Answer:
Mean = 1.6
Variance = 0.84
Standard deviation = 0.916
Explanation:
We are given the following probability distribution below;
X P(X)
![X^(2) * P(X)](https://img.qammunity.org/2021/formulas/mathematics/college/jizbbq8ovoe3f4hpno80jifkkfcwmq3735.png)
0 0.1 0 0
1 0.4 0.4 0.4
2 0.3 0.6 1.2
3 0.2 0.6 1.8
Total 1.6 3.4
Now, the mean of the probability distribution is given by;
Mean, E(X) =
= 1.6
Also, the variance of the probability distribution is given by;
Variance, V(X) =
![\sum X^(2) * P(X) - (\sum X * P(X))^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/fg1xcp1i7jn7ls5sy2trdzaa1ytxp6ucmv.png)
=
![3.4 - (1.6)^(2)](https://img.qammunity.org/2021/formulas/mathematics/college/6wucrs3iylfb7h781bte4hbx5mzzsad467.png)
= 3.4 - 2.56 = 0.84
And the standard deviation of the probability distribution is given by;
Standard deviation, S.D. (X) =
=
= 0.916.