Statistics Quiz
Practice Developing Empirical Probability Distributions Expected Value in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A spinner game is played 120 times. The observed counts are: land on 1 (18 times), land on 2 (27 times), land on 3 (33 times), land on 4 (42 times). Define the random variable X as the number landed on. Using relative frequencies from these data as estimated probabilities, what is the expected value of X based on the empirical distribution?
This quiz focuses on Developing Empirical Probability Distributions Expected Value, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.
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A spinner game is played 120 times. The observed counts are: land on 1 (18 times), land on 2 (27 times), land on 3 (33 times), land on 4 (42 times). Define the random variable X as the number landed on. Using relative frequencies from these data as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: From the 120 spins, we calculate relative frequencies: P(X=1) = 18/120 = 0.15, P(X=2) = 27/120 = 0.225, P(X=3) = 33/120 = 0.275, and P(X=4) = 42/120 = 0.35. These empirical probabilities sum to 1, confirming we've accounted for all outcomes. The expected value is E(X) = 1×0.15 + 2×0.225 + 3×0.275 + 4×0.35. Computing: 1×0.15 = 0.15, 2×0.225 = 0.45, 3×0.275 = 0.825, and 4×0.35 = 1.40. Adding these gives E(X) = 0.15 + 0.45 + 0.825 + 1.40 = 2.825, which rounds to 2.83.
A coffee shop tracked tips from 80 customers. The observed tip amounts were: \0(15customers),$1(25customers),$2(30customers),and$5(10customers).LetXbethetipamount(indollars)fromarandomlyselectedcustomer.Treattherelativefrequenciesfromthese80observationsasestimatedprobabilities.WhatistheexpectedvalueofX$ based on the empirical distribution?
Explanation: To develop an empirical probability distribution, we use relative frequencies from observations as probability estimates, then compute the expected value as a weighted average of the outcomes. For the 80 coffee shop customers, the relative frequencies are: 15/80 ≈ 0.1875 for 0,25/80=0.3125for1, 30/80 = 0.375 for 2,and10/80=0.125for5. The expected tip E[X] = 0*(0.1875) + 1*(0.3125) + 2*(0.375) + 5*(0.125) = 0 + 0.3125 + 0.75 + 0.625 = 1.6875, or approximately $1.69. This weighted average reflects the average tip amount expected from a customer based on the data. Empirical estimates like this are useful for real-world predictions but depend on sample size. The calculation highlights how higher tips, though less frequent, contribute significantly via weighting.
A school surveyed 100 students about how many siblings they have. The observed results were: 0 siblings (28 students), 1 sibling (46 students), 2 siblings (20 students), and 3 siblings (6 students). Let X be the number of siblings for a randomly selected student. Using relative frequencies from this sample as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Using empirical data to estimate probabilities involves calculating relative frequencies and then finding the expected value as a weighted average. For the 100 students, relative frequencies are: 28/100 = 0.28 for 0 siblings, 46/100 = 0.46 for 1, 20/100 = 0.2 for 2, and 6/100 = 0.06 for 3. E[X] = 0*(0.28) + 1*(0.46) + 2*(0.2) + 3*(0.06) = 0 + 0.46 + 0.4 + 0.18 = 1.04. This suggests an average of about 1.04 siblings per student in this sample. The weighted average accounts for the distribution's skew toward fewer siblings. Empirical estimates like this can inform demographic studies but may vary with larger samples.
A website recorded the number of ads clicked by visitors during a test of 200 visits. The observed counts were: 0 clicks (110 visits), 1 click (60 visits), 2 clicks (24 visits), and 3 clicks (6 visits). Let X be the number of ads clicked in a visit. Using the relative frequencies from these 200 visits as estimated probabilities (not exact), what is the expected value of X based on the empirical distribution?
Explanation: Empirical expected values use relative frequencies as probability estimates to form a weighted average of outcomes from sample data. For the 200 website visits, relative frequencies are: 110/200 = 0.55 for 0 clicks, 60/200 = 0.3 for 1, 24/200 = 0.12 for 2, and 6/200 = 0.03 for 3. E[X] = 0*(0.55) + 1*(0.3) + 2*(0.12) + 3*(0.03) = 0 + 0.3 + 0.24 + 0.09 = 0.63. This suggests an average of 0.63 ads clicked per visit based on the test. The weighting reflects the predominance of zero clicks pulling the average down. Empirical approaches like this are essential for analyzing user behavior with real data.
A coffee shop sampled 60 customers and recorded the tip amount (in dollars) left on a small order. The observed counts were: \0(15customers),$1(24customers),$2(16customers),$3(5customers).LetXbethetipamountindollars.Usingtherelativefrequenciesfromthissampleasestimatedprobabilities,whatistheexpectedvalueofX$ based on the empirical distribution?
