find the mean and standard deviation in the following probability distribution

x 1 2 3
p(x) 0.2 0.6 0.2

*i think that the mean is 2 but am lost after that when it comes to the standard deviation. please show work and calculations. the more i work with numbers the more confused i get. thank you.
mary hornbeck

To find the mean and standard deviation of a probability distribution, follow these steps:

Step 1: Multiply each value of x by its corresponding probability p(x).
x * p(x) = (1 * 0.2) + (2 * 0.6) + (3 * 0.2) = 0.2 + 1.2 + 0.6 = 2

Step 2: The mean is the sum of the above products from Step 1.
Mean (μ) = 2

Step 3: Calculate the variance.
Variance (σ²) = [(x₁ - μ)² * p(x₁)] + [(x₂ - μ)² * p(x₂)] + [(x₃ - μ)² * p(x₃)]
= [(1 - 2)² * 0.2] + [(2 - 2)² * 0.6] + [(3 - 2)² * 0.2]
= (1² * 0.2) + (0² * 0.6) + (1² * 0.2)
= 0.2 + 0 + 0.2
= 0.4

Step 4: Calculate the standard deviation.
Standard Deviation (σ) = √(Variance)
= √(0.4)
= 0.632

Therefore, the mean of the given probability distribution is 2, and the standard deviation is approximately 0.632.