Question

The variance \(\sigma^2\) of the data

Is _______.

 

Answer 1

Correct Answer: 29

Calculate sums:

\[ \sum f_i = 22, \quad \sum f_i x_i = 176, \quad \sum f_i x_i^2 = 2048. \]

Calculate the mean \( \bar{x} \):

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{176}{22} = 8. \]

Calculate the variance \( \sigma^2 \):

\[ \sigma^2 = \frac{1}{N} \sum f_i x_i^2 - \bar{x}^2, \]

where \( N = \sum f_i \).

Plugging in values:

\[ \sigma^2 = \frac{1}{22} \times 2048 - (8)^2 = \frac{2048}{22} - 64. \]

Compute:

\[ \frac{2048}{22} = 93.09090909 \quad \text{and} \quad \sigma^2 = 93.09090909 - 64 = 29.09090909. \]

Thus, the variance is: 29.09090909

Answer 2

To find the variance \(\sigma^2\) of the given data, we will use the formula for the variance of a discrete frequency distribution. We need to compute the total frequency, the mean of the data, and the sum of the squared data values multiplied by their frequencies.

Concept Used:

The variance (\(\sigma^2\)) of a discrete frequency distribution is given by the formula:

\[ \sigma^2 = \frac{\sum_{i=1}^{n} f_i x_i^2}{\sum_{i=1}^{n} f_i} - \left( \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} \right)^2 \]

This can also be written as:

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

where \(N = \sum f_i\) is the total frequency and \(\bar{x}\) is the mean of the data.

Step-by-Step Solution:

Step 1: Construct a calculation table to find \(N = \sum f_i\), \(\sum f_i x_i\), and \(\sum f_i x_i^2\).

\(x_i\)	\(f_i\)	\(f_i x_i\)	\(x_i^2\)	\(f_i x_i^2\)
0	3	\(3 \times 0 = 0\)	0	\(3 \times 0 = 0\)
1	2	\(2 \times 1 = 2\)	1	\(2 \times 1 = 2\)
5	3	\(3 \times 5 = 15\)	25	\(3 \times 25 = 75\)
6	2	\(2 \times 6 = 12\)	36	\(2 \times 36 = 72\)
10	6	\(6 \times 10 = 60\)	100	\(6 \times 100 = 600\)
12	3	\(3 \times 12 = 36\)	144	\(3 \times 144 = 432\)
17	3	\(3 \times 17 = 51\)	289	\(3 \times 289 = 867\)
Total	\(\sum f_i = 22\)	\(\sum f_i x_i = 176\)		\(\sum f_i x_i^2 = 2048\)

Step 2: Calculate the total frequency \(N\).

\[ N = \sum f_i = 3 + 2 + 3 + 2 + 6 + 3 + 3 = 22 \]

Step 3: Calculate the mean \(\bar{x}\).

From the table, we have \(\sum f_i x_i = 176\).

\[ \bar{x} = \frac{\sum f_i x_i}{N} = \frac{176}{22} = 8 \]

Step 4: Use the value of \(\sum f_i x_i^2\) from the table.

From the table, we have \(\sum f_i x_i^2 = 2048\).

Final Computation & Result:

Step 5: Substitute the calculated values into the variance formula.

\[ \sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 \] \[ \sigma^2 = \frac{2048}{22} - (8)^2 \] \[ \sigma^2 = \frac{1024}{11} - 64 \]

To subtract, we find a common denominator:

\[ \sigma^2 = \frac{1024 - 64 \times 11}{11} = \frac{1024 - 704}{11} \] \[ \sigma^2 = \frac{320}{11} \]

The variance \(\sigma^2\) of the data is \(\frac{320}{11}\) = 29.09.