Tenth Grade Mathematics: Mean of Grouped Data: Direct, Assumed Mean, and Step-Deviation Methods
In modern data science, economics, and empirical research, raw observations are rarely analyzed as isolated numbers; they are organized into continuous frequency distributions known as grouped data. When individual values are grouped into class intervals, their precise values are lost within the bins, requiring a representative midpoint—the class mark—to perform calculations. To compute the arithmetic mean of large grouped datasets efficiently, statisticians utilize three progressively powerful algebraic algorithms: the Direct Method, the Assumed Mean Method, and the Step-Deviation Method.
The Concept of Class Mark
In a continuous grouped frequency distribution, each interval has a lower class limit and an upper class limit. The central representative value of any class interval is its class mark (xi):
Class Mark Formula:
Upper Class Limit + Lower Class Limit
xi = -------------------------------------
2
The difference between the upper and lower limits of any class interval is called the class size, commonly denoted by h.
Method 1: The Direct Method
The Direct Method is the straightforward extension of the standard weighted average formula to grouped data:
Theorem (The Direct Method Formula):
Sum(fi * xi)
Mean x̄ = --------------
Sum(fi)
where fi is the frequency of the i-th class interval, and xi is its corresponding class mark.
Computational Steps:
1. For each class interval, calculate the class mark xi = (Lower + Upper) / 2.
2. Multiply each frequency fi by its class mark xi to obtain the product fi * xi.
3. Compute the sum of all frequencies: N = Sum(fi).
4. Compute the sum of all products: Sum(fi * xi).
5. Divide Sum(fi * xi) by Sum(fi).
When numbers and frequencies are small, the Direct Method is fast; however, when class marks or frequencies involve large multi-digit numbers, direct multiplication becomes prone to arithmetic calculation errors.
Method 2: The Assumed Mean Method
To prevent cumbersome multi-digit arithmetic, the Assumed Mean Method shifts the origin of the data. We select an arbitrary central class mark A (called the assumed mean) and compute the deviations di = xi - A:
Deviation Transformation:
di = xi - A
Multiplying by fi and summing: Sum(fi * di) = Sum[fi * (xi - A)] = Sum(fi * xi) - A * Sum(fi). Dividing by Sum(fi) = N: Sum(fi * di) / N = [Sum(fi * xi) / N] - A = Mean - A.
Adding A back to both sides yields the formula:
Theorem (The Assumed Mean Formula):
Sum(fi * di)
Mean x̄ = A + --------------
Sum(fi)
where di = xi - A.
Subtracting A dramatically shrinks the numbers, converting large three-digit products into small, easily manageable arithmetic operations.
Method 3: The Step-Deviation Method
When all class intervals have an identical class size h, the deviations di = xi - A are all exact multiples of h. We can scale down the data by dividing each deviation by h:
Scale Transformation:
ui = (xi - A) / h = di / h
Theorem (The Step-Deviation Formula):
Sum(fi * ui)
Mean x̄ = A + [ -------------- ] * h
Sum(fi)
where ui = (xi - A) / h, and h is the class size.
The step-deviation method reduces the computation to multiplying single-digit integers (-3, -2, -1, 0, 1, 2, 3), making it the most reliable, error-free method for large grouped datasets.
Comparison of Methods:
+-------------------+--------------------+------------------------------------+
| Method | Variable Used | Master Formula |
+-------------------+--------------------+------------------------------------+
| Direct | xi (Class mark) | x̄ = Sum(fi*xi) / Sum(fi) |
+-------------------+--------------------+------------------------------------+
| Assumed Mean | di = xi - A | x̄ = A + Sum(fi*di) / Sum(fi) |
+-------------------+--------------------+------------------------------------+
| Step-Deviation | ui = (xi - A) / h | x̄ = A + [Sum(fi*ui) / Sum(fi)] * h |
+-------------------+--------------------+------------------------------------+
Detailed Worked Example: Step-Deviation Method
Calculate the mean daily wage of 50 workers from the following distribution:
Class Interval (Daily Wages in $): 100-120 | 120-140 | 140-160 | 160-180 | 180-200
Number of Workers (Frequency fi): 12 | 14 | 8 | 6 | 10
Step 1: Class size h = 120 - 100 = 20.
