Can someone help me work this out.

Find the a) mean, b)variance and c) standard deviation for the probability distribution below

K Pr(X=k)
0 0.4
1 0.2
2 0.3
3 0.1

For mean I came up with 1.1
0(0.4)+1(0.2)+2(0.3)+3(0.1)

For variance I came up with 1.09
0.4(0-1.1)sq+0.2(1-1.1)sq+0.3(2-1.1)sq+0.1(3-1.1)sq
sq means squared

and for standard deviation I came up with 1.0440.
square root of 1.09

I did all of the steps but I am unsure.
Can someone show me theirs?

Thanks in advance.

What you did seems correct to me.

To find the mean, variance, and standard deviation for the given probability distribution, follow these steps:

a) Calculate the mean:
The mean, also called the expected value, of a probability distribution is found by multiplying each value by its corresponding probability and summing them up.

Mean = 0(0.4) + 1(0.2) + 2(0.3) + 3(0.1) = 0 + 0.2 + 0.6 + 0.3 = 1.1

So, the mean of the probability distribution is 1.1.

b) Calculate the variance:
The variance of a probability distribution measures the spread or dispersion of the distribution. To calculate the variance, first, find the squared deviation of each value from the mean by subtracting the mean from each value and squaring the result. Then multiply each squared deviation by its corresponding probability and sum them up.

Variance = 0.4(0-1.1)^2 + 0.2(1-1.1)^2 + 0.3(2-1.1)^2 + 0.1(3-1.1)^2
= 0.4(1.1)^2 + 0.2(0.1)^2 + 0.3(0.9)^2 + 0.1(1.9)^2
= 0.4(1.21) + 0.2(0.01) + 0.3(0.81) + 0.1(3.61)
= 0.484 + 0.002 + 0.243 + 0.361
= 1.09

So, the variance of the probability distribution is 1.09.

c) Calculate the standard deviation:
The standard deviation is the square root of the variance. It measures the average amount by which each value in the distribution differs from the mean.

Standard Deviation = √(Variance) = √(1.09) ≈ 1.044

Therefore, the standard deviation of the probability distribution is approximately 1.044.

By following these steps, you can verify that your calculations for mean, variance, and standard deviation are correct.