variance and standard deviation of a discrete random variable
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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