Question

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 =

Answer 1

Answer:

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

Answer 2

The mean and variance of the random variable with the given probability mass function is 1.186 and 0.198 respectively and this can be determined by using the formula of mean and variance.

Given :

$f(x) = \left(\dfrac{216}{43}\right)\times \left(\dfrac{1}{6}\right)^x$

The mean can be evaluated by using the following calculation:

$\rm E(x) = \sum x\times f(x)$

$\rm E(x) = 1\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3$

$\rm E(x) = \dfrac{36}{43}+\dfrac{12}{43}+\dfrac{3}{43}$

E(x) = 1.186

The variance can be evaluated by using the following calculation.

$\rm E(x^2)=\sum x^2 f(x)$

$\rm E(x^2) = 1^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3$

$\rm E(x^2) = \dfrac{36}{43}+\dfrac{24}{43}+\dfrac{9}{43}$

$\rm E(x^2) = 1.605$

Now, the variance is given by:

$\rm V(x) = E(x^2)-[E(x)]^2$

$\rm V(x) = 1.605-(1.186)^2$

V(x) = 0.198

The variance is 0.198 and the mean is 1.186.

For more information, refer to the link given below:

https://brainly.com/question/23910632