Statistics Quiz: Calculating Expected Value Of Random Variables

What this quiz covers

This quiz focuses on Calculating Expected Value Of Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

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.

Question 1

1. 2
2. -2
3. 1
4. 0 (correct answer)

Explanation: Expected value tells us the average net winnings over many raffle tickets. We calculate it by multiplying each outcome by its probability and adding the results. For X=-2 with probability 3/4, we get (-2)×(3/4)=-3/2. For X=6 with probability 1/4, we get 6×(1/4)=3/2. Adding these products: -3/2 + 3/2 = 0. This means over many tickets, you'd break even on average. The key insight is that expected value uses weighted averages—the more likely loss of $2 happens often enough to exactly balance the less likely win of$6.

Question 2

1. 0 (correct answer)
2. -1
3. 1
4. 2

Explanation: Expected value is the focus, representing the average net winnings X from many raffle tickets. Compute by multiplying each X by its probability and adding. For X=-1 with 3/4 and X=3 with 1/4, products are -0.75 and 0.75. Summing gives 0 dollars. In the long run, you'd break even on average. It's weighted by probabilities, not a plain average of outcomes— that's an important misconception to avoid. Use the 'value × chance' approach for each possibility.

Question 3

1. 2.666\ldots
2. 5
3. 2.6 (correct answer)
4. 3

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. In this simple game, apply it by calculating 1 × 0.3, 2 × 0.4, and 5 × 0.3. Then sum these products to find the expected points. This value represents the long-run average points per round. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 4

1. 5
2. 3.5 (correct answer)
3. 8
4. 2

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. For this ball draw, apply it by calculating 2 × (3/4) and 8 × (1/4). Then sum these products to find the expected value of the draw. This value represents the long-run average value per draw. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 5

1. 10
2. 7
3. 3.5 (correct answer)
4. 4.666\ldots

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. In this spinner game, apply it by calculating 0 × (1/2), 4 × (1/4), and 10 × (1/4). Then sum these products to find the expected prize amount. This value represents the long-run average prize per spin. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 6

1. 1 (correct answer)
2. 0.75
3. 2
4. 1.5

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. For these coin flips, apply it by calculating 0 × (1/4), 1 × (1/2), and 2 × (1/4). Then sum these products to find the expected number of heads. This value represents the long-run average number of heads per two flips. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 7

1. 2.5 (correct answer)
2. 4
3. 9
4. 3

Explanation: Expected value represents the average payout per roll over many games. To find it, multiply each payout by its probability, then sum. For X=0 (rolls 1-3), the probability is 3/6=1/2, giving 0×(1/2)=0. For X=3 (rolls 4-5), the probability is 2/6=1/3, giving 3×(1/3)=1. For X=9 (roll 6), the probability is 1/6, giving 9×(1/6)=3/2. Summing: 0 + 1 + 3/2 = 5/2 = 2.5. This $2.50 average reflects that high payouts are rare while zero payouts are common. Remember, expected value weights outcomes by their likelihood, not just averaging the possible values.

Question 8

1. 9/4
2. 3 (correct answer)
3. 1/4
4. 5

Explanation: Expected value represents the average discount you'd receive over many plays. To find it, multiply each discount amount by its probability and sum the results. For X=1 with probability 1/4, we get 1×(1/4)=1/4. For X=3 with probability 1/2, we get 3×(1/2)=3/2. For X=5 with probability 1/4, we get 5×(1/4)=5/4. Adding these products: 1/4 + 3/2 + 5/4 = 1/4 + 6/4 + 5/4 = 12/4 = 3. The expected discount of $3 makes sense—it's exactly the middle value, which happens because the probabilities are symmetric. Remember, expected value weights outcomes by likelihood, creating a balance point for the distribution.

Question 9

1. 3
2. 8/4
3. 2 (correct answer)
4. 4

Explanation: Expected value tells us the average number of exercise days per week across many students. We calculate it by multiplying each outcome by its probability, then summing. For X=0 days with probability 1/4, we get 0×(1/4)=0. For X=2 days with probability 1/2, we get 2×(1/2)=1. For X=4 days with probability 1/4, we get 4×(1/4)=1. Adding these: 0 + 1 + 1 = 2. This average of 2 days per week reflects that the middle value (2 days) is most likely. Don't confuse this with a simple average of (0+2+4)/3=2—here we get the same result, but only because the distribution is symmetric.

Question 10

1. 5
2. 4 (correct answer)
3. $15/5$
4. 3

Explanation: Expected value represents the average payout over many card draws. To find it, we multiply each payout by its probability and add the results. For X=0 with probability $\tfrac{2}{5}$, we get $0 \times (\tfrac{2}{5}) = 0$. For X=5 with probability $\tfrac{2}{5}$, we get $5 \times (\tfrac{2}{5}) = 2$. For X=10 with probability $\tfrac{1}{5}$, we get $10 \times (\tfrac{1}{5}) = 2$. Summing these products: $0 + 2 + 2 = 4$. This means over many draws, you'd average $4 per draw. Notice how the expected value of$4$ isn't one of the possible payouts—it's the weighted average that accounts for how often each outcome occurs.

