College Statistics Quiz: Discrete Random Variables And Expected Value

20 questions · exam conditions

0:00

Discrete Random Variables And Expected ValueQuestion 1 of 20

The probability distribution for a discrete random variable X is given by P(X=k) = c · k² for k = 1, 2, 3, where c is a constant. What is the expected value of X?

1/14

36/14

25/9

18/7

← Back to quizzes

College Statistics Quiz

College Statistics Quiz: Discrete Random Variables And Expected Value

Practice Discrete Random Variables And Expected Value in College Statistics 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 And Expected Value, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

All questions

Question 1

The probability distribution for a discrete random variable X is given by P(X=k) = c · k² for k = 1, 2, 3, where c is a constant. What is the expected value of X?

1/14
36/14
25/9
18/7 (correct answer)

Explanation: When working with discrete probability distributions, you must ensure the probabilities sum to 1, then use that constraint to find any unknown constants before calculating the expected value. First, find the constant c. Since probabilities must sum to 1:

P(X=1) + P(X=2) + P(X=3) = 1

c(1^2) + c(2^2) + c(3^2) = 1

c + 4c + 9c = 1

14c = 1

c = \frac{1}{14}

Now calculate the expected value using

E[X] = \sum k \cdot P(X=k)

E[X] = 1 \cdot \frac{1}{14} + 2 \cdot \frac{4}{14} + 3 \cdot \frac{9}{14}

E[X] = \frac{1}{14} + \frac{8}{14} + \frac{27}{14} = \frac{36}{14} = \frac{18}{7}

Choice A (1/14) is simply the value of the constant c, not the expected value. Choice B (36/14) represents the correct numerator but fails to reduce the fraction to lowest terms. Choice C (25/9) likely comes from incorrectly calculating the sum of squared terms or making an arithmetic error in the setup. The key strategy here is systematic execution: always find unknown constants first by using the fundamental property that probabilities sum to 1, then apply the expected value formula. Don't rush to calculate—verify your constant is correct before proceeding to the expected value calculation.

Question 2

The probability distribution for a discrete random variable X is given by P(X=k) = c · k² for k = 1, 2, 3, where c is a constant. What is the expected value of X?

1/14
36/14
25/9
18/7 (correct answer)

P(X=1) + P(X=2) + P(X=3) = 1

c(1^2) + c(2^2) + c(3^2) = 1

c + 4c + 9c = 1

14c = 1

c = \frac{1}{14}

Now calculate the expected value using

E[X] = \sum k \cdot P(X=k)

E[X] = 1 \cdot \frac{1}{14} + 2 \cdot \frac{4}{14} + 3 \cdot \frac{9}{14}

E[X] = \frac{1}{14} + \frac{8}{14} + \frac{27}{14} = \frac{36}{14} = \frac{18}{7}

Question 3

The number of typos, X, on a page of a certain book is a discrete random variable with the probability distribution P(X=k) for k=0, 1, 2, ... . The mean of X is 2.5 and the variance of X is 1.25. An editor revises the book, which reduces the number of typos on each page by half. What is the expected value of the number of typos on a revised page?

0.625
1.12
1.25 (correct answer)
2.5

Explanation: Let X be the original number of typos and Y be the number of typos on a revised page. The problem states that the revision reduces the number of typos by half, so Y = X / 2 = 0.5X. We are asked for the expected value of Y, E[Y]. Using the properties of linear transformations of random variables: E[Y] = E[0.5X] = 0.5 * E[X]. Given that E[X] = 2.5, we have: E[Y] = 0.5 * 2.5 = 1.25. The information about the variance of X is extraneous and is included to test whether the student can identify the relevant information. Distractor A is the variance of Y divided by 2: Var(Y) = Var(0.5X) = 0.5²Var(X) = 0.25 * 1.25 = 0.3125. This is not 0.625. What is 0.625? 1.25/2. This is what you would get if you thought Var(aX)=aVar(X) and then took that as the mean. Distractor B is the standard deviation of Y, approx: sqrt(0.3125) is about 0.56. So B is not that. Distractor D is the original expected value, E[X].

Question 4

A raffle sells 500 tickets at $5 each. There is one grand prize of $1000, two second-place prizes of $250, and five third-place prizes of $50. For the game to be considered 'fair,' the expected net winnings must be zero. How much should the grand prize be increased to make this raffle fair, assuming all other prizes and the ticket price remain the same?

