX has the following probability distribution: P(X=−2)=0.1P(X=-2)=0.1, P(X=1)=0.5P(X=1)=0.5, P(X=5)=0.4P(X=5)=0.4. Find the standard deviation of XX, correct to three significant figures.

  1. 2.24
  2. 2.30
  3. 2.37 (correct answer)
  4. 5.61
Explanation: First, find the expected value, E(X)E(X). E(X)=(−2)(0.1)+(1)(0.5)+(5)(0.4)=−0.2+0.5+2.0=2.3E(X) = (-2)(0.1) + (1)(0.5) + (5)(0.4) = -0.2 + 0.5 + 2.0 = 2.3 Next, find E(X2)E(X^2). E(X2)=(−2)2(0.1)+(1)2(0.5)+(5)2(0.4)=4(0.1)+1(0.5)+25(0.4)=0.4+0.5+10.0=10.9E(X^2) = (-2)^2(0.1) + (1)^2(0.5) + (5)^2(0.4) = 4(0.1) + 1(0.5) + 25(0.4) = 0.4 + 0.5 + 10.0 = 10.9 Then, find the variance, Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2. Var(X)=10.9−(2.3)2=10.9−5.29=5.61Var(X) = 10.9 - (2.3)^2 = 10.9 - 5.29 = 5.61 Finally, the standard deviation is the square root of the variance: σ=Var(X)=5.61≈2.3685...\sigma = \sqrt{Var(X)} = \sqrt{5.61} \approx 2.3685... Rounding to three significant figures gives 2.37. Distractor D (5.61) is the variance. Distractor B (2.30) is the expected value. Distractor A (2.24) results from an error in calculating E(X2)E(X^2) where (−2)2(-2)^2 is taken as -4.

Question 8

An insurance company sells a policy for $250. The policy covers an accident that has a 2% probability of occurring. If the accident occurs, the company pays out $10,000. Let XX be the company's profit from one policy. Find the expected profit, E(X)E(X), for the company per policy sold.

  1. -$200
  2. $250
  3. $245
  4. $50 (correct answer)
Explanation: When you encounter expected value problems in business contexts, you're calculating the long-run average outcome by weighing each possible result by its probability. To find the company's expected profit per policy, you need to identify all possible outcomes and their probabilities. There are two scenarios: either the accident occurs (2% chance) or it doesn't (98% chance). If no accident occurs (98% probability): The company keeps the $250 premium and pays nothing, so profit = $250. If an accident occurs (2% probability): The company receives $250 but pays out $10,000, so profit = $250 - 10,000=−10,000 = -9,750. The expected profit is: E(X)=0.98×250+0.02×(−9,750)=245−195=50E(X) = 0.98 × 250 + 0.02 × (-9,750) = 245 - 195 = 50 Therefore, the answer is D) $50. Let's examine why the other options are incorrect: A) -$200 likely comes from incorrectly focusing only on the loss scenario or making calculation errors with the probabilities. B) $250 represents the premium collected, but this ignores the potential payout entirely. This would only be correct if there were no risk of claims. C) $245 results from calculating only the "no accident" scenario (0.98 × $250) while completely ignoring the negative impact of potential payouts. Study tip: In expected value problems involving insurance or gambling, always identify every possible outcome, calculate the profit/loss for each scenario, then multiply by probabilities. The key insight is that profit equals revenue minus costs, not just revenue.

Question 9

A player pays $5 to play a game with a spinner. The spinner has three sectors: Red (P=0.5P=0.5), Blue (P=0.3P=0.3), and Green (P=0.2P=0.2). Landing on Red wins $2, and landing on Blue wins $5. What must be the prize for landing on Green for the game to be considered fair?

