Introduction
In Class 9 you found the mean, median and mode of ungrouped data. In Class 10, data is often grouped into class intervals with a frequency for each class. Because the exact observations inside a class are not available, the class mark is used as a representative value while calculating the mean.
For a class interval, class mark xi = (upper limit + lower limit) ⁄ 2. For example, the class mark of 20, 30 is 25.
Mean of Grouped Data
There are three methods for finding the mean of grouped data. All three give the same answer, you can use whichever makes the arithmetic easiest.
1. Direct method
Mean (x̄) = Σ fixi ⁄ Σ fi
where fi is the frequency of class i and xi is its class mark. Σ fi = n is the total number of observations.
This is straightforward but involves multiplying large class marks, so it can be heavy when the numbers are big.
2. Assumed mean method
Pick any class mark as the assumed mean a (choosing the middle class usually keeps numbers small). Let di = xi − a be the deviation of each class mark from a.
x̄ = a + Σ fidi ⁄ Σ fi
Because di values are smaller than xi values, this often makes calculations easier.
3. Step-deviation method
If all class intervals have the same width h, we can simplify further. Let ui = (xi − a) ⁄ h. Since h divides evenly, ui values are small integers.
x̄ = a + (Σ fiui ⁄ Σ fi) × h
where ui = (xi − a) ⁄ h and h is the class size. This is the shortest method when the class size is constant, and is the one most often expected in the exam for grouped data.
Find the mean of the following data:
Classes: 0, 10 (f=5), 10, 20 (f=8), 20, 30 (f=12), 30, 40 (f=10), 40, 50 (f=5).
Class marks xi: 5, 15, 25, 35, 45. Class size h = 10. Choose assumed mean a = 25 (middle class).
Compute ui = (xi − 25)⁄10: −2, −1, 0, 1, 2.
fiui: 5×(−2)=−10, 8×(−1)=−8, 12×0=0, 10×1=10, 5×2=10. Sum Σ fiui = 2. Total Σ fi = 40.
x̄ = 25 + (2 ⁄ 40) × 10 = 25 + 0.5 = 25.5
You can verify with the direct method: Σ fixi = 25+120+300+350+225 = 1020, and 1020⁄40 = 25.5.
All three methods give the same mean. In practice, the step-deviation method saves the most time, especially when class marks are large.
Mode of Grouped Data
The mode is the value that occurs most often. For grouped data, we first find the modal class (the class with the highest frequency) and then use a formula that estimates where the peak lies inside that class.
Mode = l + (f1 − f0) ⁄ (2f1 − f0 − f2) × h
where l = lower limit of the modal class, h = class size, f1 = frequency of the modal class, f0 = frequency of the class before the modal class, f2 = frequency of the class after the modal class.
If the highest frequency occurs in the first or last class, one of f0 or f2 is taken as 0.
Find the mode of the data:
Classes: 0, 10 (f=4), 10, 20 (f=8), 20, 30 (f=15), 30, 40 (f=10), 40, 50 (f=6).
The highest frequency is 15, so the modal class is 20, 30. Here l = 20, h = 10, f1 = 15, f0 = 8, f2 = 10.
Mode = 20 + (15 − 8) ⁄ (30 − 8 − 10) × 10 = 20 + 7⁄12 × 10 = 20 + 5.83 = 25.83 (approx.)
Remember: the grouped-data formulas in this chapter assume the classes are continuous, that is, the upper limit of one class equals the lower limit of the next. If the data is given in non-continuous form (for example, 0, 9, 10, 19), convert it to continuous form (0, 9.5, 9.5, 19.5) before applying these formulas, but such cases are rare in Class 10.
Median of Grouped Data
The median is the middle value when all observations are arranged in order. For grouped data, we use cumulative frequencies to find the class that contains the middle observation.
Median = l + (n⁄2 − cf) ⁄ f × h
where l = lower limit of the median class, n = Σ fi (total observations), cf = cumulative frequency of the class before the median class, f = frequency of the median class, h = class size. The median class is the class whose cumulative frequency first reaches or exceeds n⁄2.
Find the median of the data:
Classes: 0, 10 (f=5), 10, 20 (f=7), 20, 30 (f=10), 30, 40 (f=8), 40, 50 (f=4). n = 34.
Cumulative frequencies: 5, 12, 22, 30, 34. Need n⁄2 = 17.
The cumulative frequency first reaches 17 in the class 20, 30 (cf = 22). So median class is 20, 30. Here l = 20, cf of previous class = 12, f = 10, h = 10.
Median = 20 + (17 − 12) ⁄ 10 × 10 = 20 + 5 = 25
If n is odd, the median is still given by the same formula, n⁄2 may be a.5 value, but the same median class rule applies. When every class has equal width, the formula works cleanly.
A useful relationship that sometimes appears for reference (not required to derive in the exam) is:
3 Median = Mode + 2 Mean
This is an approximate empirical relationship. It is good for a quick check, but do not use it to calculate one average from the others in a formal question unless the question itself asks you to verify it.
Summary
- Class mark xi = (upper limit + lower limit) ⁄ 2 is used as the representative value for grouped data.
- Mean, Direct: x̄ = Σ fixi ⁄ Σ fi.
- Mean, Assumed mean: x̄ = a + Σ fidi ⁄ Σ fi, where di = xi − a.
- Mean, Step-deviation: x̄ = a + (Σ fiui ⁄ Σ fi) × h, where ui = (xi − a)⁄h. Best when class size h is constant.
- Mode: Mode = l + (f1−f0)⁄(2f1−f0−f2) × h. Modal class = class with highest frequency.
- Median: Median = l + (n⁄2 − cf)⁄f × h. Median class = first class with cumulative frequency ≥ n⁄2.
- Empirical check (approximate): 3 Median = Mode + 2 Mean.
Practice Questions
- The following table gives the number of books read by 40 students in a month. Find the mean using the step-deviation method: 0, 5 (f=4), 5, 10 (f=8), 10, 15 (f=12), 15, 20 (f=10), 20, 25 (f=6).
- Find the mode of: 0, 10 (f=6), 10, 20 (f=10), 20, 30 (f=16), 30, 40 (f=12), 40, 50 (f=8).
- Find the median of: 0–20 (f=5), 20–40 (f=10), 40–60 (f=15), 60–80 (f=8), 80–100 (f=6).
- In a grouped data, the mean is 25.5 and the mode is 25.83. Estimate the median using the empirical relationship.
- A class has marks: 10, 20 (f=3), 20, 30 (f=7), 30, 40 (f=9), 40, 50 (f=6), 50, 60 (f=5). Find the mean by the assumed mean method taking a = 35, and verify by the direct method.