Question

for a dataset with class intervals 1-10, 10-20, and 20-30 and frequencies 2, 8, and 10, with the mean of 18, the variance is:

• A. 43.725
• B. 120
• C. 110
• D. 100.01

Answer 1

Answer: (A)

43.725

Answer 2

The problem requires us to calculate the variance for a grouped dataset, given class intervals, frequencies, and the mean.

1. Determine the Midpoints ($x_i$) for each Class Interval:

The midpoint of a class interval is calculated as (Lower Limit + Upper Limit) / 2.

• For the interval 1-10: $x_1 = (1 + 10) / 2 = 5.5$
• For the interval 10-20: $x_2 = (10 + 20) / 2 = 15$
• For the interval 20-30: $x_3 = (20 + 30) / 2 = 25$

2. List the Frequencies ($f_i$):

• $f_1 = 2$
• $f_2 = 8$
• $f_3 = 10$

3. Note the Given Mean ($\bar{x}$):

• $\bar{x} = 18$

4. Calculate the Total Frequency ($\sum f_i$):

$\sum f_i = 2 + 8 + 10 = 20$

5. Calculate the Deviation from the Mean ($x_i - \bar{x}$) for each Midpoint:

• For $x_1 = 5.5$: $5.5 - 18 = -12.5$
• For $x_2 = 15$: $15 - 18 = -3$
• For $x_3 = 25$: $25 - 18 = 7$

6. Calculate the Squared Deviation ($(x_i - \bar{x})^2$) for each Midpoint:

• For $x_1 = 5.5$: $(-12.5)^2 = 156.25$
• For $x_2 = 15$: $(-3)^2 = 9$
• For $x_3 = 25$: $(7)^2 = 49$

7. Calculate the Product of Frequency and Squared Deviation ($f_i (x_i - \bar{x})^2$) for each Midpoint:

• For $x_1 = 5.5, f_1 = 2$: $2 \times 156.25 = 312.5$
• For $x_2 = 15, f_2 = 8$: $8 \times 9 = 72$
• For $x_3 = 25, f_3 = 10$: $10 \times 49 = 490$

8. Calculate the Sum of $f_i (x_i - \bar{x})^2$:

$\sum f_i (x_i - \bar{x})^2 = 312.5 + 72 + 490 = 874.5$

9. Calculate the Variance ($\sigma^2$):

The formula for the variance of grouped data is:

$$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$$

Substitute the calculated values:

$$\sigma^2 = \frac{874.5}{20} = 43.725$$

The variance of the dataset is 43.725.