  1. $10.00
  2. $12.50 (correct answer)
  3. $25.00
  4. $37.50
Explanation: A fair game means the expected profit is zero, or the expected payout is equal to the cost of playing. The cost is $5. Let GG be the prize for landing on Green. The expected payout is E(Payout)=(2)(0.5)+(5)(0.3)+(G)(0.2)E(\text{Payout}) = (2)(0.5) + (5)(0.3) + (G)(0.2). For a fair game, E(Payout)=5E(\text{Payout}) = 5. 1.0+1.5+0.2G=51.0 + 1.5 + 0.2G = 5 2.5+0.2G=52.5 + 0.2G = 5 0.2G=2.50.2G = 2.5 G=2.50.2=12.5G = \frac{2.5}{0.2} = 12.5 The prize for Green must be 12.50.DistractorC(12.50. Distractor C (25.00) incorrectly assumes the expected payout from Green alone must equal the cost (0.2G=50.2G=5). Distractor D ($37.50) results from setting the expected profit to $5 instead of $0.

Question 10

A lottery ticket has the following probability distribution for the prize, XX. P(X=0)=0.6P(X=0)=0.6, P(X=10)=0.3P(X=10)=0.3, and P(X=k)=0.1P(X=k)=0.1. The expected prize value of the ticket is $8. Find the value of the prize kk.

  1. 30
  2. 50 (correct answer)
  3. 80
  4. 110
Explanation: The expected value is given by the formula E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x). We are given E(X)=8E(X) = 8. E(X)=(0)(0.6)+(10)(0.3)+(k)(0.1)E(X) = (0)(0.6) + (10)(0.3) + (k)(0.1) 8=0+3+0.1k8 = 0 + 3 + 0.1k 8=3+0.1k8 = 3 + 0.1k Subtract 3 from both sides: 5=0.1k5 = 0.1k Divide by 0.1: k=50.1=50k = \frac{5}{0.1} = 50 Distractor C (80) results from ignoring the $10 prize (8=0.1k8 = 0.1k). Distractor D (110) results from an algebraic error (8+3=0.1k8+3=0.1k).

Question 11

An insurance company sells a policy for $250. The policy covers an accident that has a 2% probability of occurring. If the accident occurs, the company pays out $10,000. Let XX be the company's profit from one policy. Find the expected profit, E(X)E(X), for the company per policy sold.

  1. -$200
  2. $250
  3. $245
  4. $50 (correct answer)
Explanation: When you encounter expected value problems in business contexts, you're calculating the long-run average outcome by weighing each possible result by its probability. To find the company's expected profit per policy, you need to identify all possible outcomes and their probabilities. There are two scenarios: either the accident occurs (2% chance) or it doesn't (98% chance). If no accident occurs (98% probability): The company keeps the $250 premium and pays nothing, so profit = $250. If an accident occurs (2% probability): The company receives $250 but pays out $10,000, so profit = $250 - 10,000=−10,000 = -9,750. The expected profit is: E(X)=0.98×250+0.02×(−9,750)=245−195=50E(X) = 0.98 × 250 + 0.02 × (-9,750) = 245 - 195 = 50 Therefore, the answer is D) $50. Let's examine why the other options are incorrect: A) -$200 likely comes from incorrectly focusing only on the loss scenario or making calculation errors with the probabilities. B) $250 represents the premium collected, but this ignores the potential payout entirely. This would only be correct if there were no risk of claims. C) $245 results from calculating only the "no accident" scenario (0.98 × $250) while completely ignoring the negative impact of potential payouts. Study tip: In expected value problems involving insurance or gambling, always identify every possible outcome, calculate the profit/loss for each scenario, then multiply by probabilities. The key insight is that profit equals revenue minus costs, not just revenue.

Question 12

A discrete random variable XX can take the values 1, 2, 3, and 4. Its probability distribution is given by P(X=x)=kxP(X=x) = kx for some constant kk. Find the expected value of XX.

  1. 2.5
  2. 3.0 (correct answer)
  3. 3.5
  4. 4.0
Explanation: First, we must find the value of the constant kk. The sum of all probabilities must be 1. P(X=1)+P(X=2)+P(X=3)+P(X=4)=1P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 k(1)+k(2)+k(3)+k(4)=1k(1) + k(2) + k(3) + k(4) = 1 k+2k+3k+4k=1k + 2k + 3k + 4k = 1 10k=1  ⟹  k=0.110k = 1 \implies k = 0.1 Now we have the probabilities: P(X=1)=0.1,P(X=2)=0.2,P(X=3)=0.3,P(X=4)=0.4P(X=1)=0.1, P(X=2)=0.2, P(X=3)=0.3, P(X=4)=0.4. Calculate the expected value, E(X)E(X): E(X)=1(0.1)+2(0.2)+3(0.3)+4(0.4)E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) E(X)=0.1+0.4+0.9+1.6=3.0E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0 Distractor A (2.5) is the arithmetic mean of the values, ignoring probabilities. Other distractors result from calculation errors.

