Statistics for CDS

Statistics in the CDS syllabus covers the collection and tabulation of data, graphs (bar charts, pie charts, histograms and frequency polygons) and measures of central tendency. In the paper this becomes a steady handful of short questions. You find a mean from a frequency table, correct a mean after a misread value, read an angle off a pie chart, or apply the grouped-data median formula.

None of it is deep, but the grouped-data formulas are easy to half-remember. This guide gives each formula with its meaning, so that you can rebuild it rather than recall it blindly, and then works through the question types that recur.

Mean

• Raw data: mean = sum of the values ÷ number of values.
• Frequency data: mean = Σfx ÷ Σf, where x is each value and f its frequency.
• Grouped data: use the class mark (the midpoint of each class) as x. Class mark = (lower limit + upper limit) ÷ 2.

Properties that save time

• Add a constant c to every value, and the mean increases by c.
• Multiply every value by k, and the mean is multiplied by k. The median and mode behave the same way.
• The deviations from the mean always add to zero.
• Combined mean of two groups: (n₁ × mean₁ + n₂ × mean₂) ÷ (n₁ + n₂). It is not the plain average of the two means unless the groups are the same size.
• Correcting a mean: recover the total (mean × number), fix it by adding the right value and removing the wrong one, and divide again.

Median

• Arrange the data in order.
• Odd count n: the median is the ((n + 1)/2)th value.
• Even count n: the median is the average of the (n/2)th and (n/2 + 1)th values.

Grouped data

Find the cumulative frequencies and locate the class in which the cumulative frequency first reaches N/2. That is the median class. Then:

Median = L + ((N/2 − cf) ÷ f) × h

Here L is the lower limit of the median class, cf the cumulative frequency before that class, f the frequency of the class and h its width. The formula assumes the values are spread evenly across the class and moves the right fraction of the way into it.

Mode

• Raw data: the most frequent value. A data set can have more than one mode.
• Grouped data: the modal class is the one with the highest frequency. Then:

Mode = L + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h

Here f₁ is the frequency of the modal class, and f₀ and f₂ are the frequencies of the classes just before and after it.

Empirical relation

For moderately skewed data, mode ≈ 3 median − 2 mean. CDS questions treat it as exact when they give two of the three and ask for the third.

Dispersion, briefly

• Range = largest value − smallest value.
• Standard deviation measures spread around the mean. Adding a constant to every value leaves it unchanged. Multiplying every value by k multiplies it by k (by the size of k, if k is negative).

Graphs

Worked questions

Question 1: Find the mean of the following data. Values 5, 10, 15, 20 and 25 have frequencies 3, 5, 8, 3 and 1.

• Σf = 20. Σfx = 15 + 50 + 120 + 60 + 25 = 270.
• Mean = 270 ÷ 20 = 13.5.

Question 2: The mean of 40 observations was found to be 25. Later it was noticed that an observation of 45 had been read as 25. Find the correct mean.

• Wrong total = 40 × 25 = 1,000. Correct total = 1,000 − 25 + 45 = 1,020.
• Correct mean = 1,020 ÷ 40 = 25.5.

Question 3: Section A has 30 students with a mean score of 60, and Section B has 20 students with a mean score of 70. Find the combined mean.

• (30 × 60 + 20 × 70) ÷ 50 = (1,800 + 1,400) ÷ 50 = 64.
• The plain average of 60 and 70, which is 65, is the trap option.

Question 4: Find the median and mode of this distribution.

• N = 40, so N/2 = 20. Cumulative frequencies: 5, 13, 25, 35, 40. The median class is 20–30.
• Median = 20 + ((20 − 13) ÷ 12) × 10 = 20 + 5.83 = 25.83 (to two decimals).
• The modal class is also 20–30, with f₁ = 12, f₀ = 8 and f₂ = 10.
• Mode = 20 + (4 ÷ (24 − 18)) × 10 = 20 + 6.67 = 26.67.
• As a check, the mean from class marks is 1,020 ÷ 40 = 25.5, and 3 × 25.83 − 2 × 25.5 = 26.5, close to the mode, as the empirical relation predicts.

Question 5: A family's monthly budget is ₹30,000, of which ₹9,000 goes on food. What angle does food take in a pie chart?

• (9,000 ÷ 30,000) × 360° = 108°.

Question 6: The mean of 10 numbers is 20. Each number is multiplied by 3 and then 5 is added. Find the new mean.

• 20 × 3 + 5 = 65.

Practice set

1. The mean of five numbers is 20. Four of them are 15, 18, 22 and 25. Find the fifth.
2. Find the median of 4, 9, 2 and 7.
3. The mean of x, x + 2 and x + 4 is 10. Find x.
4. Find the mean of the first 10 natural numbers.
5. Find the mode of 4, 6, 6, 7, 8, 8, 8, 9.
6. In a pie chart of 500 students, the sector for sports is 72°. How many students chose sports?
7. The mean of 5 observations is 12. When one observation is removed, the mean of the rest is 10. Find the removed observation.
8. Find the median of the first 8 prime numbers.

Answers

1. 20. The total is 100, and the four sum to 80.
2. 5.5. Sorted: 2, 4, 7, 9. The middle pair averages to 5.5.
3. 8. 3x + 6 = 30.
4. 5.5. 55 ÷ 10.
5. 8. It appears three times.
6. 100. 72/360 = 1/5 of 500.
7. 20. 60 − 4 × 10.
8. 9. The primes are 2, 3, 5, 7, 11, 13, 17 and 19, so the median is (7 + 11) ÷ 2.

What to do next

• Write the grouped median and mode formulas from memory, with what each letter means
• Redo Question 4 without looking, including the cumulative frequency column
• Revise averages and mixtures for more combined-mean questions
• Use the maths revision sheet to check the rest of the maths formulas
• Solve the statistics questions from the last five CDS papers, and note which graph types they used