Answer:
Explanation:
The expected value of this distribution will be the Mean
upon calculation
For the distribution of C
1) P(x=xi) 0 1 2 3 4
0 0.20 0.20 0.20 0.20
![Mean = Pi*xi](https://img.qammunity.org/2021/formulas/mathematics/college/zch3kce3zi1r1qnwuxpvzey3il6dhj2h78.png)
= 0*0+1*0.2+2*0.2+3*0.2+4*0.2
Mean = 2
2) Standard deviation of C
![√((xi-mean)^2 *P(xi))](https://img.qammunity.org/2021/formulas/mathematics/college/7185ph9c7o9ttw3usgjcjs3cfnw101ke6g.png)
= 1.09
For the distribution of D
1) P(x=xi) 0 1 2 3 4
0 0.1 0.4 0.2 0.1
![Mean = Pi*xi](https://img.qammunity.org/2021/formulas/mathematics/college/zch3kce3zi1r1qnwuxpvzey3il6dhj2h78.png)
= 1.9
2) Standard deviation
= 0.984
Upon comparing the result we observe that
C has greater expected value than D
C has greater standard deviation than D which means it is more spreading of the given data than D.