Question 13

In a game, a player's score XX has the following probability distribution: P(X=−2)=0.1P(X=-2)=0.1, P(X=1)=0.5P(X=1)=0.5, P(X=5)=0.4P(X=5)=0.4. Find the standard deviation of XX, correct to three significant figures.

  1. 2.24
  2. 2.30
  3. 2.37 (correct answer)
  4. 5.61
Explanation: First, find the expected value, E(X)E(X). E(X)=(−2)(0.1)+(1)(0.5)+(5)(0.4)=−0.2+0.5+2.0=2.3E(X) = (-2)(0.1) + (1)(0.5) + (5)(0.4) = -0.2 + 0.5 + 2.0 = 2.3 Next, find E(X2)E(X^2). E(X2)=(−2)2(0.1)+(1)2(0.5)+(5)2(0.4)=4(0.1)+1(0.5)+25(0.4)=0.4+0.5+10.0=10.9E(X^2) = (-2)^2(0.1) + (1)^2(0.5) + (5)^2(0.4) = 4(0.1) + 1(0.5) + 25(0.4) = 0.4 + 0.5 + 10.0 = 10.9 Then, find the variance, Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2. Var(X)=10.9−(2.3)2=10.9−5.29=5.61Var(X) = 10.9 - (2.3)^2 = 10.9 - 5.29 = 5.61 Finally, the standard deviation is the square root of the variance: σ=Var(X)=5.61≈2.3685...\sigma = \sqrt{Var(X)} = \sqrt{5.61} \approx 2.3685... Rounding to three significant figures gives 2.37. Distractor D (5.61) is the variance. Distractor B (2.30) is the expected value. Distractor A (2.24) results from an error in calculating E(X2)E(X^2) where (−2)2(-2)^2 is taken as -4.

Question 14

The score, XX, in a game has the probability distribution P(X=1)=0.5P(X=1)=0.5, P(X=2)=0.3P(X=2)=0.3, P(X=3)=0.2P(X=3)=0.2. A player's prize money, MM, is calculated using the formula M=5X2+10M = 5X^2 + 10. What is the expected prize money, E(M)E(M)?

  1. 17.50
  2. 24.45
  3. 27.50 (correct answer)
  4. 82.25
Explanation: To find the expected prize money, we use the property of expectation: E(M)=E(5X2+10)=5E(X2)+10E(M) = E(5X^2 + 10) = 5E(X^2) + 10. First, we need to calculate E(X2)E(X^2). E(X2)=12(0.5)+22(0.3)+32(0.2)E(X^2) = 1^2(0.5) + 2^2(0.3) + 3^2(0.2) E(X2)=1(0.5)+4(0.3)+9(0.2)=0.5+1.2+1.8=3.5E(X^2) = 1(0.5) + 4(0.3) + 9(0.2) = 0.5 + 1.2 + 1.8 = 3.5 Now substitute this value back into the equation for E(M)E(M). E(M)=5(3.5)+10=17.5+10=27.5E(M) = 5(3.5) + 10 = 17.5 + 10 = 27.5 The expected prize money is $27.50. Distractor B (24.45) is a common error from calculating 5[E(X)]2+105[E(X)]^2 + 10 instead of 5E(X2)+105E(X^2) + 10. Distractor A (17.50) forgets to add the constant 10.

Question 15

A discrete random variable XX has the probability distribution P(X=1)=pP(X=1)=p, P(X=2)=qP(X=2)=q, and P(X=3)=0.3P(X=3)=0.3. Given that the expected value E(X)=1.9E(X) = 1.9, what is the value of pp?

