Statistics Class 10 Maths Notes

Statistics is Unit VII together with Probability, and the two together carry 11 of the 80 theory marks. The retained content is the mean, median and mode of grouped data, with a bimodal table avoided this year. Nothing else from the old syllabus — no ungrouped mean-median-mode from Class 9, no step deviation, no ogive.Every question gives you a table with a frequency column already filled in. Your job is to add the columns your chosen measure needs and then divide once.

• Mean of grouped data — by the direct method, the only method assessed.
• Median of grouped data — from the cumulative frequency column of the table.
• Mode of grouped data — from the empirical relation, using the modal class and the two classes beside it.
• A bimodal table is avoided, so the modal class is always unique.

• Exclusive intervals — 0 to 10, 10 to 20, 20 to 30. Midpoints are whole numbers and the lower boundary equals the lower limit.
• Inclusive intervals — 1 to 10, 11 to 20, 21 to 30. Midpoints are half-decimals and 0.5 must be subtracted from each lower limit to get the boundary.
• Frequency f — the number of observations falling in that class interval.
• The board often gives the table already split into 0 to 10, 10 to 20 and so on. Use that form and keep the arithmetic whole.
• Every class must have the same class size h for the mean, median and mode formulas to work as given.

Worked example — the marks obtained by 30 students of a class are grouped as below. Class intervals are exclusive, so the class size h is 10 throughout. Find the mean of the marks.

• 0 to 10 : f = 2, xᵢ = 5, fᵢxᵢ = 10
• 10 to 20 : f = 5, xᵢ = 15, fᵢxᵢ = 75
• 20 to 30 : f = 9, xᵢ = 25, fᵢxᵢ = 225
• 30 to 40 : f = 10, xᵢ = 35, fᵢxᵢ = 350
• 40 to 50 : f = 4, xᵢ = 45, fᵢxᵢ = 180
• Total : Σfᵢ = 2 + 5 + 9 + 10 + 4 = 30, and Σfᵢxᵢ = 10 + 75 + 225 + 350 + 180 = 840
• Mean = Σfᵢxᵢ ÷ Σfᵢ = 840 ÷ 30 = 28 marks

The median is the middle observation once the data is arranged in order, so you first have to know where the middle falls. For grouped data that means counting from the top class downward until you have half the observations, and the class in which you land is the median class.

• 10 to 20 : f = 4, cf = 4, xᵢ = 15, fᵢxᵢ = 60
• 20 to 30 : f = 6, cf = 10, xᵢ = 25, fᵢxᵢ = 150
• 30 to 40 : f = 10, cf = 20, xᵢ = 35, fᵢxᵢ = 350
• 40 to 50 : f = 14, cf = 34, xᵢ = 45, fᵢxᵢ = 630
• 50 to 60 : f = 10, cf = 44, xᵢ = 55, fᵢxᵢ = 550
• Total : Σfᵢ = 44, so half of 44 is 22. Scan the cf column and find the first class whose cf is 22 or more, which is 30 to 40, since its cf jumps from 20 to 34. So the median class is 30 to 40.
• Read off the values: l = 30, the cf of the preceding class = 20, f = 10, the frequency of the median class, and h = 10.
• Median = 30 + [(22 − 20)/10] × 10 = 30 + (2/10) × 10 = 30 + 2 = 32 marks
• Sanity check: the 22nd observation is the second one in the class 30 to 40, so the median must lie just above 30. It does.

Worked example — the ages, in years, of 49 people are grouped as below. All classes are exclusive with class size h = 10. Find the mode of the ages.

• 0 to 10 : f = 3, xᵢ = 5, fᵢxᵢ = 15
• 10 to 20 : f = 5, xᵢ = 15, fᵢxᵢ = 75
• 20 to 30 : f = 10, xᵢ = 25, fᵢxᵢ = 250
• 30 to 40 : f = 15, xᵢ = 35, fᵢxᵢ = 525
• 40 to 50 : f = 10, xᵢ = 45, fᵢxᵢ = 450
• 50 to 60 : f = 6, xᵢ = 55, fᵢxᵢ = 330
• Total : Σfᵢ = 49
• The highest frequency is 15, so the modal class is 30 to 40. Read off f₁ = 15, and l₁ = 30, the lower boundary of the modal class.
• The class before it is 20 to 30, so f₀ = 10. The class after it is 40 to 50, so f₂ = 10, and l₂, the lower boundary of that next class, is 40.
• Mode = 30 + [(15 − 10)/(2 × 15 − 10 − 10)] × 10 = 30 + (5/10) × 10 = 30 + 5 = 35 years

Marks are lost in this chapter for using a symbol without saying what it is, so state each one in words before you substitute. The list below covers every symbol used in the mean, median and mode formulas.

