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Mean of Grouped Data

Class 10Statistics

Mean of grouped data is the most important measure of central tendency in statistics. It is covered in CBSE Class 10 Mathematics, Chapter 14 (Statistics).



When data is presented in a grouped frequency distribution table (class intervals with frequencies), individual values are not known. The mean is calculated using the class marks (midpoints of class intervals) as representatives of each class.



Three methods are used to calculate the mean of grouped data:

  • Direct Method — best for small data sets
  • Assumed Mean Method — reduces computation effort
  • Step-Deviation Method — most efficient for equal class sizes

What is Mean of Grouped Data - Direct, Assumed Mean & Step-Deviation Methods?

Definition: The mean (or arithmetic mean) of grouped data is the average value, calculated by dividing the sum of all values by the total number of observations. Since individual values are unknown in grouped data, class marks are used as representative values.


Key terms:

  • Class interval — a range of values (e.g., 10-20, 20-30).
  • Class mark (x_i) — midpoint of a class interval = (upper limit + lower limit)/2.
  • Frequency (f_i) — number of observations in a class interval.
  • Sigma f_i — total number of observations (sum of all frequencies).
  • Assumed mean (a) — a convenient value (usually the class mark of the middle class) used to simplify calculations.
  • Deviation (d_i) — d_i = x_i - a (difference of class mark from assumed mean).
  • Step-deviation (u_i) — u_i = (x_i - a)/h, where h is the class size.

Mean of Grouped Data Formula

Method 1: Direct Method

Mean = Sigma(f_i x_i) / Sigma(f_i)


Method 2: Assumed Mean Method

Mean = a + Sigma(f_i d_i) / Sigma(f_i)

Where d_i = x_i - a, and a = assumed mean.


Method 3: Step-Deviation Method

Mean = a + [Sigma(f_i u_i) / Sigma(f_i)] x h

Where u_i = (x_i - a)/h, a = assumed mean, h = class size.


Choosing the method:

  • Use Direct Method when class marks and frequencies are small.
  • Use Assumed Mean Method when class marks are large.
  • Use Step-Deviation Method when class sizes are equal and values are large (most efficient).

Derivation and Proof

Derivation of the Assumed Mean Method:

  1. Let the true mean be x-bar = Sigma(f_i x_i) / Sigma(f_i).
  2. Let a be the assumed mean. Define d_i = x_i - a, so x_i = a + d_i.
  3. Sigma(f_i x_i) = Sigma f_i(a + d_i) = a x Sigma(f_i) + Sigma(f_i d_i).
  4. x-bar = a + Sigma(f_i d_i) / Sigma(f_i).
  5. This is the Assumed Mean formula. Hence derived.

Derivation of the Step-Deviation Method:

  1. Start from d_i = x_i - a.
  2. If all class sizes are equal (= h), then d_i is always a multiple of h.
  3. Define u_i = d_i / h = (x_i - a)/h. So d_i = h x u_i.
  4. Sigma(f_i d_i) = Sigma f_i(h x u_i) = h x Sigma(f_i u_i).
  5. x-bar = a + h x Sigma(f_i u_i) / Sigma(f_i).
  6. This is the Step-Deviation formula. Hence derived.

All three methods are algebraically equivalent and always produce the same answer.

Types and Properties

Problems on mean of grouped data in Class 10 include:

Type 1: Calculate Mean Using Direct Method

  • Construct the f_i x x_i column, sum it, and divide by Sigma(f_i).

Type 2: Calculate Mean Using Assumed Mean Method

  • Choose a suitable assumed mean (a), calculate d_i = x_i - a, find Sigma(f_i d_i).

Type 3: Calculate Mean Using Step-Deviation Method

  • Choose a, calculate u_i = (x_i - a)/h, find Sigma(f_i u_i).

Type 4: Finding Missing Frequency

  • Given the mean and all but one frequency, find the missing frequency using the mean formula.

Type 5: Combined Mean

  • Combine two groups with known means and total frequencies to find the overall mean.

Type 6: Comparison of Methods

  • Solve the same problem using all three methods and verify the same answer.

Methods

Direct Method — Step by Step:

  1. Write the class intervals and their frequencies (f_i).
  2. Find the class mark for each interval: x_i = (upper limit + lower limit)/2.
  3. Calculate f_i x x_i for each class.
  4. Sum all f_i to get Sigma(f_i) = N.
  5. Sum all f_i x x_i to get Sigma(f_i x_i).
  6. Mean = Sigma(f_i x_i) / Sigma(f_i).