  1. 0.4 (correct answer)
  2. 0.5
  3. 0.6
  4. 0.7
Explanation: We have two unknowns, pp and qq, so we need to set up a system of two linear equations. Equation 1: The sum of probabilities is 1. p+q+0.3=1  ⟹  p+q=0.7p + q + 0.3 = 1 \implies p + q = 0.7 Equation 2: The expected value is 1.9. E(X)=1(p)+2(q)+3(0.3)=1.9E(X) = 1(p) + 2(q) + 3(0.3) = 1.9 p+2q+0.9=1.9  ⟹  p+2q=1.0p + 2q + 0.9 = 1.9 \implies p + 2q = 1.0 Now we solve the system:
  1. p+q=0.7p + q = 0.7
  2. p+2q=1.0p + 2q = 1.0 Subtracting equation (1) from equation (2): (p+2q)−(p+q)=1.0−0.7  ⟹  q=0.3(p + 2q) - (p + q) = 1.0 - 0.7 \implies q = 0.3 Substitute q=0.3q=0.3 back into equation (1): p+0.3=0.7  ⟹  p=0.4p + 0.3 = 0.7 \implies p = 0.4 Verification: E(X)=1(0.4)+2(0.3)+3(0.3)=0.4+0.6+0.9=1.9E(X) = 1(0.4) + 2(0.3) + 3(0.3) = 0.4 + 0.6 + 0.9 = 1.9 ✓

Question 16

Let XX be the daily high temperature in degrees Celsius in a city, with an expected value E(X)=15E(X) = 15. The temperature in degrees Fahrenheit, YY, is given by the formula Y=1.8X+32Y = 1.8X + 32. What is the expected daily high temperature in degrees Fahrenheit?

  1. 27.0
  2. 47.0
  3. 59.0 (correct answer)
  4. 80.6
Explanation: We use the property of expected values that E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b. Here, a=1.8a = 1.8 and b=32b = 32. We are given E(X)=15E(X) = 15. Therefore, E(Y)=E(1.8X+32)=1.8⋅E(X)+32E(Y) = E(1.8X + 32) = 1.8 \cdot E(X) + 32. E(Y)=1.8(15)+32=27+32=59.0E(Y) = 1.8(15) + 32 = 27 + 32 = 59.0 Distractor A (27.0) forgets to add the constant 32. Distractor B (47.0) forgets to multiply by the coefficient 1.8. Distractor D (80.6) incorrectly uses a property of variance, a2E(X)+ba^2 E(X) + b.

Question 17

A machine produces bolts. The number of defective bolts per batch, XX, is a random variable with a variance of Var(X)=4Var(X) = 4. The cost, in dollars, associated with the defects is modelled by the equation C=10X+50C = 10X + 50. What is the variance of the cost, Var(C)Var(C)?

  1. 90
  2. 400 (correct answer)
  3. 450
  4. 1650
Explanation: We use the property of variance that Var(aX+b)=a2Var(X)Var(aX + b) = a^2 Var(X). Here, a=10a = 10 and b=50b = 50. We are given Var(X)=4Var(X) = 4. Therefore, Var(C)=Var(10X+50)=102⋅Var(X)Var(C) = Var(10X + 50) = 10^2 \cdot Var(X). Var(C)=100⋅4=400Var(C) = 100 \cdot 4 = 400 The constant b=50b=50 does not affect the variance. Distractor A (90) comes from the incorrect calculation 10⋅4+5010 \cdot 4 + 50. Distractor C (450) from 102⋅4+5010^2 \cdot 4 + 50. Distractor D (1650) from (10⋅4+50)2(10\cdot4+50)^2. A common error is to calculate a⋅Var(X)a \cdot Var(X) instead of a2⋅Var(X)a^2 \cdot Var(X), which would give 10⋅4=4010 \cdot 4 = 40.

Question 18

The score, XX, in a game has the probability distribution P(X=1)=0.5P(X=1)=0.5, P(X=2)=0.3P(X=2)=0.3, P(X=3)=0.2P(X=3)=0.2. A player's prize money, MM, is calculated using the formula M=5X2+10M = 5X^2 + 10. What is the expected prize money, E(M)E(M)?