Explanation: Using empirical distributions, we estimate probabilities from sample relative frequencies to calculate expected values as weighted averages. For 60 customers, tips were 0from15,1 from 24, 2from16,and3 from 5. Estimated probabilities: 15/60 for 0, 24/60 for 1, 16/60 for 2, 5/60 for 3. E[X] = 0*(15/60) + 1*(24/60) + 2*(16/60) + 3*(5/60) = 71/60 ≈ 1.1833. This average tip amount is based on observed data. Empirical methods show that such estimates approximate true values and rely on weighted averages, potentially changing with new samples.
A teacher recorded how many times a student raised their hand during each of 40 class periods. The observed counts were: 0 times (9 periods), 1 time (14 periods), 2 times (11 periods), 3 times (6 periods). Let X be the number of hand-raises in a class period. Using the relative frequencies from these 40 periods as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Empirical expected values are computed by treating relative frequencies as probability estimates and finding the weighted average of possible values. Over 40 class periods, hand raises were 0 times in 9 periods, 1 time in 14, 2 times in 11, and 3 times in 6. The probabilities are estimated as 9/40 for 0, 14/40 for 1, 11/40 for 2, and 6/40 for 3. E[X] = 0*(9/40) + 1*(14/40) + 2*(11/40) + 3*(6/40) = 54/40 = 1.35. This is the average number of hand raises per period from the sample. The approach emphasizes empirical estimation, where the expected value is the sample mean, a weighted average that may vary with different data.
A café offers a punch card promotion. Over 160 customer visits, the number of punches earned per visit is recorded as: 0 punches (32 visits), 1 punch (80 visits), 2 punches (40 visits), 3 punches (8 visits). Let X be the number of punches earned on a randomly selected visit. Using the relative frequencies from this data as estimates of the probabilities, what is the expected value of X based on the empirical distribution?
Explanation: We develop an empirical probability distribution by using relative frequencies from the punch card data. Frequencies: 32/160 for 0, 80/160 for 1, 40/160 for 2, 8/160 for 3. The expected value is a weighted average: E[X] = 0*(32/160) + 1*(80/160) + 2*(40/160) + 3*(8/160). This equals (80 + 80 + 24)/160 = 184/160 = 1.15. Alternatively, divide total punches by visits. This method emphasizes empirical estimation over theoretical assumptions. It serves as an approximation to the true average punches per visit.
A researcher surveys 120 commuters about how many days per week they work from home. The observed counts are: 0 days (30), 1 day (24), 2 days (36), 3 days (18), 4 days (12). Define the random variable X as the number of work-from-home days for a randomly selected commuter. Treat the relative frequencies from this survey as estimated probabilities. What is the expected value of X based on the empirical distribution?
Explanation: Empirical probability distributions allow us to estimate expected values from observed data by using relative frequencies as probability approximations. For this survey, relative frequencies are 30/120 for 0, 24/120 for 1, 36/120 for 2, 18/120 for 3, and 12/120 for 4. The expected value is a weighted average: E[X] = 0*(30/120) + 1*(24/120) + 2*(36/120) + 3*(18/120) + 4*(12/120). This computes to (24 + 72 + 54 + 48)/120 = 198/120 = 1.65. You can also think of it as the total work-from-home days divided by the number of commuters. This method emphasizes that the estimate is based on the sample and serves as an approximation to the true population expected value. Larger samples would yield more reliable estimates.
A carnival game was played 200 times. The outcomes and counts were: win 0(92times),win2 (68 times), win 5(40times).LetXbetheamountwon(indollars)onasingleplay.Usingtheempiricaldistributionformedbyconvertingcountstorelativefrequencies(sotheprobabilitiesareestimatesbasedontheobserveddata),whatistheexpectedvalueofX$ based on the empirical distribution?
Explanation: To find the expected value using empirical data, we first convert the observed counts to relative frequencies (empirical probabilities). The relative frequencies are: P(X=0) = 92/200 = 0.46, P(X=2) = 68/200 = 0.34, and P(X=5) = 40/200 = 0.20. The expected value is calculated as a weighted average: E(X) = 0×0.46 + 2×0.34 + 5×0.20. This gives us E(X) = 0 + 0.68 + 1.00 = 1.68. The empirical distribution shows that players most often win nothing, sometimes win 2,andoccasionallywin5, resulting in an average win of $1.68 per play.
