Find the Variance 4 , 6 , 8 , 12 , 15 , 17 , 10 , 24

$4$ , $6$ , $8$ , $12$ , $15$ , $17$ , $10$ , $24$

Step 1

The mean of a set of numbers is the sum divided by the number of terms.

$\overline{x} = \frac{4 + 6 + 8 + 12 + 15 + 17 + 10 + 24}{8}$

Step 2

Simplify the numerator.

Add $4$ and $6$ .

$\overline{x} = \frac{10 + 8 + 12 + 15 + 17 + 10 + 24}{8}$

Add $10$ and $8$ .

$\overline{x} = \frac{18 + 12 + 15 + 17 + 10 + 24}{8}$

Add $18$ and $12$ .

$\overline{x} = \frac{30 + 15 + 17 + 10 + 24}{8}$

Add $30$ and $15$ .

$\overline{x} = \frac{45 + 17 + 10 + 24}{8}$

Add $45$ and $17$ .

$\overline{x} = \frac{62 + 10 + 24}{8}$

Add $62$ and $10$ .

$\overline{x} = \frac{72 + 24}{8}$

Add $72$ and $24$ .

$\overline{x} = \frac{96}{8}$

Step 3

Divide $96$ by $8$ .

$\overline{x} = 12$

Step 4

Set up the formula for variance. The variance of a set of values is a measure of the spread of its values.

$s^{2} = \sum_{i = 1}^{n} \frac{{(x_{i} - x_{a v g})}^{2}}{n - 1}$

Step 5

Set up the formula for variance for this set of numbers.

$s = \frac{{(4 - 12)}^{2} + {(6 - 12)}^{2} + {(8 - 12)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Step 6

Simplify the result.

Simplify the numerator.

Subtract $12$ from $4$ .

$s = \frac{{(- 8)}^{2} + {(6 - 12)}^{2} + {(8 - 12)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $- 8$ to the power of $2$ .

$s = \frac{64 + {(6 - 12)}^{2} + {(8 - 12)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $6$ .

$s = \frac{64 + {(- 6)}^{2} + {(8 - 12)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $- 6$ to the power of $2$ .

$s = \frac{64 + 36 + {(8 - 12)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $8$ .

$s = \frac{64 + 36 + {(- 4)}^{2} + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $- 4$ to the power of $2$ .

$s = \frac{64 + 36 + 16 + {(12 - 12)}^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $12$ .

$s = \frac{64 + 36 + 16 + 0^{2} + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raising $0$ to any positive power yields $0$ .

$s = \frac{64 + 36 + 16 + 0 + {(15 - 12)}^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $15$ .

$s = \frac{64 + 36 + 16 + 0 + 3^{2} + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $3$ to the power of $2$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + {(17 - 12)}^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $17$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 5^{2} + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $5$ to the power of $2$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 25 + {(10 - 12)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $10$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 25 + {(- 2)}^{2} + {(24 - 12)}^{2}}{8 - 1}$

Raise $- 2$ to the power of $2$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 25 + 4 + {(24 - 12)}^{2}}{8 - 1}$

Subtract $12$ from $24$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 25 + 4 + 12^{2}}{8 - 1}$

Raise $12$ to the power of $2$ .

$s = \frac{64 + 36 + 16 + 0 + 9 + 25 + 4 + 144}{8 - 1}$

Add $64$ and $36$ .

$s = \frac{100 + 16 + 0 + 9 + 25 + 4 + 144}{8 - 1}$

Add $100$ and $16$ .

$s = \frac{116 + 0 + 9 + 25 + 4 + 144}{8 - 1}$

Add $116$ and $0$ .

$s = \frac{116 + 9 + 25 + 4 + 144}{8 - 1}$

Add $116$ and $9$ .

$s = \frac{125 + 25 + 4 + 144}{8 - 1}$

Add $125$ and $25$ .

$s = \frac{150 + 4 + 144}{8 - 1}$

Add $150$ and $4$ .

$s = \frac{154 + 144}{8 - 1}$

Add $154$ and $144$ .

$s = \frac{298}{8 - 1}$

Subtract $1$ from $8$ .

$s = \frac{298}{7}$

Step 7

Approximate the result.

$s^{2} \approx 42.5714$

Find the Variance 4 , 6 , 8 , 12 , 15 , 17 , 10 , 24

Related Questions

Support LCMGCF.com by disabling your adblocker.