  1. 17.50
  2. 24.45
  3. 27.50 (correct answer)
  4. 82.25
Explanation: To find the expected prize money, we use the property of expectation: E(M)=E(5X2+10)=5E(X2)+10E(M) = E(5X^2 + 10) = 5E(X^2) + 10. First, we need to calculate E(X2)E(X^2). E(X2)=12(0.5)+22(0.3)+32(0.2)E(X^2) = 1^2(0.5) + 2^2(0.3) + 3^2(0.2) E(X2)=1(0.5)+4(0.3)+9(0.2)=0.5+1.2+1.8=3.5E(X^2) = 1(0.5) + 4(0.3) + 9(0.2) = 0.5 + 1.2 + 1.8 = 3.5 Now substitute this value back into the equation for E(M)E(M). E(M)=5(3.5)+10=17.5+10=27.5E(M) = 5(3.5) + 10 = 17.5 + 10 = 27.5 The expected prize money is $27.50. Distractor B (24.45) is a common error from calculating 5[E(X)]2+105[E(X)]^2 + 10 instead of 5E(X2)+105E(X^2) + 10. Distractor A (17.50) forgets to add the constant 10.

Question 19

The discrete random variable XX has probability distribution P(X=0)=0.6P(X=0)=0.6 and P(X=a)=0.4P(X=a)=0.4, where a>0a>0. The variance of XX is given as Var(X)=3.84Var(X) = 3.84. Find the value of aa.

  1. 2.0
  2. 3.2
  3. 4.0 (correct answer)
  4. 9.6
Explanation: First, find expressions for E(X)E(X) and E(X2)E(X^2) in terms of aa. E(X)=0(0.6)+a(0.4)=0.4aE(X) = 0(0.6) + a(0.4) = 0.4a E(X2)=02(0.6)+a2(0.4)=0.4a2E(X^2) = 0^2(0.6) + a^2(0.4) = 0.4a^2 Now use the variance formula: Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2. 3.84=0.4a2−(0.4a)23.84 = 0.4a^2 - (0.4a)^2 3.84=0.4a2−0.16a23.84 = 0.4a^2 - 0.16a^2 3.84=0.24a23.84 = 0.24a^2 a2=3.840.24=16a^2 = \frac{3.84}{0.24} = 16 Since a>0a>0, we take the positive square root: a=16=4a = \sqrt{16} = 4. Distractor B (3.2) results from forgetting to square the 0.4 inside the bracket: 3.84=0.4a2−0.4a23.84 = 0.4a^2 - 0.4a^2 which doesn't work, or some other algebra error. Distractor D (9.6) comes from 3.84/0.43.84/0.4.

Question 20

A machine produces bolts. The number of defective bolts per batch, XX, is a random variable with a variance of Var(X)=4Var(X) = 4. The cost, in dollars, associated with the defects is modelled by the equation C=10X+50C = 10X + 50. What is the variance of the cost, Var(C)Var(C)?

  1. 90
  2. 400 (correct answer)
  3. 450
  4. 1650
Explanation: We use the property of variance that Var(aX+b)=a2Var(X)Var(aX + b) = a^2 Var(X). Here, a=10a = 10 and b=50b = 50. We are given Var(X)=4Var(X) = 4. Therefore, Var(C)=Var(10X+50)=102⋅Var(X)Var(C) = Var(10X + 50) = 10^2 \cdot Var(X). Var(C)=100⋅4=400Var(C) = 100 \cdot 4 = 400 The constant b=50b=50 does not affect the variance. Distractor A (90) comes from the incorrect calculation 10⋅4+5010 \cdot 4 + 50. Distractor C (450) from 102⋅4+5010^2 \cdot 4 + 50. Distractor D (1650) from (10⋅4+50)2(10\cdot4+50)^2. A common error is to calculate a⋅Var(X)a \cdot Var(X) instead of a2⋅Var(X)a^2 \cdot Var(X), which would give 10⋅4=4010 \cdot 4 = 40.