A die-like game piece was rolled 80 times. The observed results were: 1 (10 times), 2 (14 times), 3 (16 times), 4 (12 times), 5 (18 times), 6 (10 times). Let X be the face value rolled. Using the relative frequencies from these data as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: From 80 rolls, we calculate empirical probabilities: P(X=1) = 10/80, P(X=2) = 14/80, P(X=3) = 16/80, P(X=4) = 12/80, P(X=5) = 18/80, and P(X=6) = 10/80. The expected value is: E(X) = 1×(10/80) + 2×(14/80) + 3×(16/80) + 4×(12/80) + 5×(18/80) + 6×(10/80). Computing: E(X) = (10 + 28 + 48 + 48 + 90 + 60)/80 = 284/80 = 3.55. This empirical average is close to the theoretical 3.5 for a fair die, with slight variation due to sampling.
A streaming service surveyed 250 users about how many ads they saw in the first 10 minutes of a show. Observed counts: 0 ads (40 users), 1 ad (85 users), 2 ads (90 users), 3 ads (35 users). Let X be the number of ads seen by a randomly selected surveyed user. Using the relative frequencies from the survey as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: For the streaming service survey with 250 users, we use relative frequencies to estimate the probability of seeing each number of ads. The empirical probabilities are: P(0 ads) = 40/250, P(1 ad) = 85/250, P(2 ads) = 90/250, and P(3 ads) = 35/250. The expected value is calculated as E(X) = 0×(40/250) + 1×(85/250) + 2×(90/250) + 3×(35/250). Computing the non-zero terms: 85/250 + 2×90/250 + 3×35/250 = 85/250 + 180/250 + 105/250. Summing: (85 + 180 + 105)/250 = 370/250 = 1.48. This matches the marked answer, confirming that users see an average of 1.48 ads based on the empirical distribution.
A basketball player’s made free throws are recorded over 30 practice shots. The observed outcomes were: miss (0 points) occurred 8 times and make (1 point) occurred 22 times. Let X be the points scored on a randomly selected shot (0 or 1). Using relative frequencies from the observed data as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: The expected value estimates the average points per shot using empirical probabilities from the practice data. Relative frequencies are 8/30 ≈ 0.267 for 0 and 22/30 ≈ 0.733 for 1. Compute the weighted average: 0 × 0.267 + 1 × 0.733 ≈ 0.733. This is a weighted average where each outcome is multiplied by its relative frequency. Equivalently, total points divided by shots: (0×8 + 1×22)/30 = 22/30 ≈ 0.733. Such an empirical estimate depends on the observed shots and may change with more practice.
A streaming service samples 150 users and records how many ads they saw in the first 10 minutes. The observed counts were: 0 ads (45 users), 1 ad (60 users), 2 ads (30 users), 3 ads (15 users). Let X be the number of ads seen. Using relative frequencies from this sample as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Expected value empirically estimates the average ads seen using relative frequencies from the sample. These are 45/150 = 0.3 for 0, 60/150 = 0.4 for 1, 30/150 = 0.2 for 2, and 15/150 = 0.1 for 3. The weighted average is 0 × 0.3 + 1 × 0.4 + 2 × 0.2 + 3 × 0.1 = 1.1. This weights each count by its proportion in the data. Alternatively, total ads divided by users: (0×45 + 1×60 + 2×30 + 3×15)/150 = 165/150 = 1.1. This estimate is sample-based and could differ with another group of users.
A website tracked the number of pages visited by a user during 100 sessions. The observed counts were: 1 page (25 sessions), 2 pages (40 sessions), 3 pages (20 sessions), 4 pages (15 sessions). Let X be the number of pages visited in a randomly selected session. Using the empirical distribution from the observed relative frequencies (so probabilities are estimates), what is the expected value of X based on the empirical distribution?
Explanation: From 100 sessions, we determine empirical probabilities: P(X=1) = 25/100 = 0.25, P(X=2) = 40/100 = 0.40, P(X=3) = 20/100 = 0.20, and P(X=4) = 15/100 = 0.15. The expected value is the weighted average: E(X) = 1×0.25 + 2×0.40 + 3×0.20 + 4×0.15. This equals E(X) = 0.25 + 0.80 + 0.60 + 0.60 = 2.25 pages. The empirical distribution shows that users most commonly visit 2 pages per session, with an average of 2.25 pages overall.
A teacher recorded the number of pets owned by each of 50 students. The observed counts were: 0 pets (12 students), 1 pet (20 students), 2 pets (13 students), 3 pets (5 students). Let X be the number of pets for a randomly selected student from this sample. Using the empirical distribution (relative frequencies) as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: To find the expected number of pets, we first determine the empirical probabilities from the 50 students: P(X=0) = 12/50 = 0.24, P(X=1) = 20/50 = 0.40, P(X=2) = 13/50 = 0.26, and P(X=3) = 5/50 = 0.10. The expected value is calculated as: E(X) = 0×0.24 + 1×0.40 + 2×0.26 + 3×0.10. This gives us E(X) = 0 + 0.40 + 0.52 + 0.30 = 1.22 pets. The empirical distribution shows that most students have 1 pet, with the average being slightly above 1 pet per student.
