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# If xi ’s are the mid points of the class intervals of grouped data, f_{i} ’s are the corresponding frequencies and x̄ is the mean, then (f_{i}x_{i} - x̄) is equal to

a. 0

b. -1

c. 1

d. 2

**Solution:**

We know that

Mean (x) = Sum of all the observations/ Number of observations

Here

x = (f_{1}x_{1} + f_{2}x_{2} + …..+ f_{n}x_{n}) / f_{1} + f_{2} +……+ f_{n}

x = Σf_{i}x_{i} / Σf_{i}, Σf_{i} = n

x = Σf_{i}x_{i} / n

n x = Σf_{i}x_{i} --- (1)

So we get

Σ (f_{i}x_{i} - x) = (f_{1}x_{1} - x) + (f_{2}x_{2} - x)+ …..+ (f_{n}x_{n} - x)

Σ (f_{i}x_{i} - x) = (f_{1}x_{1} + f_{2}x_{2} + …..+ f_{n}x_{n}) - (x + x +….n times)

We can write it as

Σ (f_{i}x_{i} - x) = Σf_{i}x_{i} - nx

Σ(f_{i}x_{i} - x) = nx - nx (From equation 1)

Σ(f_{i}x_{i} - x) = 0

Therefore, \((f_{i}x_{i}-\overline{x})\) is equal to 0.

**✦ Try This: **Calculate mean \(\overline{x}\) when Σf_{i}x_{i} = 100 and Σf_{i} = 20.

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 14

**NCERT Exemplar Class 10 Maths Exercise 13.1**** Problem 3**

## If x_{i} ’s are the mid points of the class intervals of grouped data, f_{i} ’s are the corresponding frequencies and x̄ is the mean, then (f_{i}x_{i} - x̄) is equal to a. 0, b. -1, c. 1, d. 2

**Summary:**

If x_{i} ’s are the mid points of the class intervals of grouped data, f_{i} ’s are the corresponding frequencies and x̄ is the mean, then (f_{i}x_{i} - x̄) is equal to 0

**☛ Related Questions:**

- In the formula x̄ = a + h fi ui/ fi, for finding the mean of grouped frequency distribution, ui = a. . . . .
- The abscissa of the point of intersection of the less than type and of the more than type cumulative . . . .
- For the following distribution: Class 0-5 5-10 10-15 15-20 20-25 Frequency 10 15 12 20 9 the sum of . . . .

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