$250
$500
$750 (correct answer)
$1000

Explanation: First, calculate the expected value of the prize money (the gross winnings). Let X be the prize money. P(X=$1000) = 1/500 P(X=$250) = 2/500 P(X=$50) = 5/500 P(X=$0) = 492/500 E[Prize] = 1000(1/500) + 250(2/500) + 50(5/500) + 0(492/500) E[Prize] = (1000 + 500 + 250) / 500 = 1750 / 500 = $3.50. The cost to play is $5. The current expected net winnings are E[Net] = E[Prize] - Cost = $3.50 -

5.00 = –

1.50. For a fair game, the expected net winnings must be $0. This means the expected prize money must equal the cost of the ticket, which is $5. The current expected prize money is $3.50. The total expected prize money needs to increase by $5.00 - $3.50 = $1.50. Let A be the amount added to the grand prize. The new expected prize value will be E[New Prize] = E[Old Prize] + A * P(Grand Prize). We need the total expected prize to be $5. So, the total expected prize pool must increase from $1750 to $5 * 500 = $2500. The required increase in the total prize pool is $2500 - $1750 = $750. Since this increase is applied only to the single grand prize, the grand prize itself must be increased by $750. Alternatively, let A be the increase to the grand prize. The new grand prize is 1000+A. We want the new expected prize to be $5. (1000+A)(1/500) + 250(2/500) + 50(5/500) = 5 (1000 + A + 500 + 250) / 500 = 5 1750 + A = 2500 A = $750. Distractor A, $250, is the value of a second-place prize. Distractor B, $500, might be chosen if one miscalculates the total prize fund. Distractor D, $1000, is the original grand prize value.

Question 5

A baker makes a batch of 12 special cookies. 4 of the cookies have a secret spicy ingredient. A customer buys 3 cookies, selected at random. If the customer gets exactly one spicy cookie, they win a prize. What is the expected number of spicy cookies the customer will get?

1/3
13/33
1.25
1 (correct answer)

Explanation: This is a hypergeometric distribution problem, where you're sampling without replacement from a finite population with two types of items. When you see questions about selecting items from a fixed group containing different categories, think about expected value for the category of interest. To find the expected number of spicy cookies, you need to calculate the probability of getting 0, 1, 2, or 3 spicy cookies, multiply each by the number of spicy cookies, then sum these products. The probabilities are:

P(0 spicy) = $\frac{\binom{4}{0}\binom{8}{3}}{\binom{12}{3}} = \frac{56}{220}$
P(1 spicy) = $\frac{\binom{4}{1}\binom{8}{2}}{\binom{12}{3}} = \frac{112}{220}$
P(2 spicy) = $\frac{\binom{4}{2}\binom{8}{1}}{\binom{12}{3}} = \frac{48}{220}$
P(3 spicy) = $\frac{\binom{4}{3}\binom{8}{0}}{\binom{12}{3}} = \frac{4}{220}$

Expected value =

0 \times \frac{56}{220} + 1 \times \frac{112}{220} + 2 \times \frac{48}{220} + 3 \times \frac{4}{220} = \frac{220}{220} = 1

However, there's a shortcut: for hypergeometric distributions,

E[X] = n \times \frac{K}{N}

where n = sample size (3), K = spicy cookies (4), N = total cookies (12). So

E[X] = 3 \times \frac{4}{12} = 1

. Choice A (1/3) incorrectly divides spicy cookies by total cookies. Choice B (13/33) appears to be a calculation error. Choice C (1.25) might come from incorrectly using 5/4 somewhere in the calculation. Remember: for sampling problems, use the hypergeometric expected value formula

n \times \frac{K}{N}

as your shortcut.

Question 6

An insurance company offers a one-year policy on a smartphone. The company charges a premium of $100. The policy pays out $500 if the phone is lost or stolen, which occurs with a probability of 0.05. It pays out $200 for screen damage, which occurs with a probability of 0.15. These events are mutually exclusive. If no claim is filed, the company pays out $0. What is the expected profit for the insurance company on one policy?

$45 (correct answer)
$55
$100
$445

Explanation: The insurance company's profit is the premium received minus any payout. Let P be the profit and X be the payout amount. P = $100 - X. The expected profit is E[P] = E[$100 - X] = $100 - E[X]. First, we need to calculate the expected payout, E[X]. The possible payout amounts are $500, $200, and $0. The probabilities are: P(X=$500) = 0.05 (lost/stolen) P(X=$200) = 0.15 (screen damage) P(X=$0) = 1 - 0.05 - 0.15 = 0.80 (no claim) Now, calculate E[X]: E[X] = (

500)(0.05) + (

200)(0.15) + ($0)(0.80) E[X] = $25 + $30 + $0 = $55. Finally, calculate the expected profit: E[P] = $100 - E[X] = $100 - $55 = $45. Distractor B is the expected payout, not the expected profit. Distractor C is the premium charged, representing the profit if there are never any claims. Distractor D is the profit in the best-case scenario for the company ($500 payout) minus the premium, but that's a miscalculation; it's $500 - $55, which is not meaningful.

