The sample they gave was [0,0,0,1,1,1,2,2,3,3,3,4,4,5,6,7,10,15,20,40].
Start by defining the metrics and computing them from the sample, then plot them on a number line to visualize the skew. Explain the theoretical relationship mean > median > mode for right-skewed data, and clarify why median and p95 can be whole numbers due to the discrete nature of counts.
Pro tip: Emphasize that for count data, percentiles are often computed using nearest-rank or interpolation methods that can yield whole numbers, and relate this to product metrics like '95% of days have ≤ X comments'.
Calculate the mean, median, and 95th percentile from the sample data. For the 95th percentile, specify the method (e.g., nearest-rank or linear interpolation) and note that it may be a whole number.
Place the mean, median, and p95 on a number line to show their relative positions. Highlight that for right-skewed data, the mean is typically the largest, followed by the median, and the mode is the smallest.
Discuss why mean > median > mode holds for right-skewed count data: the long right tail pulls the mean upward, while the median and mode are less affected. Mention that for discrete data, the mode may not always be less than the median, but the general tendency holds.
Explain that because comment counts are discrete, the median and percentiles can be whole numbers. The median is the middle value (or average of two middle values) of an ordered list of integers, and percentiles are often defined as the smallest value such that at least k% of observations are ≤ that value.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
CLT kicks in: sample mean is approximately Normal(3.0, (4/sqrt(50))^2).
First, state that by the Central Limit Theorem, the sample mean is approximately normally distributed with mean equal to the population mean and standard error equal to the population standard deviation divided by the square root of the sample size. Then compute the mean, standard error, and 95th percentile of the sample mean distribution using the normal approximation, and compare this p95 to the individual-level p95 (which is simply the population p95, assuming normality).
Pro tip: Emphasize that the CLT applies to the sampling distribution of the mean, not the individual data, and that the standard error shrinks by the square root of n. When comparing p95s, highlight that the sample mean's p95 is much closer to the mean than the individual p95, illustrating reduced variability.
By the Central Limit Theorem, for a sufficiently large sample size (n=50), the sample mean is approximately normally distributed, regardless of the population distribution.
The mean of the sample mean distribution is the population mean (3.0). The standard error is the population standard deviation divided by the square root of the sample size: 4.0 / sqrt(50) ≈ 0.5657.
For a normal distribution, the 95th percentile is mean + 1.645 * standard error. So, 3.0 + 1.645 * 0.5657 ≈ 3.93.
The individual-level p95 is the population p95, which under normality is 3.0 + 1.645 * 4.0 ≈ 9.58. The sample mean's p95 (≈3.93) is much lower, showing that the mean of 50 users is far less variable than individual user comments.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the sampling distribution of the mean and the formula for standard error (SE = σ/√n), then compute the new SE when n increases from 50 to 400. Explain how the mean remains unbiased, while the median and p95 of the sampling distribution become more concentrated around the true mean, with the p95 shifting closer to the mean due to reduced variance.
Pro tip: Emphasize that the standard error decreases by a factor of √(n2/n1), so quadrupling the sample size halves the SE. Also note that for skewed distributions, the median may still differ from the mean, but the sampling distribution of the mean approaches normality (CLT), making the mean, median, and p95 converge.
State that the sampling distribution of the sample mean has standard error SE = σ/√n, where σ is the population standard deviation. Assume σ is known or estimated from the sample.
Calculate the original SE with n=50 and the new SE with n=400. Show that SE_new = SE_old * √(50/400) = SE_old / √8 ≈ SE_old / 2.828, so the SE decreases by a factor of about 2.83.
Explain that the mean of the sampling distribution remains equal to the population mean (unbiased). The median and p95 are measures of the distribution's center and tail; as n increases, the distribution becomes narrower, so the median converges to the mean, and the p95 moves closer to the mean (since the tail is compressed).
If the sampling distribution is approximately normal, the p95 is at mean + 1.645*SE. With smaller SE, the p95 shifts downward toward the mean. The median equals the mean for symmetric distributions; for skewed distributions, the median may still differ but the difference shrinks as n grows due to CLT.
Conclude that larger sample sizes reduce uncertainty, making estimates more precise. This affects confidence intervals and p-values, and highlights the importance of sample size planning in experiments.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.