Statistics
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
The class 10 Statistics chapter is about finding the mean, median and mode when data has been grouped into class intervals, using class marks, cumulative frequencies, and the median class and modal class.
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This chapter is about central tendency for grouped data. A raw table is compressed into class intervals, so the individual marks or measurements are no longer known. The mean is the balancing point, the median is the middle value when the distribution is ordered, and the mode is the most repeated value. To handle grouped data, the chapter uses class marks, frequencies and cumulative frequencies. Each measure asks a different question, and each has a particular interval that defines it: the median class and the modal class.
The mean can be found by the direct method, x̄ = (Σfᵢxᵢ)/(Σfᵢ). When the values are large, choosing an assumed mean a and working with deviations makes the arithmetic easier. The assumed-mean method still gives the true mean because the average deviation is added back to a. If all class intervals have the same width h, the step-deviation method divides every deviation by h before averaging and then multiplies by h. All three methods are algebraically identical, so they always agree.
The median depends on cumulative frequencies. Add the class frequencies in order until the cumulative frequency first reaches or exceeds N/2; that class is the median class. The median is then placed inside that class by interpolation. The formula uses the lower limit of the median class, the cumulative frequency just before it, and the frequency inside it. A common mistake is to use the cumulative frequency of the median class itself instead of the cumulative frequency of the class before it.
The mode is found from the modal class, which is simply the class interval with the greatest frequency. The formula interpolates inside the modal class using the frequencies of the previous class and the next class. A high-frequency class is not necessarily near the middle, so the modal class and median class may be different. The main difficulty in this chapter is choosing the correct class first and then substituting l, h, f and cf without error.
Key terms
- Grouped data
- Observations sorted into class intervals, with only the frequency of each interval known. Individual values inside an interval are not preserved.
- Class mark or midpoint xᵢ
- The value (lower limit + upper limit)/2 for a class. It represents every observation in that class when calculating the mean.
- Class size h
- The width of a class interval, usually upper limit minus lower limit for continuous classes. It is used in the step-deviation, median and mode formulas.
- Cumulative frequency
- A running total obtained by adding the class frequencies one by one from the lowest class. It is needed to locate the median class.
- Median class
- The class interval whose cumulative frequency is the first to reach or exceed N/2, where N is the total frequency. The median is then found inside this class.
- Modal class
- The class interval with the highest frequency. The mode is calculated by interpolating inside this class using the frequencies of the previous and next classes.
- Assumed mean method
- A shortcut for the mean in which a convenient class mark a is chosen as the origin, deviations dᵢ = xᵢ − a are used, and the average deviation is added back to a.
- Step-deviation method
- An assumed-mean method used when all class sizes are equal. Each deviation is divided by the class size h, making the numbers even smaller, and the result is multiplied by h at the end.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Mean of grouped data by the direct method | x̄ = (Σfᵢxᵢ)/(Σfᵢ) | x̄ is the mean, xᵢ is the class mark of the ith class, fᵢ is its frequency, and Σfᵢ is the total number of observations N. |
| Mean by the assumed-mean method | x̄ = a + (Σfᵢdᵢ)/(Σfᵢ), where dᵢ = xᵢ − a | a is the assumed mean, dᵢ is the deviation of the class mark xᵢ from a, and N = Σfᵢ is the total frequency. |
| Mean by the step-deviation method | x̄ = a + h(Σfᵢuᵢ)/(Σfᵢ), where uᵢ = (xᵢ − a)/h | h is the equal class size, uᵢ is the step deviation, and N = Σfᵢ is the total frequency. |
| Median of grouped data | Median = l + (h(n/2 − cf))/f | l is the lower limit of the median class, cf is the cumulative frequency of the class before the median class, f is the frequency of the median class, h is its class size, and n is the total frequency. |
| Mode of grouped data | Mode = l + (h(f₁ − f₀))/(2f₁ − f₀ − f₂) | l is the lower limit of the modal class, f₁ is its frequency, f₀ is the frequency of the previous class, f₂ is the frequency of the next class, and h is the class size. |
Practice questions with answers
1. Find the mean of this distribution by the direct method: 0–10: 3, 10–20: 5, 20–30: 8, 30–40: 3, 40–50: 1.
The class marks are 5, 15, 25, 35 and 45. N = 3+5+8+3+1 = 20, and Σfᵢxᵢ = 3×5 + 5×15 + 8×25 + 3×35 + 1×45 = 440. So x̄ = 440/20 = 22.
