Math Problem Statement

Consider the following data:

x 1 2 3 4 5 P(X=x) 0.3 0.1 0.2 0.1 0.3

Step 4 of 5: Find the value of P(X≤2) . Round your answer to one decimal place.

Solution

To find P(X2)P(X \leq 2), we need to add the probabilities of X=1X = 1 and X=2X = 2, since P(X2)P(X \leq 2) represents the probability of XX being less than or equal to 2.

From the data:

  • P(X=1)=0.3P(X = 1) = 0.3
  • P(X=2)=0.1P(X = 2) = 0.1

So, P(X2)=P(X=1)+P(X=2)=0.3+0.1=0.4P(X \leq 2) = P(X = 1) + P(X = 2) = 0.3 + 0.1 = 0.4

Thus, the value of P(X2)P(X \leq 2) is 0.4.

Would you like to go through further steps or need any more details?

Here are some related questions for further practice:

  1. What is the value of P(X>3)P(X > 3)?
  2. How do you calculate the expected value E(X)E(X) for this distribution?
  3. What is the probability of X=4X = 4?
  4. How do you calculate P(X3)P(X \geq 3)?
  5. If the probability of X=3X = 3 increased to 0.3, how would that affect the total probability?

Tip: When calculating cumulative probabilities, always remember to sum the individual probabilities for all outcomes up to and including the given value.

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Math Problem Analysis

Mathematical Concepts

Probability
Cumulative Probability

Formulas

P(X ≤ x) = Σ P(X = x_i) for x_i ≤ x

Theorems

Basic Probability Theorem

Suitable Grade Level

Grades 7-9