Find the Variance {66,46,53,50,52,47,45,49}

${66, 46, 53, 50, 52, 47, 45, 49}$

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

$\overline{x} = \frac{66 + 46 + 53 + 50 + 52 + 47 + 45 + 49}{8}$

Simplify the numerator.

Add $66$ and $46$ .

$\overline{x} = \frac{112 + 53 + 50 + 52 + 47 + 45 + 49}{8}$

Add $112$ and $53$ .

$\overline{x} = \frac{165 + 50 + 52 + 47 + 45 + 49}{8}$

Add $165$ and $50$ .

$\overline{x} = \frac{215 + 52 + 47 + 45 + 49}{8}$

Add $215$ and $52$ .

$\overline{x} = \frac{267 + 47 + 45 + 49}{8}$

Add $267$ and $47$ .

$\overline{x} = \frac{314 + 45 + 49}{8}$

Add $314$ and $45$ .

$\overline{x} = \frac{359 + 49}{8}$

Add $359$ and $49$ .

$\overline{x} = \frac{408}{8}$

Divide $408$ by $8$ .

$\overline{x} = 51$

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

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

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

$s = \frac{{(66 - 51)}^{2} + {(46 - 51)}^{2} + {(53 - 51)}^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Simplify the result.

Simplify the numerator.

Subtract $51$ from $66$ .

$s = \frac{15^{2} + {(46 - 51)}^{2} + {(53 - 51)}^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Raise $15$ to the power of $2$ .

$s = \frac{225 + {(46 - 51)}^{2} + {(53 - 51)}^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $46$ .

$s = \frac{225 + {(- 5)}^{2} + {(53 - 51)}^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

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

$s = \frac{225 + 25 + {(53 - 51)}^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $53$ .

$s = \frac{225 + 25 + 2^{2} + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

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

$s = \frac{225 + 25 + 4 + {(50 - 51)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $50$ .

$s = \frac{225 + 25 + 4 + {(- 1)}^{2} + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Raise $- 1$ to the power of $2$ .

$s = \frac{225 + 25 + 4 + 1 + {(52 - 51)}^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $52$ .

$s = \frac{225 + 25 + 4 + 1 + 1^{2} + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

One to any power is one.

$s = \frac{225 + 25 + 4 + 1 + 1 + {(47 - 51)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $47$ .

$s = \frac{225 + 25 + 4 + 1 + 1 + {(- 4)}^{2} + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

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

$s = \frac{225 + 25 + 4 + 1 + 1 + 16 + {(45 - 51)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $45$ .

$s = \frac{225 + 25 + 4 + 1 + 1 + 16 + {(- 6)}^{2} + {(49 - 51)}^{2}}{8 - 1}$

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

$s = \frac{225 + 25 + 4 + 1 + 1 + 16 + 36 + {(49 - 51)}^{2}}{8 - 1}$

Subtract $51$ from $49$ .

$s = \frac{225 + 25 + 4 + 1 + 1 + 16 + 36 + {(- 2)}^{2}}{8 - 1}$

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

$s = \frac{225 + 25 + 4 + 1 + 1 + 16 + 36 + 4}{8 - 1}$

Add $225$ and $25$ .

$s = \frac{250 + 4 + 1 + 1 + 16 + 36 + 4}{8 - 1}$

Add $250$ and $4$ .

$s = \frac{254 + 1 + 1 + 16 + 36 + 4}{8 - 1}$

Add $254$ and $1$ .

$s = \frac{255 + 1 + 16 + 36 + 4}{8 - 1}$

Add $255$ and $1$ .

$s = \frac{256 + 16 + 36 + 4}{8 - 1}$

Add $256$ and $16$ .

$s = \frac{272 + 36 + 4}{8 - 1}$

Add $272$ and $36$ .

$s = \frac{308 + 4}{8 - 1}$

Add $308$ and $4$ .

$s = \frac{312}{8 - 1}$

Subtract $1$ from $8$ .

$s = \frac{312}{7}$

Approximate the result.

$s^{2} \approx 44.5714$

Find the Variance {66,46,53,50,52,47,45,49}

Related Questions

Support LCMGCF.com by disabling your adblocker.