• fᵢ — the frequency of the i-th class, that is the number of observations in that class interval.
• xᵢ — the midpoint or class mark of the i-th class, the average of its two limits.
• Σfᵢ — the sum of all the frequencies, which is also the total number of observations, usually written N.
• Σfᵢxᵢ — the sum of the products of each frequency with its midpoint.
• cf — the cumulative frequency of a class, the running total of the frequencies of that class and every class above it.
• h — the class size or class width, the difference between the upper and the lower limit.
• l — the lower class boundary of the class you are working with.
• l₁ — the lower class boundary of the modal class.
• f₁ — the frequency of the modal class.
• f₀ — the frequency of the class immediately before the modal class.
• f₂ — the frequency of the class immediately after the modal class.
• N/2 — half the total number of observations, which tells you where the middle observation lies.

The modal class is not always the middle class of the table, and it need not sit in the centre of the distribution. Here the table has only three rows and the mode still comes out correctly.

• 20 to 30 : f = 9, xᵢ = 25, fᵢxᵢ = 225
• 30 to 40 : f = 15, xᵢ = 35, fᵢxᵢ = 525
• 40 to 50 : f = 6, xᵢ = 45, fᵢxᵢ = 270
• Total : Σfᵢ = 30
• Modal class is 30 to 40 with f₁ = 15, so l₁ = 30, f₀ = 9 from the class 20 to 30, and f₂ = 6 from the class 40 to 50.
• Mode = 30 + [(15 − 9)/(2 × 15 − 9 − 6)] × 10 = 30 + (6/15) × 10 = 30 + 4 = 34 years

A table with two classes sharing the highest frequency is bimodal, and in that situation there is no single modal class, so the empirical relation has no unambiguous answer. The 2024-25 syllabus says to avoid a bimodal table, which means you will never be asked one.

• A unimodal table has one class whose frequency is strictly greater than every other frequency. That class is the modal class.
• If you ever meet two classes tied for the highest frequency, that table is bimodal and no unique mode exists from the formula.
• Since the board avoids such tables, the f₁ − f₀ term and the 2f₁ − f₀ − f₂ term are always well defined and the denominator is never zero.
• Also confirm that the empirical relation always returns a value inside the modal class, that is between l₁ and l₁ + h.

The 2024-25 paper is 80 marks in 3 hours with 38 questions. Section A has 18 MCQs and 2 assertion-reason questions of 1 mark each, Section B has 5 very short answer questions of 2 marks, Section C has 6 short answer questions of 3 marks, Section D has 4 long answer questions of 5 marks, and Section E has 3 case-study questions of 4 marks, with internal choice in two questions each of Sections B, C and D. No calculator is allowed, so choose tables where the arithmetic comes out exact.A three-mark question usually asks for the mean of grouped data, because it is one column addition and one division. A five-mark question can ask for the median or the mode, which need the class identified and the values read off.

• The mean of the following grouped data: classes 0 to 10, 10 to 20 and 20 to 30 with frequencies 4, 6 and 10.
• Find the median of grouped data with a total frequency of 44, given that the median class is 30 to 40 with a preceding cf of 20.
• Find the mode when the modal class is 40 to 50, f₀ = 12, f₁ = 16, f₂ = 12 and h = 10. Answer 40 + (4/8) × 10 = 45.
• Which of the following is the correct method for the mean of grouped data in this syllabus? Direct method.
• Assertion-reason: the median class is the class whose cumulative frequency first reaches N/2 — assert, then give the reason.
• MCQ trap: the median always lies inside the median class and the mode always lies inside the modal class; a result outside either means the table was misread.
• MCQ trap: the mean of grouped data is not necessarily equal to the median, and neither is necessarily equal to the mode.