From the given discrete distribution, we have that:
[tex]\mu = 2.13, \sigma^2 = 0.61, \sigma = 0.78[/tex]
What are the mean, the variance and the standard deviation of a discrete distribution?
- The mean of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.
- The variance is given by the sum of the difference squared between each observation and the mean, divided by the number of values.
- The standard deviation is the square root of the variance.
In this problem, the distribution is:
[tex]P(X = 1) = \frac{2}{8}[/tex]
[tex]P(X = 2) = \frac{3}{8}[/tex]
[tex]P(X = 3) = \frac{3}{8}[/tex]
Hence, the mean is:
[tex]\mu = 1\frac{2}{8} + 2\frac{3}{8} + 3\frac{3}{8} = \frac{17}{8} = 2.13[/tex]
The variance is:
[tex]\sigma^2 = (1-2.13)^2\frac{2}{8} + (2-2.13)^2\frac{3}{8} + (3-2.13)^2\frac{3}{8} = 0.61[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{0.61} = 0.78[/tex]
More can be learned about discrete distributions at https://brainly.com/question/24855677