2. Using the assumed mean a = 35, find the mean for the distribution 0–10: 3, 10–20: 5, 20–30: 8, 30–40: 3, 40–50: 1.
Class marks are 5, 15, 25, 35 and 45. The deviations from 35 are −30, −20, −10, 0 and 10. Σfᵢdᵢ = 3(−30) + 5(−20) + 8(−10) + 3(0) + 1(10) = −260. With N = 20, x̄ = 35 + (−260)/20 = 35 − 13 = 22.
3. For a grouped distribution, N = 40, Σfᵢuᵢ = −8, assumed mean a = 25 and class size h = 5. Find the mean by the step-deviation method.
x̄ = a + h(Σfᵢuᵢ)/N = 25 + 5×(−8/40) = 25 − 1 = 24. The mean is 24.
4. Find the median of the distribution: 0–10: 5, 10–20: 8, 20–30: 12, 30–40: 10, 40–50: 5.
N = 40, so N/2 = 20. Cumulative frequencies are 5, 13, 25, 35 and 40, so the median class is 20–30. With l = 20, cf = 13, f = 12 and h = 10, median = 20 + 10(20−13)/12 = 20 + 70/12 = 25.83.
5. The less-than cumulative frequencies for marks are 4, 12, 26, 44 and 50 for the intervals 0–10, 10–20, 20–30, 30–40 and 40–50. Find the median mark.
The class frequencies are 4, 8, 14, 18 and 6. N = 50, so N/2 = 25. The median class is 20–30 because the cumulative frequency 26 first passes 25. For this class, l = 20, cf = 12, f = 14 and h = 10, so median = 20 + 10(25−12)/14 = 20 + 130/14 = 29.29.
6. The mean of the following distribution is 20. Find the missing frequency k: 0–10: 4, 10–20: 6, 20–30: k, 30–40: 2, 40–50: 1.
Class marks are 5, 15, 25, 35 and 45. Total N = 13 + k, and Σfᵢxᵢ = 4×5 + 6×15 + 25k + 2×35 + 1×45 = 225 + 25k. Since mean = 20, (225 + 25k)/(13 + k) = 20. Solving gives 225 + 25k = 260 + 20k, so k = 7.
7. For the distribution 0–10: 5, 10–20: 9, 20–30: 15, 30–40: 6, 40–50: 4, find the mode.
The highest frequency is 15 in the class 20–30, so l = 20, f₁ = 15, f₀ = 9, f₂ = 6 and h = 10. Mode = 20 + 10(15−9)/(30−9−6) = 20 + 60/15 = 24.
8. A distribution has N = 48, median class 20–30, cumulative frequency before this class 15, frequency of this class 10. Find the median.
N/2 = 24. For the median class, l = 20, cf = 15, f = 10 and h = 10. Median = 20 + 10(24−15)/10 = 20 + 9 = 29.
9. For a distribution with N = 100, the cumulative frequencies of five classes are 18, 45, 78, 95 and 100. Which class is the median class?
Here N/2 = 50. The first cumulative frequency that reaches or exceeds 50 is 78, so the median class is the third class interval.
10. A class has the highest frequency. Does it always contain the median? Explain.
No. The modal class is the class with the greatest frequency, while the median class is the class whose cumulative frequency first reaches or exceeds N/2. The two classes may be the same, but the modal class need not lie at the middle of the distribution.
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