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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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Find the mean marks of the following distribution: Marks(20, 30, 40, 50), No. of students(4, 6, 8, 2).
32
34
36
38
34
Use the deviation method: Assume the mean is 30. Calculate deviations d=x−30: (-10, 0, 10, 20). Weighted sum: 4(−10)+6(0)+8(10)+2(20)=−40+0+80+40=80. Divide by total students (20): 2080=4. Add to assumed mean: 30+4=34.
Marks (x): 20, 30, 40, 50; Number of students (f): 4, 6, 8, 2.
Mean(x)=∑fi∑fixi
Use the deviation method: Assume the mean is 30. Calculate deviations d=x−30: (-10, 0, 10, 20). Weighted sum: 4(−10)+6(0)+8(10)+2(20)=−40+0+80+40=80. Divide by total students (20): 2080=4. Add to assumed mean: 30+4=34.
Students often add the marks values directly (20+30+40+50=140) and divide by the number of categories (4), ignoring the frequency weightings.
Calculate sum of frequencies
Sum the total number of students by adding the frequencies.
∑fi=4+6+8+2=20
Calculate sum of product of frequency and marks
Multiply each mark xi by its corresponding frequency fi and sum them up.
∑fixi=(20×4)+(30×6)+(40×8)+(50×2)=80+180+320+100=680
Calculate mean
Apply the mean formula using the calculated sums.
x=20680=34
B is correct because the weighted mean of the distribution calculated using the formula ∑fi∑fixi is 34.
This concept of weighted averages is identical to the method used to calculate the Combined Mean in statistics or the Center of Mass in Physics.