(I give 100 points)
Determine the variance and standard deviation of the given discrete probability distribution.
v P(v)
0 0.15
1 0.25
2 0.36
3 0.18
4 0.04
5 0.02
Determine the variance and standard deviation of the given discrete probability distribution.
v P(v)
0 0.15
1 0.25
2 0.36
3 0.18
4 0.04
5 0.02
Answers & Comments
Verified answer
Answer:
To find the variance and standard deviation, we need to first calculate the expected value of the distribution:
E(X) = Σ [P(v) x v] for all values of v
E(X) = (0 x 0.15) + (1 x 0.25) + (2 x 0.36) + (3 x 0.18) + (4 x 0.04) + (5 x 0.02)
E(X) = 1.78
The variance can be calculated as follows:
Var(X) = Σ [P(v) x (v - E(X))^2] for all values of v
Var(X) = (0.15 x (0 - 1.78)^2) + (0.25 x (1 - 1.78)^2) + (0.36 x (2 - 1.78)^2) + (0.18 x (3 - 1.78)^2) + (0.04 x (4 - 1.78)^2) + (0.02 x (5 - 1.78)^2)
Var(X) = 1.26
Therefore, the variance of the distribution is 1.26.
The standard deviation can be found by taking the square root of the variance:
SD(X) = √Var(X)
SD(X) = √1.26
SD(X) ≈ 1.12
Therefore, the standard deviation of the distribution is approximately 1.12.
Expected value or mean (μ) = Σ [P(v) × v]
= (0.15 × 0) + (0.25 × 1) + (0.36 × 2) + (0.18 × 3) + (0.04 × 4) + (0.02 × 5)
= 2.07
Now, we can use the formula below to find the variance:
Variance (σ^2) = Σ [P(v) × (v - μ)^2]
σ^2 = (0.15 × (0 - 2.07)^2) + (0.25 × (1 - 2.07)^2) + (0.36 × (2 - 2.07)^2) + (0.18 × (3 - 2.07)^2) + (0.04 × (4 - 2.07)^2) + (0.02 × (5 - 2.07)^2)
σ^2 = 1.4451
Finally, we can find the standard deviation by taking the square root of the variance:
Standard deviation (σ) = sqrt(σ^2)
σ = sqrt(1.4451)
σ ≈ 1.2016
Therefore, the variance is 1.4451 and the standard deviation is approximately 1.2016.