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Sagot :
Answer:
[tex]c = \frac{1}{12}[/tex]
The mean of the distribution is 0.25.
The variance of the distribution is of 0.6875.
Step-by-step explanation:
Probability density function:
For it to be a probability function, the sum of the probabilities must be 1. The probabilities are 3c, 3c and 6c, so:
[tex]3c + 3c + 6c = 1[/tex]
[tex]12c = 1[/tex]
[tex]c = \frac{1}{12}[/tex]
So the probability distribution is:
[tex]P(X = -1) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25[/tex]
[tex]P(X = 0) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25[/tex]
[tex]P(X = 1) = 6c = 6\frac{1}{12} = \frac{1}{2} = 0.5[/tex]
Mean:
Sum of each outcome multiplied by its probability. So
[tex]E(X) = -1(0.25) + 0(0.25) + 1(0.5) = -0.25 + 0.5 = 0.25[/tex]
The mean of the distribution is 0.25.
Variance:
Sum of the difference squared between each value and the mean, multiplied by its probability. So
[tex]V^2(X) = 0.25(-1-0.25)^2 + 0.25(0 - 0.25)^2 + 0.5(1 - 0.25)^2 = 0.6875[/tex]
The variance of the distribution is of 0.6875.
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