If A Random Variable X With The Following Distribution Has Mean Value Of 0.6, Find -1 ON 0.2 2 0,1 3
The given distribution has mean value of 0.6. The values of m and n can have multiple solutions. sum of the probabilities P(X<=1) is 0.5, variance of X is 2.11, expected value of Y=3X+1 is 4.9, and variance of Y is 19.0.
The given distribution can be represented as follows:
X -1 0.2 2 0.1 3
P(X) 0.2 0.2 0 0.1 0.5
To find the mean value of X, we use the formula:
E(X) = Σ [ xi * P(X = xi)
E(X) = (-1 * 0.2) + (0.2 * 0.2) + (2 * 0) + (0.1 * 0.1) + (3 * 0.5)
E(X) = 1.3
Since E(X) = 0.6, we can set up the following equation:
E(X) = Σ [ xi * P(X = xi) ] = 0.6
(-1 * 0.2) + (0.2 * 0.2) + (2 * 0) + (0.1 * 0.1) + (3 * 0.5) = 0.6
Simplifying this equation gives us:
0.1 + 1.5 = m * n
m * n = 1.6
We can choose any values of m and n that satisfy this equation. For example, we can choose m = 1 and n = 1.6, or m = 2 and n = 0.8.
P(X <= 1) is the sum of the probabilities of all values of X less than or equal to 1:
P(X <= 1) = P(X = -1) + P(X = 0.2) + P(X = 0.1)
P(X <= 1) = 0.2 + 0.2 + 0.1
P(X <= 1) = 0.5
The variance of X can be found using the formula:
Var(X) = E(X^2) – [E(X)]^2
To find E(X^2), we use the formula:
E(X^2) = Σ [ xi^2 * P(X = xi) ]
E(X^2) = (-1)^2 * 0.2 + 0.2^2 * 0.2 + 2^2 * 0 + 0.1^2 * 0.1 + 3^2 * 0.5
E(X^2) = 4.3
Then, we can calculate the variance:
Var(X) = E(X^2) – [E(X)]^2
Var(X) = 4.3 – 1.3^2
Var(X) = 2.11
The expected value of Y = 3X + 1 can be found using the formula:
E(Y) = E(3X + 1) = 3E(X) + 1
E(Y) = 3(1.3) + 1
E(Y) = 4.9
To find the variance of Y, we use the formula:
Var(Y) = Var(3X + 1) = 9Var(X)
Var(Y) = 9(2.11)
Var(Y) = 19.0
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