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 0.2 1 0.3 2 0.2 3 0.2 4 0.1
Transcript of Variance and standard deviation of a discrete random variable
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
π 2=β [π₯2 βπ (π₯ )]βπ2
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
π 2=β [π₯2 βπ (π₯ )]βπ2
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
π 2=β [π₯2 βπ (π₯ )]βπ2
First find the mean.
π=β [π₯ βπ (π₯ )]
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
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x)0 0.21 0.32 0.23 0.24 0.1
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x)0 0.2 0(0.2) = 01 0.32 0.23 0.24 0.1
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x)0 0.2 0(0.2) = 01 0.3 1(0.3) = 0.32 0.23 0.24 0.1
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]Create a column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]
Sum the column of xβP(x)
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]
Sum the column of xβP(x)
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Create a column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Sum the column of x2βP(x)
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
π 2=β [π₯2 βπ (π₯ )]βπ2
Sum the column of x2βP(x)Ξ£[x2βP(x)]=4.5
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
4.5 β 1.72
1.61
Ξ£[x2βP(x)]=4.5
Variance
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
4.5 β 1.72
1.61
Ξ£[x2βP(x)]=4.5
Variance
π=βπ2
Example 7: Find the variance and standard deviation of the probability distribution.
X P(x) xβP(x) x2 x2βP(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
π=β [π₯ βπ (π₯ )]=1.7
Ξ£[xβP(x)]=1.7
4.5 β 1.72
1.61
Ξ£[x2βP(x)]=4.5
Variance
π=βπ2=β1.61β1.27
Standard Deviation