Step 2: Calculate class marks xi:
100-120: xi = 110
120-140: xi = 130
140-160: xi = 150 <--- Select Assumed Mean A = 150
160-180: xi = 170
180-200: xi = 190
Step 3: Calculate ui = (xi - 150) / 20:
(110 - 150) / 20 = -2
(130 - 150) / 20 = -1
(150 - 150) / 20 = 0
(170 - 150) / 20 = 1
(190 - 150) / 20 = 2
Step 4: Compute products fi * ui:
12 * (-2) = -24
14 * (-1) = -14
8 * ( 0) = 0
6 * ( 1) = 6
10 * ( 2) = 20
Step 5: Sum columns:
Sum(fi) = 12 + 14 + 8 + 6 + 10 = 50
Sum(fi * ui) = -24 - 14 + 0 + 6 + 20 = -38 + 26 = -12
Step 6: Apply Step-Deviation Formula:
x̄ = A + [Sum(fi*ui) / Sum(fi)] * h
x̄ = 150 + [-12 / 50] * 20
x̄ = 150 - 240 / 50 = 150 - 4.8 = 145.20 dollars.
The mean daily wage of the workers is $145.20.
Chapter Exercises
Exercise 1: The following table shows the marks obtained by 30 students in a mathematics examination: Marks 10-25 (2 students), 25-40 (3 students), 40-55 (7 students), 55-70 (6 students), 70-85 (6 students), and 85-100 (6 students). Find the mean marks using the Direct Method.
Exercise 2: For the data in Exercise 1, calculate the mean marks using the Assumed Mean Method by choosing A = 62.5, and verify that both methods yield identical results.
Exercise 3: The mean of the following frequency distribution is 18. Find the missing frequency f: Class 11-13 (frequency 3), 13-15 (frequency 6), 15-17 (frequency 9), 17-19 (frequency 13), 19-21 (frequency f), 21-23 (frequency 5), and 23-25 (frequency 4).
Solutions and Explanations
Solution 1: We calculate class marks xi and products fi * xi: For 10-25: xi = 17.5, fi = 2, fixi = 35. For 25-40: xi = 32.5, fi = 3, fixi = 97.5. For 40-55: xi = 47.5, fi = 7, fixi = 332.5. For 55-70: xi = 62.5, fi = 6, fixi = 375. For 70-85: xi = 77.5, fi = 6, fixi = 465. For 85-100: xi = 92.5, fi = 6, fixi = 555. Sum of frequencies: Sum(fi) = 2 + 3 + 7 + 6 + 6 + 6 = 30. Sum of products: Sum(fixi) = 35 + 97.5 + 332.5 + 375 + 465 + 555 = 1860. Mean x̄ = Sum(fixi) / Sum(fi) = 1860 / 30 = 62. The mean mark obtained by the students is 62.
Solution 2: Using the Assumed Mean A = 62.5. Calculate deviations di = xi - 62.5: For 17.5: di = -45, fidi = 2 * (-45) = -90. For 32.5: di = -30, fidi = 3 * (-30) = -90. For 47.5: di = -15, fidi = 7 * (-15) = -105. For 62.5: di = 0, fidi = 6 * 0 = 0. For 77.5: di = 15, fidi = 6 * 15 = 90. For 92.5: di = 30, fidi = 6 * 30 = 180. Sum of products: Sum(fidi) = -90 - 90 - 105 + 0 + 90 + 180 = -15. Total frequency Sum(fi) = 30. Mean x̄ = A + Sum(fidi) / Sum(fi) = 62.5 + (-15 / 30) = 62.5 - 0.5 = 62. Both methods yield an identical mean of 62, confirming the mathematical equivalence of the methods.
Solution 3: Class marks xi are: 12, 14, 16, 18, 20, 22, 24. Frequencies fi are: 3, 6, 9, 13, f, 5, 4. Compute products fi * xi: 312 = 36; 614 = 84; 916 = 144; 1318 = 234; f20 = 20f; 522 = 110; 424 = 96. Total frequency Sum(fi) = 3 + 6 + 9 + 13 + f + 5 + 4 = 40 + f. Total sum of products: Sum(fixi) = 36 + 84 + 144 + 234 + 110 + 96 + 20f = 704 + 20f. Given that the mean x̄ = 18: x̄ = Sum(fixi) / Sum(fi) ==> 18 = (704 + 20f) / (40 + f). Cross-multiply: 18 * (40 + f) = 704 + 20f ==> 720 + 18f = 704 + 20f ==> 720 - 704 = 20f - 18f ==> 16 = 2*f ==> f = 8. Thus, the missing frequency is 8.