← All topicsπŸ“‹ Mean of Grouped DataMaths
Mean, Median & Mode Β· Topic 4 of 6

πŸ“‹ Mean of Grouped Data

1 exam question type, fully solved

When data comes as a frequency distribution β€” values (or class intervals) paired with how many times each occurs β€” finding the mean by raw addition is slow. Two simple shortcuts keep the numbers tiny.

Three ways to the same mean

For class intervals, first replace each class by its midpoint .

β€’ Direct method: β€” multiply each value by its frequency, add up, divide by the total frequency.

β€’ Assumed-mean method (faster with large midpoints): pick a central value , let , then .

β€’ Step-deviation method (fastest when all classes share a width ): let , then .

All three give the identical answer; the deviation methods just shrink the arithmetic. Sign check: if (or ) is negative the mean is below ; if positive, above. Common slips: using class limits instead of midpoints, dividing by the number of classes instead of , or forgetting to multiply by in step-deviation.

πŸ“ŠStatistics ExplorerEnter up to 10 numbers

Using 8 values (max 10)

Mean=5.5Median=6
β€” Meanβ€” Median● Mode (larger dot)● Other values
Sorted Array
23457779
Mean
(4 + 7 + 2 + 9 + 7 + 3 + 7 + 5) / 8
5.5
Median
Avg of positions 4 and 5 = (5 + 7) / 2
6
Mode
7 (unimodal)
Distribution insight: The mean is smaller than the median β€” the data is left-skewed (pulled by small outliers).
Empirical Check: Mode = 3Γ—Median βˆ’ 2Γ—Mean
3Γ—6 βˆ’ 2Γ—5.5 = 7 (actual mode: 7)

With , and , the mean is .

πŸ“Practice Questions

Q1For grouped data the mean uses x equal to each class's:

Q2Assumed-mean method: A = 40, Ξ£fd = 60, Ξ£f = 30. The mean is:

Q3Step-deviation: A = 50, h = 5, Ξ£fu = 20, Ξ£f = 50. The mean is:

πŸ“š

Real exam questions β€” Mean of Grouped Data

1 question types Β· 5 solved examples from real SSC papers

Represent each class by its midpoint, then use Mean=A+βˆ‘fdβˆ‘f\text{Mean}=A+\dfrac{\sum f d}{\sum f} (assumed mean) or A+hβˆ‘fuβˆ‘fA+h\dfrac{\sum f u}{\sum f} (step deviation) to keep the numbers small.

How to solve this type
Direct method: Mean=βˆ‘fxβˆ‘f\text{Mean}=\dfrac{\sum f x}{\sum f}, where xx is the class MIDPOINT (mid =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}) and ff its frequency.
Assumed-mean method (faster with big midpoints): pick a central value AA, let d=xβˆ’Ad=x-A, then Mean=A+βˆ‘fdβˆ‘f\text{Mean}=A+\dfrac{\sum f d}{\sum f}.
Step-deviation method (fastest when classes are equally wide hh): let u=xβˆ’Ahu=\dfrac{x-A}{h}, then Mean=A+hΓ—βˆ‘fuβˆ‘f\text{Mean}=A+h\times\dfrac{\sum f u}{\sum f}.
All three give the same answer β€” the deviation methods just shrink the arithmetic. If βˆ‘fd\sum f d (or βˆ‘fu\sum f u) is negative, the mean is below AA; if positive, above.

Find the mean of the frequency distribution x = 5, 10, 15, 20, 25 with frequencies f = 2, 4, 5, 3, 1.

A15B14.5C13D14

Find the mean using the assumed-mean method with A = 25: values x = 10, 20, 30, 40, 50 and frequencies f = 1, 3, 5, 7, 4 (Ξ£f = 20).

A35B34C33D36

Find the mean of the table: scores 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 with frequencies 2, 4, 12, 21, 6, 3, 2.

A35.8B33.4C32.6D34.2

Using the assumed-mean method with A = 30, Ξ£fd = βˆ’60 and n = 20, find the mean.

A27B24C26D28

In a frequency distribution the assumed mean is 100, class width h = 10, Ξ£fu = βˆ’50 and Ξ£f = 100. Find the mean (step-deviation method).

A92B93C94D95