Assumed Mean Method — Step by Step:

  1. Find class marks (x_i).
  2. Choose an assumed mean (a) — usually the class mark of the middle class or highest-frequency class.
  3. Calculate d_i = x_i - a for each class.
  4. Calculate f_i x d_i for each class.
  5. Sum to get Sigma(f_i d_i).
  6. Mean = a + Sigma(f_i d_i) / Sigma(f_i).

Step-Deviation Method — Step by Step:

  1. Find class marks (x_i) and class size h.
  2. Choose an assumed mean (a).
  3. Calculate u_i = (x_i - a)/h for each class.
  4. Calculate f_i x u_i for each class.
  5. Sum to get Sigma(f_i u_i).
  6. Mean = a + [Sigma(f_i u_i) / Sigma(f_i)] x h.

Tips:

  • Choose the assumed mean close to the expected mean (usually the middle class mark).
  • The step-deviation method works only when class sizes are equal.
  • Always verify: the mean should lie within the range of the data.

Solved Examples

Example 1: Direct Method

Problem: Find the mean of the following data using the direct method:

Class IntervalFrequency (f_i)
0-105
10-208
20-3015
30-4012
40-5010

Solution:

Classf_ix_if_i x_i
0-105525
10-20815120
20-301525375
30-401235420
40-501045450
Total501390

Mean = 1390/50 = 27.8.

Answer: The mean is 27.8.

Example 2: Assumed Mean Method

Problem: Find the mean of the following data using the assumed mean method:

ClassFrequency
100-12012
120-14014
140-1608
160-1806
180-20010

Solution:

Take assumed mean a = 150 (class mark of middle class 140-160).

Classf_ix_id_i = x_i - 150f_i d_i
100-12012110-40-480
120-14014130-20-280
140-160815000
160-180617020120
180-2001019040400
Total50-240

Mean = 150 + (-240)/50 = 150 - 4.8 = 145.2.

Answer: The mean is 145.2.

Example 3: Step-Deviation Method

Problem: Find the mean using the step-deviation method:

ClassFrequency
50-1006
100-15010
150-20020
200-2508
250-3006

Solution:

h = 50, a = 175 (class mark of 150-200).

Classf_ix_iu_i = (x_i - 175)/50f_i u_i
50-100675-2-12
100-15010125-1-10
150-2002017500
200-250822518
250-3006275212
Total50-2

Mean = 175 + (-2/50) x 50 = 175 - 2 = 173.

Answer: The mean is 173.

Example 4: Finding Missing Frequency

Problem: The mean of the following distribution is 64. Find the value of p.

ClassFrequency
20-403
40-60p
60-8020
80-10012
100-1205

Solution:

  • Sigma(f_i x_i) = 3(30) + p(50) + 20(70) + 12(90) + 5(110) = 90 + 50p + 1400 + 1080 + 550 = 3120 + 50p
  • Sigma(f_i) = 40 + p

Mean = 64:

  • 64 = (3120 + 50p)/(40 + p)
  • 2560 + 64p = 3120 + 50p
  • 14p = 560
  • p = 40

Answer: The missing frequency p = 40.

Example 5: Student Marks Data

Problem: The marks obtained by 40 students are as follows. Find the mean marks.

MarksNumber of Students
0-204
20-406
40-6010
60-8012
80-1008

Solution (Direct Method):

Marksf_ix_if_i x_i
0-2041040
20-40630180
40-601050500
60-801270840
80-100890720
Total402280

Mean = 2280/40 = 57.

Answer: The mean marks are 57.

Example 6: Comparing All Three Methods

Problem: Verify that all three methods give the same mean for: 10-20 (3), 20-30 (5), 30-40 (7), 40-50 (5).


Direct Method:

  • Sigma(f_i x_i) = 3(15) + 5(25) + 7(35) + 5(45) = 45 + 125 + 245 + 225 = 640
  • Sigma(f_i) = 20
  • Mean = 640/20 = 32

Assumed Mean Method (a = 35):

  • Sigma(f_i d_i) = 3(-20) + 5(-10) + 7(0) + 5(10) = -60 - 50 + 0 + 50 = -60
  • Mean = 35 + (-60)/20 = 35 - 3 = 32

Step-Deviation Method (a = 35, h = 10):

  • Sigma(f_i u_i) = 3(-2) + 5(-1) + 7(0) + 5(1) = -6 - 5 + 0 + 5 = -6
  • Mean = 35 + (-6/20) x 10 = 35 - 3 = 32

Answer: All three methods give mean = 32. Verified.

Example 7: Daily Wages Data (Step-Deviation)

Problem: Find the mean daily wages using the step-deviation method:

Wages (Rs.)Workers
200-25010
250-30015
300-35020
350-40025
400-45018
450-50012

Solution:

h = 50, a = 375 (class mark of 350-400).

