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If the mean of the following data is 61, then the value of X is:
| xi | 55 | 60 | 70 | 95 |
|---|---|---|---|---|
| fi | 48 | X | 31 | 23 |
50
48
42
55
48
Since the mean is 61, use deviation method: 48(55-61) + X(60-61) + 31(70-61) + 23(95-61) = 0. Solving: 48(-6) - X + 31(9) + 23(34) = 0, which simplifies to -288 - X + 279 + 782 = 0.
Mean = 61, data values xi = {55, 60, 70, 95}, frequencies fi = {48, X, 31, 23}
x=∑fi∑fixi
Since the mean is 61, use deviation method: 48(55-61) + X(60-61) + 31(70-61) + 23(95-61) = 0. Solving: 48(-6) - X + 31(9) + 23(34) = 0, which simplifies to -288 - X + 279 + 782 = 0.
Many students forget to include the unknown frequency X in the denominator sum (Σfi = 48 + X + 31 + 23).
Calculate the sum of frequencies
The sum of all frequencies is ∑fi=48+X+31+23=102+X.
∑fi=102+X
Calculate the sum of products
The sum of products of xi and fi is ∑fixi=(55×48)+(60×X)+(70×31)+(95×23). Calculating these: 2640+60X+2170+2185=6995+60X.
∑fixi=6995+60X
Apply mean formula
Using x=61, we have 61=102+X6995+60X. Multiply both sides by (102+X) to get 61(102+X)=6995+60X.
6222+61X=6995+60X
Solve for X
Isolate X: 61X−60X=6995−6222, which gives X=773.
X=773
B is correct because the algebraic solution of the mean formula yields X = 773.
This type of problem frequently appears in Data Interpretation sections to test algebraic manipulation of statistical averages.