Flashcards · 4040 · Topic 5

Statistical measures, their interpretation and appropriate use flashcards

21 cards on Topic 5, subtopics 5.1 to 5.4. Pick a subtopic or study them all.

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Where is the median in a list of n values, and in a grouped table?: List or ungrouped table: the (n + 1)/2 th value of the ordered data. Grouped table: the n/2 th value, found by interpolation.
What is the formula for the median of grouped data?: Median ≈ L + ((n/2 − F) ÷ f) × w, where L is the lower boundary of the median class, F the cumulative frequency before it, f its frequency and w its width.
Leaves: 0–<10 (3), 10–<20 (9), 20–<30 (6), 30–<40 (2). Estimate the mean and median: Mean ≈ (15 + 135 + 150 + 70) ÷ 20 = 370 ÷ 20 = 18.5 cm. Median: 10th value, 3 below 10 → 10 + (7/9) × 10 = 17.8 cm.
Classes 5 ≤ x < 10 (frequency 15) and 10 ≤ x < 20 (frequency 24). Which is the modal class?: Densities: 15 ÷ 5 = 3 and 24 ÷ 10 = 2.4, so the modal class is 5 ≤ x < 10, even though 24 is the larger frequency.
The value 40 is added to 6, 7, 7, 8, 9. What happens to the mean, median and mode?: Mean 7.4 → 77 ÷ 6 = 12.83. Median 7 → 7.5. Mode stays 7. One extreme value moves the mean a lot and the median only a little.
How do you choose between the mean, median and mode?: Categories → mode. Skewed data or outliers → median. Roughly symmetrical numerical data, or when totals matter → mean. Justify with the figure and a reason tied to the data.

What are the variance and the standard deviation?: Variance: the mean of the squared distances from the mean, in squared units. Standard deviation: the square root of the variance, in the same units as the data.
What is the squares form of the variance, for a list and for a frequency table?: List: Σx² ÷ n − x̄². Table: Σfx² ÷ Σf − x̄², with x the class midpoint for grouped data (an estimate). Here the divisor is n (or Σf).
For ten values, Σx = 60 and Σx² = 400. Find the variance and standard deviation: Mean = 6. Variance = 400 ÷ 10 − 6² = 40 − 36 = 4. Standard deviation = 2.
What is the difference between Σx² and (Σx)²?: Σx²: square each value, then add. (Σx)²: add, then square. Also, Σ(x − x̄)² is positive, but Σ(x − x̄) is always zero.
Team A has standard deviation 4.1 points and team B 7.3 points, with equal means. Which is more consistent?: Team A, because its smaller standard deviation means its scores lie closer to the mean.

What are the quartiles and the P-th percentile as percentiles?: Q1 is the 25th percentile, the median the 50th and Q3 the 75th. The P-th percentile is the value with P% of the data at or below it.
n = 84. Which cumulative frequencies give Q1, the median, Q3 and the 40th percentile?: Q1: 84/4 = 21. Median: 42. Q3: 63. 40th percentile: 0.4 × 84 = 33.6.
22 values are at or below 30, and the class 30 < x ≤ 40 holds 16. Estimate the value with 30 at or below it: Value = lower boundary + (target − CF before) ÷ class frequency × width = 30 + (30 − 22) ÷ 16 × 10 = 35.
How do you find the value that marks off the top 10% or the highest 20 values?: Top 10%: the 90th percentile, at 0.9n. Highest 20 of n values: read at cumulative frequency n − 20.
What is the difference between a percentile and a percentage?: A percentile is a value in the data. A percentage is a share of the data.

How do you get Σx and Σx² for a set from its mean and standard deviation?: Σx = n × mean, and Σx² = n × (σ² + mean²). Example: 5 values with mean 6 and σ = 2 → Σx² = 5 × (4 + 36) = 200.
Ten values have mean 8 and thirty values have mean 12. Find the mean of all forty: (10 × 8 + 30 × 12) ÷ 40 = 440 ÷ 40 = 11. Weight each mean by its group size.
How do you find the standard deviation of two combined sets?: Add the n, Σx and Σx² of both sets. Combined σ = √(total Σx² ÷ total n − combined mean²). Never average the two standard deviations.
Set P: 5 values, mean 6, σ = 2. Set Q: 3 values, mean 10, σ = 1. Find the combined σ: Σx² = 200 + 303 = 503; Σx = 60; n = 8; mean 7.5. σ² = 503 ÷ 8 − 7.5² = 62.875 − 56.25 = 6.625, so σ = 2.57 (3 s.f.).
Twelve values with mean 30 join n values with mean 40. The combined mean is 36. Find n: 360 + 40n = 36(12 + n) = 432 + 36n, so 4n = 72 and n = 18.