Question 7

A discrete random variable X has a mean of E[X] = 5 and a variance of Var(X) = 4. A second random variable Y is defined as Y = 10 - 2X. What are the expected value and standard deviation of Y?

E[Y] = 0, SD(Y) = 4 (correct answer)
E[Y] = 0, SD(Y) = 8
E[Y] = 10, SD(Y) = -4
E[Y] = 20, SD(Y) = 8

Explanation: We use the properties of linear transformations of random variables. For the expected value: E[aX + b] = aE[X] + b In this case, Y = -2X + 10, so a = -2 and b = 10. E[Y] = -2E[X] + 10 = -2(5) + 10 = -10 + 10 = 0. For the variance and standard deviation: Var(aX + b) = a²Var(X) SD(aX + b) = |a|SD(X) First, find the standard deviation of X: SD(X) = √Var(X) = √4 = 2. Now, find the standard deviation of Y: SD(Y) = |–2|SD(X) = 2 * 2 = 4. So, E[Y] = 0 and SD(Y) = 4. Distractor B incorrectly calculates SD(Y) as |a|Var(X) or a²SD(X). Distractor C incorrectly calculates E[Y] and also provides an impossible negative standard deviation, which results from not taking the absolute value of 'a'. Distractor D incorrectly calculates E[Y] as 10 - 2(5) but makes a sign error, and incorrectly calculates SD(Y).

Question 8

A manufacturer produces widgets in batches of 100. The probability that any given widget is defective is 0.03, independently of other widgets. The company incurs a cost of $2 for each defective widget and makes a profit of $0.50 for each non-defective widget in a batch. What is the expected profit for a batch of 100 widgets?

$45.50
$42.50 (correct answer)
$48.50
$44.00

Explanation: Let D be the number of defective widgets in a batch of 100, and let G be the number of non-defective (good) widgets. D follows a binomial distribution with n=100 and p=0.03. The expected number of defective widgets is E[D] = np = 100(0.03) = 3. The number of good widgets is G = 100 - D. The expected number of good widgets is E[G] = 100 - E[D] = 100 - 3 = 97. The total profit is a function of D and G: Profit = $0.50 * G - $2.00 * D. The expected profit is E[Profit] = E[$0.50 * G - $2.00 * D] = $0.50 * E[G] - $2.00 * E[D]. E[Profit] = $0.50 * (97) - $2.00 * (3) = $48.50 - $6.00 = $42.50. Let me re-read the question. Ah, it's a cost FOR each defective, and profit FOR each non-defective. My setup is correct. Let's re-calculate. E[D]=3. E[G]=97. Profit = 0.50G - 2D. E[Profit] = 0.50E[G] - 2E[D] = 0.5097 - 23 = 48.50 - 6.00 = 42.50. So B is the correct answer. Let's check my logic and distractors. Let's define profit per item. Let Xi be the profit from the i-th widget. Xi = 0.50 if not defective (p=0.97), Xi = -2.00 if defective (p=0.03). E[Xi] = (0.50)(0.97) + (-2.00)(0.03) = 0.485 - 0.06 = 0.425. For a batch of 100, the total expected profit is 100 * E[Xi] = 100 * 0.425 = $42.50. The answer is indeed B. What could be distractors? D: 0.50*(100) - 2.00*(3) = 50 - 6 = 44. This assumes profit on all 100 and subtracts cost for defectives. This is a plausible misreading.
A: 0.50*(97) - 2.00*(0.03) = 48.50 - 0.06 = 48.44. Not very plausible. C: 0.50*(97) = 48.50. This ignores the cost of defectives. Ok, so B=

42.50 is correct. D=

44.00 is a good distractor. C=$48.50 is another good distractor. Let's make a distractor A. What if they calculate E[D] and E[G] and add? 3+97=100. Not helpful. What if they add the profit and cost? 0.597 + 23 = 48.5+6 = 54.5. That's a good one. So, let's make the options: A: $54.50, B: $42.50, C: $48.50, D: $44.00. This is a better set. I will choose B as correct. Okay, I've got a good question but my original answer key was wrong. Let me fix that. The correct answer is $42.50. Let's re-write the explanation. Let X be a random variable for the profit from a single widget. X =

