Example 7: Find the variance and standard deviation of the probability distribution.X P(x)0 0.21 0.32 0.23 0.24 0.1

Example 7: Find the variance and standard deviation of the probability distribution.X P(x)0 0.21 0.32 0.23 0.24 0.1

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.X P(x)0 0.21 0.32 0.23 0.24 0.1

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

First find the mean.

Example 7: Find the variance and standard deviation of the probability distribution.X P(x)0 0.21 0.32 0.23 0.24 0.1

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

First find the mean.

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]

Example 7: Find the variance and standard deviation of the probability distribution.X P(x)0 0.21 0.32 0.23 0.24 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.21 0.32 0.23 0.24 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.32 0.23 0.24 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.23 0.24 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.24 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]Create a column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]

Sum the column of x∙P(x)

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]

Sum the column of x∙P(x)

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.32 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.64 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2

0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 12 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 43 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 94 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Create a column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Sum the column of x2∙P(x)

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 )]−𝜇2

Sum the column of x2∙P(x)Σ[x2∙P(x)]=4.5

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

4.5 – 1.72

1.61

Σ[x2∙P(x)]=4.5

Variance

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

4.5 – 1.72

1.61

Σ[x2∙P(x)]=4.5

Variance

𝜎=√𝜎2

Example 7: Find the variance and standard deviation of the probability distribution.

X P(x) x∙P(x) x2 x2∙P(x)0 0.2 0(0.2) = 0 02 = 0 0(0.2) = 01 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.32 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.83 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.84 0.1 4(0.1) = 0.4 42 = 16 16(0.1) = 1.6

𝜇=∑ [𝑥 ∙𝑃 (𝑥 )]=1.7

Σ[x∙P(x)]=1.7

4.5 – 1.72

1.61

Σ[x2∙P(x)]=4.5

Variance

𝜎=√𝜎2=√1.61≈1.27

Standard Deviation