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# In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula x = a + f_{i}d_{i}/f_{i}

where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer

**Solution:**

We know that

It is not required that assumed mean should be the mid points of the classes

a can be taken as any value which is easy for simplification

Therefore, the statement is not correct.

**✦ Try This: **The probability of getting a bad egg in a lot of 600 is 0.025. The number of bad eggs in the lot is

It is given that

Total number of eggs = 600

Probability of getting a bad egg P (E) = 0.025

Consider x as the number of bad eggs

The formula to find the probability is

P (E) = Number of bad eggs/ Total number of eggs

Substituting the values

0.025 = x/600

By further calculation

25/1000 = x/600

x = 25/1000 × 600

x = 1/40 × 600

x = 15

Therefore, the number of bad eggs in the lot is 15.

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

**NCERT Exemplar Class 10 Maths Exercise 13.2**** Problem 2**

## In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula x = a + f_{i}d_{i}/f_{i} where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer

**Summary:**

The statement “In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula x̄ = a + f_{i}d_{i}/f_{i} where a is the assumed mean. a must be one of the mid-points of the classes” is not correct

**☛ Related Questions:**

- Is it true to say that the mean, mode and median of grouped data will always be different? Justify y . . . .
- Will the median class and modal class of grouped data always be different? Justify your answer
- In a family having three children, there may be no girl, one girl, two girls or three girls. So, the . . . .

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