Wagesf_ix_iu_if_i u_i
200-25010225-3-30
250-30015275-2-30
300-35020325-1-20
350-4002537500
400-45018425118
450-50012475224
Total100-38

Mean = 375 + (-38/100) x 50 = 375 - 19 = 356.

Answer: The mean daily wage is Rs. 356.

Example 8: Combined Mean of Two Groups

Problem: Group A has 30 students with mean marks 60. Group B has 20 students with mean marks 75. Find the combined mean.


Solution:

Given:

  • n_1 = 30, x-bar_1 = 60
  • n_2 = 20, x-bar_2 = 75

Combined mean:

  • = (n_1 x x-bar_1 + n_2 x x-bar_2) / (n_1 + n_2)
  • = (30 x 60 + 20 x 75) / (30 + 20)
  • = (1800 + 1500) / 50
  • = 3300 / 50 = 66

Answer: The combined mean is 66.

Real-World Applications

Economics:

  • Calculating average income, expenditure, or GDP growth from grouped data collected across income brackets.

Education:

  • Finding average marks of students when marks are grouped in intervals (0-20, 20-40, etc.).

Demographics:

Quality Control:

  • Manufacturing data on product weights/dimensions is grouped. Mean calculations check if production meets specifications.

Healthcare:

  • Patient data (blood pressure, BMI) is grouped. Mean values help assess population health trends.

Government:

  • Grouped data on rainfall, temperature, or crop yield is summarised using means for policy decisions.

Key Points to Remember

  • Mean of grouped data uses class marks (midpoints) as representatives of each class interval.
  • Three methods: Direct, Assumed Mean, and Step-Deviation. All give the same result.
  • Class mark x_i = (upper limit + lower limit)/2.
  • Choose assumed mean (a) as the class mark of the class with highest frequency or the middle class.
  • Step-deviation method works only when all class sizes are equal.
  • The mean may not be a whole number even if all data values are whole numbers.
  • The mean is affected by extreme values (unlike median).
  • For finding missing frequency, set up the mean equation and solve for the unknown.
  • Always verify: the mean should lie within the range of the data.
  • Combined mean of two groups: (n_1 x-bar_1 + n_2 x-bar_2) / (n_1 + n_2).

Practice Problems

  1. Find the mean of: 0-10 (7), 10-20 (10), 20-30 (15), 30-40 (8), 40-50 (10) using the direct method.
  2. Find the mean of: 500-600 (4), 600-700 (7), 700-800 (12), 800-900 (5), 900-1000 (2) using the assumed mean method.
  3. Find the mean of: 25-35 (6), 35-45 (10), 45-55 (8), 55-65 (12), 65-75 (4) using the step-deviation method.
  4. If the mean of the distribution 10-20 (5), 20-30 (p), 30-40 (10), 40-50 (7), 50-60 (8) is 36, find p.
  5. The mean of 60 observations in a grouped frequency table is 28.5. Find the sum of all observations.
  6. A data set has class intervals 0-5, 5-10, 10-15, 15-20 with frequencies 3, 8, 12, 7. Find the mean using all three methods and verify.

Frequently Asked Questions

Q1. What is the mean of grouped data?

The mean of grouped data is the average value calculated using class marks (midpoints of class intervals) and their frequencies. Mean = Sigma(f_i x_i) / Sigma(f_i).

Q2. What are the three methods to calculate mean of grouped data?

Direct Method (Sigma f_i x_i / Sigma f_i), Assumed Mean Method (a + Sigma f_i d_i / Sigma f_i), and Step-Deviation Method (a + [Sigma f_i u_i / Sigma f_i] x h).

Q3. What is a class mark?

The class mark is the midpoint of a class interval, calculated as (upper limit + lower limit)/2. For example, the class mark of 20-30 is (20+30)/2 = 25.

Q4. How do you choose the assumed mean?

Choose the class mark of the middle class or the class with the highest frequency. Any value can be chosen — it does not affect the final answer.

Q5. When should the step-deviation method be used?

When all class intervals have equal width (h) and the class marks are large. It reduces computation by working with smaller numbers (u_i).

Q6. Do all three methods give the same answer?

Yes. All three methods are algebraically equivalent and always give the same mean.

Q7. Can the mean lie outside the data range?

No. The mean always lies between the smallest and largest class marks. If your answer is outside this range, there is a calculation error.

Q8. What is the deviation d_i in the assumed mean method?

d_i = x_i - a, where x_i is the class mark and a is the assumed mean. It represents how far each class mark is from the assumed mean.

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