A delivery company tracks how many packages arrive late each day for 100 days. The observed counts are: 0 late (40 days), 1 late (35 days), 2 late (15 days), 3 late (7 days), 4 late (3 days). Let X be the number of late packages on a randomly selected day. Using the relative frequencies from these 100 days as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Empirical estimation uses sample relative frequencies to approximate the probability distribution for late packages. Frequencies: 40/100 for 0, 35/100 for 1, 15/100 for 2, 7/100 for 3, 3/100 for 4. E[X] is the weighted average: 0*(40/100) + 1*(35/100) + 2*(15/100) + 3*(7/100) + 4*(3/100). This computes to (35 + 30 + 21 + 12)/100 = 98/100 = 0.98. It's also the total late packages divided by days tracked. This highlights the role of weighted averages in expected value calculations. The estimate is data-driven and may refine with additional observations.
A researcher tested a small robot 120 times and recorded whether it completed a task successfully. The observed outcomes were: Success (78 trials) and Failure (42 trials). The robot earns 4 points for Success and loses 1 point for Failure. Let X be the points earned on a trial. Using the relative frequencies from the 120 trials as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Empirical expected values are estimated by weighting outcomes by their relative frequencies to form an average. In 120 robot trials, success (4 points) occurred 78 times and failure (-1 point) 42 times. Estimated probabilities: 78/120 for 4, 42/120 for -1. E[X] = 4*(78/120) + (-1)*(42/120) = 270/120 = 2.25. This is the average points per trial from the sample. The method highlights empirical probability estimation and the concept of weighted averages, which approximate true expectations and may vary with different data sets.
A snack machine is tested for 50 purchases. The observed outcomes were: no snack dispensed (0 snacks) occurred 4 times, 1 snack occurred 43 times, and 2 snacks occurred 3 times. Let X be the number of snacks dispensed per purchase. Using the observed relative frequencies as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Using empirical data, the expected value estimates the average snacks dispensed per purchase. Relative frequencies are 4/50 = 0.08 for 0, 43/50 = 0.86 for 1, and 3/50 = 0.06 for 2. The weighted average is 0 × 0.08 + 1 × 0.86 + 2 × 0.06 = 0.98. This method weights outcomes by their observed probabilities from the sample. Alternatively, it's the total snacks divided by purchases: (0×4 + 1×43 + 2×3)/50 = 49/50 = 0.98. This estimate depends on the tested purchases and could differ in another trial.
A spinner is spun 60 times with these observed results: outcome 1 occurred 9 times, outcome 2 occurred 21 times, outcome 3 occurred 18 times, and outcome 4 occurred 12 times. Let X be the number landed on. Using the observed relative frequencies as estimated probabilities, what is the expected value of X based on the empirical distribution?
Explanation: Expected value is the anticipated average result over many spins, estimated here using empirical probabilities from the data. The relative frequencies are 9/60 = 0.15 for 1, 21/60 = 0.35 for 2, 18/60 = 0.3 for 3, and 12/60 = 0.2 for 4. Compute the weighted average: 1 × 0.15 + 2 × 0.35 + 3 × 0.3 + 4 × 0.2 = 2.55. This reflects the average outcome per spin based on the observed distribution. Alternatively, it's the total of all outcomes divided by 60: (1×9 + 2×21 + 3×18 + 4×12)/60 = 153/60 = 2.55. Such an estimate relies on the sample and could change with more spins.
A student plays a carnival game 100 times and records outcomes: win \0occurred52times,win$2occurred33times,andwin$5occurred15times.LetXbethedollaramountwononaplay.Usingrelativefrequenciesfromtheobserveddataasestimatedprobabilities(whichmayvarywithadifferentsample),whatistheexpectedvalueofX$ based on the empirical distribution?
Explanation: The expected value represents the long-run average outcome of the game based on repeated plays. To estimate it empirically, we use the relative frequencies from the observed data as probabilities for each possible winnings amount. These are calculated as 52/100 = 0.52 for 0,33/100=0.33for2, and 15/100 = 0.15 for $5. The expected value is then the weighted average: 0 × 0.52 + 2 × 0.33 + 5 × 0.15 = 1.41. This calculation is equivalent to the total winnings divided by the number of plays, which is (0×52 + 2×33 + 5×15)/100 = 141/100 = 1.41. Note that this is an empirical estimate and may differ with another sample of 100 plays.