Algebra Examples

$1$ , $1.12$ , $4$ , $1.28$ , $4.1$ , $5.3$

Step 1

Find the mean.

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Step 1.1

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

$\overline{x} = \frac{1 + 1.12 + 4 + 1.28 + 4.1 + 5.3}{6}$

Step 1.2

Simplify the numerator.

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Step 1.2.1

Add $1$ and $1.12$ .

$\overline{x} = \frac{2.12 + 4 + 1.28 + 4.1 + 5.3}{6}$

Step 1.2.2

Add $2.12$ and $4$ .

$\overline{x} = \frac{6.12 + 1.28 + 4.1 + 5.3}{6}$

Step 1.2.3

Add $6.12$ and $1.28$ .

$\overline{x} = \frac{7.4 + 4.1 + 5.3}{6}$

Step 1.2.4

Add $7.4$ and $4.1$ .

$\overline{x} = \frac{11.5 + 5.3}{6}$

Step 1.2.5

Add $11.5$ and $5.3$ .

$\overline{x} = \frac{16.8}{6}$

Step 1.3

Divide $16.8$ by $6$ .

$\overline{x} = 2.8$

Step 1.4

Divide.

$\overline{x} = 2.8$

Step 2

Simplify each value in the list.

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Step 2.1

Convert $1$ to a decimal value.

$1$

Step 2.2

Convert $1.12$ to a decimal value.

$1.12$

Step 2.3

Convert $4$ to a decimal value.

$4$

Step 2.4

Convert $1.28$ to a decimal value.

$1.28$

Step 2.5

Convert $4.1$ to a decimal value.

$4.1$

Step 2.6

Convert $5.3$ to a decimal value.

$5.3$

Step 2.7

The simplified values are $1, 1.12, 4, 1.28, 4.1, 5.3$ .

$1, 1.12, 4, 1.28, 4.1, 5.3$

Step 3

Set up the formula for sample standard deviation. The standard deviation of a set of values is a measure of the spread of its values.

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

Step 4

Set up the formula for standard deviation for this set of numbers.

$s = \sqrt{\frac{{(1 - 2.8)}^{2} + {(1.12 - 2.8)}^{2} + {(4 - 2.8)}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5

Simplify the result.

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Step 5.1

Subtract $2.8$ from $1$ .

$s = \sqrt{\frac{{(- 1.8)}^{2} + {(1.12 - 2.8)}^{2} + {(4 - 2.8)}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.2

Raise $- 1.8$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + {(1.12 - 2.8)}^{2} + {(4 - 2.8)}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.3

Subtract $2.8$ from $1.12$ .

$s = \sqrt{\frac{3.24 + {(- 1.68)}^{2} + {(4 - 2.8)}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.4

Raise $- 1.68$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + 2.8224 + {(4 - 2.8)}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.5

Subtract $2.8$ from $4$ .

$s = \sqrt{\frac{3.24 + 2.8224 + {1.2}^{2} + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.6

Raise $1.2$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + {(1.28 - 2.8)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.7

Subtract $2.8$ from $1.28$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + {(- 1.52)}^{2} + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.8

Raise $- 1.52$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + 2.3104 + {(4.1 - 2.8)}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.9

Subtract $2.8$ from $4.1$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + 2.3104 + {1.3}^{2} + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.10

Raise $1.3$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + 2.3104 + 1.69 + {(5.3 - 2.8)}^{2}}{6 - 1}}$

Step 5.11

Subtract $2.8$ from $5.3$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + 2.3104 + 1.69 + {2.5}^{2}}{6 - 1}}$

Step 5.12

Raise $2.5$ to the power of $2$ .

$s = \sqrt{\frac{3.24 + 2.8224 + 1.44 + 2.3104 + 1.69 + 6.25}{6 - 1}}$

Step 5.13

Add $3.24$ and $2.8224$ .

$s = \sqrt{\frac{6.0624 + 1.44 + 2.3104 + 1.69 + 6.25}{6 - 1}}$

Step 5.14

Add $6.0624$ and $1.44$ .

$s = \sqrt{\frac{7.5024 + 2.3104 + 1.69 + 6.25}{6 - 1}}$

Step 5.15

Add $7.5024$ and $2.3104$ .

$s = \sqrt{\frac{9.8128 + 1.69 + 6.25}{6 - 1}}$

Step 5.16

Add $9.8128$ and $1.69$ .

$s = \sqrt{\frac{11.5028 + 6.25}{6 - 1}}$

Step 5.17

Add $11.5028$ and $6.25$ .

$s = \sqrt{\frac{17.7528}{6 - 1}}$

Step 5.18

Subtract $1$ from $6$ .

$s = \sqrt{\frac{17.7528}{5}}$

Step 5.19

Divide $17.7528$ by $5$ .

$s = \sqrt{3.55056}$

Step 6

The standard deviation should be rounded to one more decimal place than the original data. If the original data were mixed, round to one decimal place more than the least precise.

$1.9$