I'm sorry, but you didn't provide a specific question. However, I can guide you on how to solve the problem based on the information you provided.

First, let's complete the table:

For xP(x), multiply each x value by its corresponding P(x) value:

x = 0, P(x) = 0.12, so xP(x) = 0*0.12 = 0
x = 1, P(x) = 0.25, so xP(x) = 1*0.25 = 0.25
x = 2, P(x) = 0.18, so xP(x) = 2*0.18 = 0.36
x = 3, P(x) = 0.35, so xP(x) = 3*0.35 = 1.05
x = 4, P(x) = 0.10, so xP(x) = 4*0.10 = 0.40

Next, we calculate the expected value (mean), 𝜇 = 𝐸(𝑋) = ∑[𝑥𝑃(𝑥)]. Add up all the values of xP(x) we just calculated:

𝜇 = 0 + 0.25 + 0.36 + 1.05 + 0.40 = 2.06

For the variance, we need to calculate (𝑥 − 𝜇)2𝑃(𝑥) for each x, then sum them up:

For x = 0, (𝑥 − 𝜇)2𝑃(𝑥) = (0 - 2.06)^2 * 0.12 = 0.51
For x = 1, (𝑥 − 𝜇)2𝑃(𝑥) = (1 - 2.06)^2 * 0.25 = 0.28
For x = 2, (𝑥 − 𝜇)2𝑃(𝑥) = (2 - 2.06)^2 * 0.18 = 0.002
For x = 3, (𝑥 − 𝜇)2𝑃(𝑥) = (3 - 2.06)^2 * 0.35 = 0.29
For x = 4, (𝑥 − 𝜇)2𝑃(𝑥) = (4 - 2.06)^2 * 0.10 = 0.37

Add these up to get the variance:

Var(X) = 0.51 + 0.28 + 0.002 + 0.29 + 0.37 = 1.46

Finally, the standard deviation is the square root of the variance:

Standard deviation = sqrt(1.46) = 1.21 (rounded to two decimal places)

Please note that these calculations are based on the assumption that the provided probabilities sum up to 1, which is a requirement for a probability distribution.

Question

I'm sorry, but you didn't provide a specific question. However, I can guide you on how to solve the problem based on the information you provided.

First, let's complete the table:

For xP(x), multiply each x value by its corresponding P(x) value:

x = 0, P(x) = 0.12, so xP(x) = 0*0.12 = 0
x = 1, P(x) = 0.25, so xP(x) = 1*0.25 = 0.25
x = 2, P(x) = 0.18, so xP(x) = 2*0.18 = 0.36
x = 3, P(x) = 0.35, so xP(x) = 3*0.35 = 1.05
x = 4, P(x) = 0.10, so xP(x) = 4*0.10 = 0.40

Next, we calculate the expected value (mean), 𝜇 = 𝐸(𝑋) = ∑[𝑥𝑃(𝑥)]. Add up all the values of xP(x) we just calculated:

𝜇 = 0 + 0.25 + 0.36 + 1.05 + 0.40 = 2.06

For the variance, we need to calculate (𝑥 − 𝜇)2𝑃(𝑥) for each x, then sum them up:

For x = 0, (𝑥 − 𝜇)2𝑃(𝑥) = (0 - 2.06)^2 * 0.12 = 0.51
For x = 1, (𝑥 − 𝜇)2𝑃(𝑥) = (1 - 2.06)^2 * 0.25 = 0.28
For x = 2, (𝑥 − 𝜇)2𝑃(𝑥) = (2 - 2.06)^2 * 0.18 = 0.002
For x = 3, (𝑥 − 𝜇)2𝑃(𝑥) = (3 - 2.06)^2 * 0.35 = 0.29
For x = 4, (𝑥 − 𝜇)2𝑃(𝑥) = (4 - 2.06)^2 * 0.10 = 0.37

Add these up to get the variance:

Var(X) = 0.51 + 0.28 + 0.002 + 0.29 + 0.37 = 1.46

Finally, the standard deviation is the square root of the variance:

Standard deviation = sqrt(1.46) = 1.21 (rounded to two decimal places)

Please note that these calculations are based on the assumption that the provided probabilities sum up to 1, which is a requirement for a probability distribution.

Knowee AI · Accepted Answer