Math Problem Statement

solution for question 20

Solution for question 20 based on the provided probability distribution table.

Solution

The table you've provided represents a discrete probability distribution for the random variable xx. The values of xx range from 0 to 5, with corresponding probabilities P(x)P(x) for each value.

Here is the data:

xx012345
P(x)P(x)0.100.410.190.150.100.05

This is a discrete probability distribution, which follows these key properties:

  1. The sum of all probabilities should be 1.
  2. Each probability is between 0 and 1 (inclusive).

Sum of Probabilities:

0.10+0.41+0.19+0.15+0.10+0.05=1.000.10 + 0.41 + 0.19 + 0.15 + 0.10 + 0.05 = 1.00

Thus, the given distribution is valid.

If you're asked to compute any related questions (such as finding the expected value, variance, or probabilities for certain ranges of xx), please let me know!

Would you like further details or help with a specific calculation? Here are some related questions:

  1. How do you calculate the expected value of this distribution?
  2. How do you compute the variance for this distribution?
  3. What is the probability that xx is greater than 2?
  4. Can you explain the concept of a discrete probability distribution in more detail?
  5. How would you apply the probability distribution in real-world scenarios?

Tip: Always ensure the sum of the probabilities in a discrete probability distribution equals 1, as it validates the distribution!

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Math Problem Analysis

Mathematical Concepts

Probability
Discrete Probability Distribution
Statistics

Formulas

Sum of probabilities: ΣP(x) = 1
Expected value (mean): E(X) = Σ[x * P(x)]
Variance: Var(X) = Σ[(x - μ)^2 * P(x)]

Theorems

Law of Total Probability

Suitable Grade Level

Grades 9-12