0.50 with probability 0.97 (non-defective) and X = -

2.00 with probability 0.03 (defective). The expected profit per widget is E[X] = (0.50)(0.97) + (-2.00)(0.03) = 0.485 - 0.06 = $0.425. For a batch of 100 widgets, the total expected profit is 100 * E[X] = 100 * $0.425 = $42.50. Alternatively, let D be the number of defective widgets in the batch. D ~ Binomial(100, 0.03). The expected number of defective widgets is E[D] = 100(0.03) = 3. The number of non-defective widgets is 100-D. The expected number of non-defective widgets is E[100-D] = 100 - E[D] = 97. The total profit is P = 0.50(100-D) - 2.00(D). The expected profit is E[P] = 0.50 * E[100-D] - 2.00 * E[D] = 0.50(97) - 2.00(3) = 48.50 - 6.00 = $42.50. Distractor D ($44.00) results from calculating the profit on all 100 items and then subtracting the expected cost: 0.50(100) - 2.00(3) =

44. Distractor C (

48.50) is the expected revenue from non-defective items, ignoring the cost of defective items. Distractor A ($45.50) might arise from another calculation error, for instance, 0.50(100-3) - 2(0.03) = 48.5 - 0.06. Let's use that. It's a plausible error.

Question 9

A quality control process involves inspecting a random sample of 5 items from a large batch. The probability of an item being defective is 0.1. The inspection process costs $10 per sample. If one or more defective items are found, the entire batch is rejected, which incurs an additional cost of $100. What is the expected total cost for inspecting one sample?

$40.95
$50.95 (correct answer)
$59.05
$110.00

Explanation: Let C be the total cost. The base cost is $10. There is an additional cost of $100 if at least one defective item is found. Let X be the number of defective items in the sample of 5. X follows a binomial distribution with n=5 and p=0.1. The total cost C can be expressed as: C = $10 (base cost) C = $10 + $100 = $110 (if X ≥ 1) The expected cost is E[C] = $10 * P(X=0) + $110 * P(X ≥ 1). We can simplify this using P(X ≥ 1) = 1 - P(X=0). E[C] = $10 * P(X=0) + $110 * (1 - P(X=0)) E[C] = 10P(X=0) + 110 - 110P(X=0) = 110 - 100*P(X=0). Now, we need to calculate P(X=0), the probability of finding zero defective items. P(X=0) = (⁵₀) * (0.1)⁰ * (0.9)⁵ = 1 * 1 * 0.59049 = 0.59049. Now, substitute this back into the formula for E[C]: E[C] = 110 - 100 * (0.59049) = 110 - 59.049 = $50.951. Rounded to the nearest cent, the expected cost is $50.95. Distractor A is P(X ≥ 1) * 100 = (1-0.59049)*100 = 40.951. This is the expected value of the rejection cost only. Distractor C is P(X=0)*100, which is $59.05, a misinterpretation of the cost structure. Distractor D is the total cost if a rejection occurs, not the expected cost.

Question 10

$45 (correct answer)
$55
$100
$445

500)(0.05) + (

Question 11

$45.50
$42.50 (correct answer)
$48.50
$44.00

42.50 is correct. D=

0.50 with probability 0.97 (non-defective) and X = -

44. Distractor C (

Question 12

A biologist is studying a species of fish. The number of eggs, X, laid by a female is a random variable with E[X] = 200 and SD(X) = 20. The number of eggs that hatch, Y, is typically 40% of the eggs laid, minus a loss of 5 eggs due to environmental factors. Thus, Y = 0.4X - 5. What is the standard deviation of Y, the number of eggs that hatch?

3
8 (correct answer)
75
400

Explanation: This problem involves finding the standard deviation of a linear transformation of a random variable. The transformation is Y = aX + b, with a = 0.4 and b = -5. The rule for the variance of a transformed variable is Var(aX + b) = a²Var(X). The rule for the standard deviation is SD(aX + b) = |a|SD(X). We are given SD(X) = 20. SD(Y) = |0.4| * SD(X) = 0.4 * 20 = 8. Distractor A is the result of an incorrect transformation, perhaps sqrt(0.4Var(X) - 5) which is not meaningful, or Var(Y) - E[Y]^2. Let's check. Var(X) = 20^2 = 400. Var(Y) = 0.4^2 * 400 = 0.16 * 400 = 64. So SD(Y) = sqrt(64)=8. Correct. Distractor A (3) might come from 0.420 - 5. This is incorrectly applying the expectation rule to the standard deviation. Distractor C is the expected value of Y: E[Y] = 0.4*E[X] - 5 = 0.4(200) - 5 = 80 - 5 = 75. Distractor D is the variance of X (20² = 400).

