Math Problem Statement

                 x

15     

22     

34     

40     

P(X = x)

0.18     

0.30     

0.29     

0.23          a.

Is this a valid probability distribution?       multiple choice

Yes

No     Round answers to two decimal places.   b.

What is the probability that the random variable X is less than 35?      Probability      c.

What is the probability that the random variable X is between 13 and 28?      Probability      d.

What is the probability that the random variable X is greater than 21?      Probability   

Solution

Let's break down the problem and solve it step by step.

Part (a) — Is this a valid probability distribution?

A valid probability distribution must satisfy two conditions:

  1. Each probability must be between 0 and 1, inclusive.
  2. The sum of all probabilities must equal 1.

Given the probabilities:

P(X=x)=0.18,0.30,0.29,0.23P(X = x) = 0.18, 0.30, 0.29, 0.23

Let's check the sum of these probabilities:

0.18+0.30+0.29+0.23=1.000.18 + 0.30 + 0.29 + 0.23 = 1.00

Since the sum is 1 and all the individual probabilities are between 0 and 1, this is a valid probability distribution.

So, the answer to part (a) is:
Yes.


Part (b) — What is the probability that the random variable XX is less than 35?

The values of XX less than 35 are X=15X = 15 and X=22X = 22. We need to sum their probabilities:

P(X<35)=P(X=15)+P(X=22)=0.18+0.30=0.48P(X < 35) = P(X = 15) + P(X = 22) = 0.18 + 0.30 = 0.48

So, the probability that XX is less than 35 is:
0.48.


Part (c) — What is the probability that the random variable XX is between 13 and 28?

The values of XX between 13 and 28 are X=15X = 15 and X=22X = 22. We need to sum their probabilities:

P(13<X<28)=P(X=15)+P(X=22)=0.18+0.30=0.48P(13 < X < 28) = P(X = 15) + P(X = 22) = 0.18 + 0.30 = 0.48

So, the probability that XX is between 13 and 28 is:
0.48.


Part (d) — What is the probability that the random variable XX is greater than 21?

The values of XX greater than 21 are X=22X = 22, X=34X = 34, and X=40X = 40. We need to sum their probabilities:

P(X>21)=P(X=22)+P(X=34)+P(X=40)=0.30+0.29+0.23=0.82P(X > 21) = P(X = 22) + P(X = 34) + P(X = 40) = 0.30 + 0.29 + 0.23 = 0.82

So, the probability that XX is greater than 21 is:
0.82.


Let me know if you'd like more details or clarifications! Here are 5 related questions you might find interesting:

  1. What conditions must a probability distribution always satisfy?
  2. How would you calculate the expected value (mean) of this probability distribution?
  3. What is the probability that XX is exactly equal to 34?
  4. How would the probabilities change if the sum was greater than 1?
  5. How do you interpret the sum of probabilities in a discrete distribution?

Tip: Always check if the sum of probabilities equals 1 when dealing with distributions!

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Math Problem Analysis

Mathematical Concepts

Probability
Discrete Random Variables
Probability Distribution

Formulas

P(X = x) for a probability distribution must sum to 1
Sum of probabilities for intervals

Theorems

Probability Distribution Theorem

Suitable Grade Level

Grades 9-12