The document shows the steps to calculate the variance and standard deviation of a probability distribution. It involves creating columns for the random variable x, the probability P(x), the products x*P(x) and x^2*P(x). The mean is calculated as the sum of x*P(x). The variance is calculated as the sum of x^2*P(x) - the mean squared.

1. 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

2. 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
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

3. 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
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2
First find the mean.

4. 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
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2
First find the mean.
𝜇 = 𝑥 ∙ 𝑃(𝑥)

5. 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
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

6. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2
1 0.3
2 0.2
3 0.2
4 0.1
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

7. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3
2 0.2
3 0.2
4 0.1
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

8. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2
3 0.2
4 0.1
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

9. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2
4 0.1
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

10. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

11. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Create a column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

12. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Sum the column of x∙P(x)
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

13. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥)
Sum the column of x∙P(x)
Σ[x∙P(x)]=1.7
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

14. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2 = 𝑥2 ∙ 𝑃(𝑥) − 𝜇2

15. Example 7: Find the variance and standard
deviation of the probability distribution.
X P(x) x∙P(x)
0 0.2 0(0.2) = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

16. 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
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

17. 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 = 0
1 0.3 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

18. 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 = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

19. 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 = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

20. 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 = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

21. 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 = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2

22. 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 = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

23. 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
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

24. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

25. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

26. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

27. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 0.1 4(0.1) = 0.4 42 = 16
𝜇 = 𝑥 ∙ 𝑃(𝑥) = 1.7
Σ[x∙P(x)]=1.7
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

28. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Create a column of x2∙P(x)

29. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Sum the column of x2∙P(x)

30. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
Sum the column of x2∙P(x)
Σ[x2∙P(x)]=4.5

31. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
= 4.5 – 1.72
= 1.61
Σ[x2∙P(x)]=4.5
Variance

32. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
= 4.5 – 1.72
= 1.61
Σ[x2∙P(x)]=4.5
Variance
𝜎 = 𝜎2

33. 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) = 0
1 0.3 1(0.3) = 0.3 12 = 1 1(0.3) = 0.3
2 0.2 2(0.2) = 0.4 22 = 4 4(0.2) = 0.8
3 0.2 3(0.2) = 0.6 32 = 9 9(0.2) = 1.8
4 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
𝜎2
= 𝑥2
∙ 𝑃(𝑥) − 𝜇2
= 4.5 – 1.72
= 1.61
Σ[x2∙P(x)]=4.5
Variance
𝜎 = 𝜎2 = 1.61 ≈ 1.27
Standard Deviation