Determine the mean and variance of the random variable with the following probability mass function. f(x) = (216/43)(1/6)^x, x = 1, 2, 3 Round your answers to three decimal places (e. g. 98.765). Mean = Variance =

The mean of function provided is 1.186.

The variance of the provided f(x) is 0.198

Step-by-step explanation:

It is provided that the probability mass function is,

f(x)= (214/43)×(1/6)ˣ; x=1,2,3

The mean is calculated as,

E(X)=∑  x × f(x)

        x

=1×(216/43)×(1/6)¹ + 2 × (216/43)×(1/6)² × 3 × (216/43)×(1/6)³

=36/43 + 12/43  +3/43

 =1.186

The mean of function provided is 1.186

Explanation | Common mistakes | Hint for next step

The expected value of the probability mass function,f(x)= (216/43×(1/6)ˣ

 is 1.1861.186 .

Step 2 of 2

To calculate the variance, first calculate  E(X²)=∑ x² × f(x)

= 1² ×(216/43) × (1/6)¹ + 2² × (216/43) × (1/6)² × 3² × (216/43) ×(1/6)³

=36/43 +24/43 +9/43

=1.605

The variance is calculated as,

V(X) =E(X²) - [E(X)]²

=1.605 -(1.186)²

= 0.198

The variance of the provided f(x) is 0.198

Explanation | Common mistakes

The variance of function f(x)=(216/43) × (1/6)ˣ ; x =1,2,3 is 0.198

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