Question 11

1. 16/3
2. 4
3. 10
4. 9/2 (correct answer)

Explanation: Expected value represents the average payout over many games. We calculate by multiplying each payout by its probability: $2 with probability 1/2 gives 2×(1/2)=1;$4 with probability 1/4 gives 4×(1/4)=1; $10 with probability 1/4 gives 10×(1/4)=5/2. Adding these: 1 + 1 + 5/2 = 2 + 5/2 = 4/2 + 5/2 = 9/2 dollars. This means you'd average$4.50 per game in the long run. Notice how the high $10 payout, despite happening only 1/4 of the time, significantly raises the average above what you might guess. Remember: multiply value by chance, then sum.

Question 12

1. 4 (correct answer)
2. 3
3. 8
4. 12

Explanation: Expected value tells us the average points we'd score over many games. We calculate it by multiplying each point value by its probability, then summing. For 0 points with probability 1/4: 0×(1/4)=0; for 4 points with probability 1/2: 4×(1/2)=2; for 8 points with probability 1/4: 8×(1/4)=2. Adding these products: 0 + 2 + 2 = 4 points. This means in the long run, you'd average 4 points per game. Don't confuse this with the simple average (0+4+8)/3=4—that ignores how likely each outcome is. The weighted average properly accounts for the fact that 4 points happens twice as often as the others.

Question 13

1. 1
2. 2
3. 5
4. 1/2 (correct answer)

Explanation: Expected value shows the average outcome including both wins and losses. Here, losing $1 happens with probability 3/4, while winning$5 happens with probability 1/4. We calculate: (-1)×(3/4) = -3/4 for the loss, and 5×(1/4) = 5/4 for the win. Adding these: -3/4 + 5/4 = 2/4 = 1/2. This means on average, you'd gain $0.50 per raffle ticket. Many people mistakenly think negative outcomes should be ignored, but expected value includes all possibilities. The positive result shows this raffle is actually favorable despite losing most of the time—the occasional big win outweighs frequent small losses.

Question 14

1. 3.5 (correct answer)
2. 2.5
3. 3
4. 6

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. For this die roll, apply it by calculating 1 × (1/6), 2 × (1/6), 3 × (1/6), 4 × (1/6), 5 × (1/6), and 6 × (1/6). Then sum these products to find the expected number shown. This value represents the long-run average result per roll. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 15

1. 1
2. 3
3. 4
4. 2 (correct answer)

Explanation: Expected value is the key concept, serving as the predicted average value of a random variable like the number X on a drawn card. Compute it by multiplying each X value by its probability and adding the results. Here, for X=1 with probability 1/2, X=2 with 1/3, and X=5 with 1/6, the products are 0.5, about 0.67, and about 0.83. Adding them gives exactly 2. This interprets as the long-run average number if drawing many times. Don't confuse it with a simple average of the values; it's weighted by their chances. Use the 'value × chance' mindset for similar problems.

Question 16

1. 0.7
2. 0.5
3. 1.4 (correct answer)
4. 2

Explanation: The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. In this survey, apply it by calculating 0 × 0.5, 2 × 0.3, and 4 × 0.2. Then sum these products to find the expected number of messages. This value represents the long-run average messages per hour. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.

Question 17

1. 20 (correct answer)
2. 15
3. 25
4. 30

Explanation: The concept is expected value, the average minutes spent reading per day across many surveyed students. Compute by multiplying each X by its probability and summing. For this, 10 times 1/4, 20 times 1/2, and 30 times 1/4. Adding gives 2.5 + 10 + 7.5 = 20 minutes. This interprets as an expected 20 minutes of reading on average per day. A misconception is simply averaging 10, 20, 30 to get 20 (which matches here due to symmetry), but generally, it's weighted, and asymmetry would differ. Strategy: always multiply 'value times chance' and add up.

Question 18

1. 12
2. 5
3. 4 (correct answer)
4. 1

Explanation: Expected value is the key concept, giving the average points you'd expect from many spins. Compute it by multiplying each X value by its probability and adding the results. Here, 1 times 1/3 is about 0.33, 4 times 1/3 is about 1.33, and 7 times 1/3 is about 2.33. Adding them yields 4 points. Over many spins, you'd average 4 points each time. Don't confuse this with the simple average of outcomes; it's weighted equally here since probabilities are equal, but generally it's a weighted average. Use the 'value times chance' approach for similar problems.

Question 19

1. 7/2 (correct answer)
2. 9/2
3. 3
4. 0

Explanation: Expected value represents the average tokens won per round over many plays. To find it, multiply each outcome by its probability and sum the results. For X=0 with probability 1/6, we get 0×(1/6)=0. For X=3 with probability 1/2, we get 3×(1/2)=3/2. For X=6 with probability 1/3, we get 6×(1/3)=2. Summing: 0 + 3/2 + 2 = 3/2 + 4/2 = 7/2. The expected value of 7/2 tokens reflects that winning 3 tokens is most likely, while the extremes are less common. This weighted average approach gives us the true long-run average, unlike simply averaging 0, 3, and 6.

Question 20

1. 2
2. 3
3. 7
4. 2.8 (correct answer)

Explanation: Expected value represents average lunch-buying days per week over many students. Multiply: 0 times 1/5 is 0, 2 times 2/5 is 0.8, 5 times 2/5 is 2. Sum to 2.8 days. Long-run average is 2.8 days. Not simple average of 0,2,5 (2.33), but weighted. Strategy: value times chance, add up.