standard deviation of 2, 4, 4, 4, 5, 5, 7, 9

The mean of the 8 values is their sum divided by 8:

\frac{2 + 4 + 4 + 4 + 5 + 5 + 7 + 9}{8} = 5

The variance is the mean of the squared deviations from the mean

5

: their sum is divided by the number of values, 8 (by 7 for a sample, asked for with "sample standard deviation"), and the standard deviation is its square root

Let's calculate:

\sqrt{\frac{{(2 - 5)}^{2} + {(4 - 5)}^{2} + {(4 - 5)}^{2} + {(4 - 5)}^{2} + {(5 - 5)}^{2} + {(5 - 5)}^{2} + {(7 - 5)}^{2} + {(9 - 5)}^{2}}{8}}

ⓘ Start by calculating what is inside the parentheses:

2 - 5

=

-3

,

4 - 5

=

-1

,

5 - 5

=

0

,

7 - 5

=

2

,

9 - 5

=

4

\sqrt{\frac{{(-3)}^{2} + {(-1)}^{2} + {(-1)}^{2} + {(-1)}^{2} + 0^{2} + 0^{2} + 2^{2} + 4^{2}}{8}}

ⓘ Calculate powers before multiplications, additions and subtractions. A power with a positive integer exponent is a repeated product:

{(-3)}^{2}

=

(-3) \times (-3)

=

9

,

{(-1)}^{2}

=

(-1) \times (-1)

=

1

,

2^{2}

=

2 \times 2

=

4

,

4^{2}

=

4 \times 4

=

16

■Adding 0 changes nothing:

0 + 0 + 2^{2}

=

2^{2}

\sqrt{\frac{9 + 1 + 1 + 1 + 4 + 16}{8}}

ⓘ The numerator and the denominator are computed before the division:

9 + 1

=

10

\sqrt{\frac{10 + 1 + 1 + 4 + 16}{8}}

ⓘ The numerator and the denominator are computed before the division:

10 + 1

=

11

\sqrt{\frac{11 + 1 + 4 + 16}{8}}

ⓘ The numerator and the denominator are computed before the division:

11 + 1

=

12

\sqrt{\frac{12 + 4 + 16}{8}}

ⓘ The numerator and the denominator are computed before the division:

12 + 4

=

16

\sqrt{\frac{16 + 16}{8}}

ⓘ The numerator and the denominator are computed before the division:

16 + 16

=

32

\sqrt{\frac{32}{8}}

ⓘ Simplify the fraction by dividing the numerator and the denominator by

8

, their greatest common divisor:

\frac{32}{8}

=

4

\sqrt{4}

ⓘ The square root of a nonnegative real is defined as the nonnegative number whose square gives that real number:

\sqrt{4}

=

2

since

2^{2}

=

4

▶Standard deviation:

2

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