SOLUTION: Problem 2 (13%) Consider the random variable X and its probability mass function: 𝑋: 70 80 90 𝑃(𝑋 = 𝑥): 0.3 0.45 0.25 a- Find the expected value of this random

Question 1182767: Problem 2 (13%)
Consider the random variable X and its probability mass function:
𝑋: 70 80 90
𝑃(𝑋 = 𝑥): 0.3 0.45 0.25
a- Find the expected value of this random variable.
b- Find the cumulative distribution of this random.
c- Calculate P(𝑋 ≥ 80) in two methods: Use the cumulative distribution and the probability mass function.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

Given table

X	70	80	90
P(X)	0.3	0.45	0.25

Form a new row that consists of multiplying the X and P(X) values
Example: 70*0.3 = 21 in the first column

X	70	80	90
P(X)	0.3	0.45	0.25
X*P(X)	21	36	22.5

Add up everything in that third row: 21+36+22.5 = 79.5
This is the expected value.

Answer: 79.5

==============================================================================================

Part (b)

Given table

X	70	80	90
P(X)	0.3	0.45	0.25

Let's define a function C(X) such that it computes the cumulative probability up to and including that X value.

We'll define C(X) like so
C(70) = P(70) = 0.3
C(80) = P(70)+P(80) = 0.3+0.45 = 0.75
C(90) = P(70)+P(80)+P(90) = 0.3+0.45+0.25 = 1
So again, we define where k is a value in the domain of X.

So we have

X	70	80	90
P(X)	0.3	0.45	0.25
C(X)	0.3	0.75	1

==============================================================================================

Part (c)

Given table

X	70	80	90
P(X)	0.3	0.45	0.25

We see that the probability of getting 80 or higher, based on that table above, is 0.45+0.25 = 0.70
Simply add the probability values for X = 80 or larger.
There's a 70% chance of this happening.

Or, we could note that C(70) = 0.30
From this, we could then say
P(𝑋 ≥ 80) = 1 - C(70)
P(𝑋 ≥ 80) = 1 - 0.30
P(𝑋 ≥ 80) = 0.70

This works because
P(70) + P(80) + P(90) = 1
The "P(80) + P(90)" portion is what we want while C(70) represents P(70)

In other words,
P(70) + P(80) + P(90) = 1
P(70) + [ P(80) + P(90) ] = 1
C(70) + P(𝑋 ≥ 80) = 1
P(𝑋 ≥ 80) = 1 - C(70)
P(𝑋 ≥ 80) = 1 - 0.3
P(𝑋 ≥ 80) = 0.7

Answer: 0.70

RELATED QUESTIONS
A discrete random variable X has probability mass function P(x) = a(2/5)^x, x=0,1,2,3...... (answered by ikleyn)

SUPPOSE that X Is a discrete random variable with p(X+0)=.25,P(X+2)=.125 AND... (answered by robertb)

Let X be a discrete random variable whose values lie in {0, 1, 2, ...} with probability... (answered by Saffana)

A discrete random variable X has the probability mass function P(x) = 4x-x^2/a for x = 0, (answered by greenestamps)

The sample space of a random experiment is {a, b, c, d, e, f}, and each outcome is... (answered by CPhill,ikleyn)

Let X be a continuous random variable. Can the function f(x) = 2^(-x) for x > 0 and 0... (answered by greenestamps)

Let X be a continuous random variable. Can the function {{{ f(x)=2^(-x) }}} for x > 0 and (answered by Edwin McCravy)

Let X be a discrete random variable whose values lie in {0, 1, 2, ...} with probability... (answered by ikleyn)

(4 points) The joint probability mass function of X and Y is given by p(1,1)=0.45... (answered by Boreal)