Developing Empirical Probability Distributions, Expected Value
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Statistics › Developing Empirical Probability Distributions, Expected Value
A coffee shop tracked tips from 80 customers. The observed tip amounts were: $\$0$ (15 customers), $$1$ (25 customers), $\$2$ (30 customers), and $$5$ (10 customers). Let $X$ be the tip amount (in dollars) from a randomly selected customer. Treat the relative frequencies from these 80 observations as estimated probabilities. What is the expected value of $X$ based on the empirical distribution?
$1.75$
$1.88$
$2.13$
$1.69$
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 = 0.3125 for $1, 30/80 = 0.375 for $2, and 10/80 = 0.125 for $5. 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?
$1.00$
$1.04$
$0.64$
$104$
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?
$126$
$0.60$
$0.63$
$1.50$
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: $\$0$ (15 customers), $$1$ (24 customers), $\$2$ (16 customers), $$3$ (5 customers). Let $X$ be the tip amount in dollars. Using the relative frequencies from this sample as estimated probabilities, what is the expected value of $X$ based on the empirical distribution?
$3.00$
$1.183\overline{3}$
$1.25$
$71$
Explanation
Using empirical distributions, we estimate probabilities from sample relative frequencies to calculate expected values as weighted averages. For 60 customers, tips were $0 from 15, $1 from 24, $2 from 16, and $3 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 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?
$270$
$2.25$
$2.50$
$3.00$
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 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?
$54$
$1.35$
$1.40$
$1.50$
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 basketball player practiced free throws for 200 shots. The observed outcomes were: Made (146 shots) and Missed (54 shots). Define a scoring variable $X$ as follows: $X=2$ for a made free throw and $X=0$ for a missed free throw. Using the relative frequencies from these 200 shots as estimated probabilities, what is the expected value of $X$ based on the empirical distribution?
$292$
$1.46$
$0.73$
$2.00$
Explanation
Empirical expected values involve estimating probabilities from frequencies and computing a weighted average of outcomes. For 200 free throws, made (2 points) 146 times and missed (0 points) 54 times. Estimated probabilities: 146/200 for 2, 54/200 for 0. E[X] = 2*(146/200) + 0*(54/200) = 292/200 = 1.46. This is the average points per shot from the data. The approach emphasizes empirical estimation, where the expected value is the sample mean, a weighted average that approximates the true value and may vary across samples.
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?
$1.20$
$184$
$1.15$
$1.00$
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?
$2.00$
$1.80$
$1.65$
$198$
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 quality-control inspector checks 80 items and records the number of defects on each item. The observed counts are: 0 defects (50 items), 1 defect (20 items), 2 defects (8 items), 3 defects (2 items). Let $X$ be the number of defects on a randomly selected item. Using the relative frequencies from this sample as estimated probabilities (they may vary from sample to sample), what is the expected value of $X$ based on the empirical distribution?
$1.50$
$0.60$
$0.53$
$42$
Explanation
Empirical estimation of expected value relies on sample relative frequencies to approximate probabilities. For defects: 50/80 for 0, 20/80 for 1, 8/80 for 2, 2/80 for 3. E[X] is the weighted average: 0*(50/80) + 1*(20/80) + 2*(8/80) + 3*(2/80) = (20 + 16 + 6)/80 = 42/80 = 0.525. Rounding to two decimal places gives approximately 0.53. This is equivalent to total defects divided by items inspected. Such an approach underscores the use of data-driven weighted averages for estimation. It provides a practical way to predict average defects in future items based on the sample.