Question 13

$40.95
$50.95 (correct answer)
$59.05
$110.00

Question 14

0.625
1.12
1.25 (correct answer)
2.5

Question 15

3
8 (correct answer)
75
400

Question 16

1/3
13/33
1.25
1 (correct answer)

P(0 spicy) = $\frac{\binom{4}{0}\binom{8}{3}}{\binom{12}{3}} = \frac{56}{220}$
P(1 spicy) = $\frac{\binom{4}{1}\binom{8}{2}}{\binom{12}{3}} = \frac{112}{220}$
P(2 spicy) = $\frac{\binom{4}{2}\binom{8}{1}}{\binom{12}{3}} = \frac{48}{220}$
P(3 spicy) = $\frac{\binom{4}{3}\binom{8}{0}}{\binom{12}{3}} = \frac{4}{220}$

Expected value =

0 \times \frac{56}{220} + 1 \times \frac{112}{220} + 2 \times \frac{48}{220} + 3 \times \frac{4}{220} = \frac{220}{220} = 1

However, there's a shortcut: for hypergeometric distributions,

E[X] = n \times \frac{K}{N}

where n = sample size (3), K = spicy cookies (4), N = total cookies (12). So

E[X] = 3 \times \frac{4}{12} = 1

n \times \frac{K}{N}

as your shortcut.

Question 17

$250
$500
$750 (correct answer)
$1000

5.00 = –

Question 18

A mail-order company finds that the number of orders it receives in a day, X, is a discrete random variable with a mean of 50 and a variance of 16. Each order has an average processing cost of $10, and there is a fixed daily operational cost of $200. What is the expected value of the total daily cost?

$500
$660
$700 (correct answer)
$900

Explanation: Let X be the number of orders. The total daily cost, C, can be modeled as a linear function of X. The cost for processing X orders is 10X. The fixed cost is $200. So, the total cost is C = 10X + 200. We need to find the expected value of C, E[C]. Using the properties of expectation: E[C] = E[10X + 200] = 10 * E[X] + 200. We are given that E[X] = 50. E[C] = 10 * (50) + 200 = 500 + 200 = $700. The variance of X (16) is extra information designed to test whether the student correctly identifies the necessary information. Distractor A is just the variable part of the cost (10 * E[X]). Distractor B might arise from incorrectly incorporating the variance: 1050 + 1610 = 500 + 160 = 660. Distractor D might arise from adding the fixed cost to the mean before multiplying: 10 * (50 + 200) = 2500, or some other combination of the numbers.

Question 19

$500
$660
$700 (correct answer)
$900

Question 20

A player rolls two standard, six-sided dice. Let the random variable W be the larger of the two numbers shown. If the numbers are the same, W is that number. What is the expected value of W?

3.5
4.0
4.47 (correct answer)
4.92

Explanation: To find E[W], we must first find the probability distribution of W. There are 36 equally likely outcomes. We can count the outcomes for each value of W.

W=1: (1,1) -> 1 outcome. P(W=1)=1/36.
W=2: (1,2), (2,1), (2,2) -> 3 outcomes. P(W=2)=3/36.
W=3: (1,3), (3,1), (2,3), (3,2), (3,3) -> 5 outcomes. P(W=3)=5/36.
W=4: (1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4) -> 7 outcomes. P(W=4)=7/36.
W=5: (1,5)...(5,5) -> 9 outcomes. P(W=5)=9/36.
W=6: (1,6)...(6,6) -> 11 outcomes. P(W=6)=11/36. (Check: 1+3+5+7+9+11 = 36).

Now, calculate E[W] = Σw · P(W=w). E[W] = 1(1/36) + 2(3/36) + 3(5/36) + 4(7/36) + 5(9/36) + 6(11/36) E[W] = (1 + 6 + 15 + 28 + 45 + 66) / 36 = 161 / 36 ≈ 4.47. Distractor A is the expected value of a single die roll. Distractor B is the median of the possible outcomes {1,2,3,4,5,6} but also the median of this distribution. Distractor D could arise from a counting error, perhaps by double-counting pairs like (1,2) and (2,1) or other misinterpretations.