Math Problem Statement

Let be a random variable with the following probability distribution. Value x of X P(X=x) 0 0.10 10 0.25 20 0.05 30 0.15 40 0.30 50 0.15 Complete the following. (If necessary, consult a list of formulas.)

(a) Find the expectation E (X) of X.

(b) Find the variance Var (X) of X.

Solution

Let's calculate the expectation (mean) E(X)E(X) and the variance Var(X)\text{Var}(X) of the random variable XX based on the given probability distribution.

(a) Find the expectation E(X)E(X) of XX:

The expectation E(X)E(X) is given by the formula:

E(X)=xxP(X=x)E(X) = \sum_x x \cdot P(X = x)

Using the given probability distribution:

E(X)=(00.10)+(100.25)+(200.05)+(300.15)+(400.30)+(500.15)E(X) = (0 \cdot 0.10) + (10 \cdot 0.25) + (20 \cdot 0.05) + (30 \cdot 0.15) + (40 \cdot 0.30) + (50 \cdot 0.15)

Let's compute this:

E(X)=(0)+(2.5)+(1.0)+(4.5)+(12.0)+(7.5)=27.5E(X) = (0) + (2.5) + (1.0) + (4.5) + (12.0) + (7.5) = 27.5

So, the expectation E(X)=27.5E(X) = 27.5.


(b) Find the variance Var(X)\text{Var}(X) of XX:

The variance Var(X)\text{Var}(X) is calculated using the formula:

Var(X)=E(X2)[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2

First, we need to calculate E(X2)E(X^2):

E(X2)=xx2P(X=x)E(X^2) = \sum_x x^2 \cdot P(X = x)

So:

E(X2)=(020.10)+(1020.25)+(2020.05)+(3020.15)+(4020.30)+(5020.15)E(X^2) = (0^2 \cdot 0.10) + (10^2 \cdot 0.25) + (20^2 \cdot 0.05) + (30^2 \cdot 0.15) + (40^2 \cdot 0.30) + (50^2 \cdot 0.15)

Let's compute this:

E(X2)=(0)+(1000.25)+(4000.05)+(9000.15)+(16000.30)+(25000.15)E(X^2) = (0) + (100 \cdot 0.25) + (400 \cdot 0.05) + (900 \cdot 0.15) + (1600 \cdot 0.30) + (2500 \cdot 0.15)

E(X2)=(0)+(25)+(20)+(135)+(480)+(375)=1035E(X^2) = (0) + (25) + (20) + (135) + (480) + (375) = 1035

Now, using the variance formula:

Var(X)=E(X2)[E(X)]2=1035(27.5)2\text{Var}(X) = E(X^2) - [E(X)]^2 = 1035 - (27.5)^2

Var(X)=1035756.25=278.75\text{Var}(X) = 1035 - 756.25 = 278.75

Thus, the variance Var(X)=278.75\text{Var}(X) = 278.75.


Summary:

  • (a) E(X)=27.5E(X) = 27.5
  • (b) Var(X)=278.75\text{Var}(X) = 278.75

Would you like more details or have any questions?

Here are 5 related questions you can explore:

  1. How is the expectation affected if all probabilities are doubled?
  2. What is the standard deviation σ(X)\sigma(X) based on this variance?
  3. How would you compute the cumulative distribution function (CDF) for this data?
  4. What is the skewness of this distribution?
  5. How does the variance change if the value of X=50X = 50 is reduced to X=45X = 45?

Tip: Variance is always a non-negative value because it represents the spread of the data around the mean.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Theory
Statistics
Expectation
Variance

Formulas

Expectation: E(X) = Σx * P(X = x)
Variance: Var(X) = E(X^2) - [E(X)]^2

Theorems

-

Suitable Grade Level

Grades 10-12