Question

Determine the mean and variance of the random variable with the following probability mass function.

f (x) = (1/2)(x/5), x = 1, 2, 3, 4

Determine:
a. P(X = 2)
b. P(X = 3)
c. P(X > 2.5)
d. P(X = 1)
e. Mean
f. Variance.

Answer 1

The probabilities for the given values of X are computed using the provided probability mass function. The mean and variance of X are then calculated by respectively summarizing the value times probability and squared deviation from the mean times probability for all possible values of X.

Step-by-step explanation:

The student is tasked with determining the mean and variance of a random variable X, given its probability mass function f(x) = (1/2)(x/5) where x takes on values 1, 2, 3, 4. Additionally, probabilities for specific values of X are also inquired.

• a. P(X = 2): This is found by substituting x = 2 into the probability mass function, f(2) = (1/2)(2/5) = 1/5.
• b. P(X = 3): Similarly, substitute x = 3 to find f(3) = (1/2)(3/5) = 3/10.
• c. P(X > 2.5): Since X can only take integer values, P(X > 2.5) is the sum of probabilities that X can take values greater than 2.5, i.e., P(X = 3) + P(X = 4).
• d. P(X = 1): Substitute x = 1 into the probability mass function to get f(1) = (1/2)(1/5) = 1/10.
• e. Mean: Calculate by summing over all possible values of x multiplied by their respective probabilities.
• f. Variance: After finding the mean μ, calculate variance by summing (x - μ)²f(x) over all possible x values.

Calculations for Mean and Variance:

1. First, compute the mean (μ) using the formula μ = ∑(x × f(x)) for all values of x.
2. Next, calculate the variance (σ²) using the formula σ² = ∑((x - μ)² × f(x)) for all values of x.

Answer 2

a. P(X = 2) = 0.2

b. P(X = 3) = 0.3

c. P(X > 2.5) = 0.7

d. P(X = 1) = 0.1

e. Mean = 3

f. Variance = 1

Step-by-step explanation:

As given,

Probability mass function (pmf) =
`((1)/(2))((x)/(5) ) = (x)/(10)`

Now,

a. P(X = 2) =
`(2)/(10)` = 0.2

b. P(X = 3) =
`(3)/(10)` = 0.3

c. P(X > 2.5) = P(X = 3) + P(X = 4) =
`(3)/(10) + (4)/(10)` = 0.3 + 0.4 = 0.7

d. P(X = 1) =
`(1)/(10)` = 0.1

e. Mean = E(X) =
`1.(1)/(10) + 2.(2)/(10) + 3.(3)/(10) + 4.(4)/(10)` = 0.1 + 0.4 + 0.9 + 1.6 = 3

f. Variance = E(X²) - [ E(X) ]² =
`1^(2) .(1)/(10) + 2^(2) .(2)/(10) + 3^(2) .(3)/(10) + 4^(2) .(4)/(10)` - [3]²

= 0.1 + 0.8 + 2.7 + 6.4 - 9

= 10 - 9 = 1

∴ we get

a. P(X = 2) = 0.2

b. P(X = 3) = 0.3

c. P(X > 2.5) = 0.7

d. P(X = 1) = 0.1

e. Mean = 3

f. Variance = 1