Solution:
Let X be the number on each ball. The probability distribution is:
now, the mean is
![\mu=\sum_^X\text{ . P\lparen X})](https://img.qammunity.org/2023/formulas/mathematics/college/ulzbmj0iduk2zyhgfqq43ggb15xxa2hgbm.png)
According to the data, this Mean would be:
![\mu=\sum_^X\text{ . P\lparen X})\text{ }=\text{ 3 . }(2)/(5)\text{ }+4\text{ . }(1)/(5)\text{ }+5\text{ . }(2)/(5)\text{ }=4](https://img.qammunity.org/2023/formulas/mathematics/college/iqxpod9nmym9be75guxb6jzstz0f8fdz5y.png)
So, we get that the Mean is:
![\mu=4](https://img.qammunity.org/2023/formulas/mathematics/college/tytdxg9ldmhmz8x05lf18mdhhybtu7b96t.png)
Now, the variance is
![\sigma\text{ }=\text{ }\sum_^\lbrack X^2\text{ . P\lparen X})\rbrack-\mu^2](https://img.qammunity.org/2023/formulas/mathematics/college/e289r8mcxx86siijbu0ral3uzsll2ae61z.png)
According to the data of the problem, we get that the variance is:
![\sigma=\text{ }\lbrack\text{3}^2\text{ . }(2)/(5)\text{ }+4^2\text{ . }(1)/(5)\text{ }+5^2\text{ . }(2)/(5)\text{ }\rbrack\frac{}{}-4^2](https://img.qammunity.org/2023/formulas/mathematics/college/nj84czcfo9imsijb00hpd1mtqi71j4kpef.png)
this is equivalent to:
![\sigma=(4)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/xdtbm3eszdlfk7orhxx4kaexxywc3nmjz4.png)
Thus, the standard deviation would be:
![\sqrt{(4)/(5)}=0.894](https://img.qammunity.org/2023/formulas/mathematics/college/5i0h0l0evko2e8ubwd0nxrph04h3gmwt4i.png)
Then, we can conclude that the correct answer is:
Variance:
![(4)/(5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/svm48olph5an1udemt4a3hvizqbjvb4y4g.png)
Standard deviation:
![0.894](https://img.qammunity.org/2023/formulas/mathematics/college/okgs9mv3mccdqbtgelzwizs5bcj5lgheom.png)