Math Problem Statement

The probability density function of a random variable X is given as: X : 0, 1, 2, 3, 4, 5, 6 P(X = x) : k, 3k, 5k, 7k, 9k, 11k, 13k. Find k, P(X < 4), P(3 < X ≤ 6).

Solution

The problem asks us to determine three things based on the probability density function (PDF) of a random variable XX:

  1. Find kk
  2. Find P(X<4)P(X < 4)
  3. Find P(3<X6)P(3 < X \leq 6)

The values of XX and the corresponding probabilities are given in the following form:

X:0,1,2,3,4,5,6X: 0, 1, 2, 3, 4, 5, 6 P(X=x):k,3k,5k,7k,9k,11k,13kP(X = x): k, 3k, 5k, 7k, 9k, 11k, 13k

Step 1: Find kk

Since the total probability of a discrete random variable must sum to 1, we can set up the following equation:

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=1P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 1

Substituting the probabilities in terms of kk:

k+3k+5k+7k+9k+11k+13k=1k + 3k + 5k + 7k + 9k + 11k + 13k = 1

Simplifying the left-hand side:

(1+3+5+7+9+11+13)k=1(1 + 3 + 5 + 7 + 9 + 11 + 13)k = 1

49k=149k = 1

Solving for kk:

k=149k = \frac{1}{49}

Step 2: Find P(X<4)P(X < 4)

This is the probability that XX takes a value less than 4, i.e., the sum of probabilities for X=0,1,2,3X = 0, 1, 2, 3:

P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Substituting the values in terms of kk:

P(X<4)=k+3k+5k+7k=(1+3+5+7)k=16kP(X < 4) = k + 3k + 5k + 7k = (1 + 3 + 5 + 7)k = 16k

Now, substitute k=149k = \frac{1}{49}:

P(X<4)=16×149=1649P(X < 4) = 16 \times \frac{1}{49} = \frac{16}{49}

Step 3: Find P(3<X6)P(3 < X \leq 6)

This is the probability that XX is between 3 and 6, inclusive of 6, i.e., P(4X6)P(4 \leq X \leq 6). This involves summing the probabilities for X=4,5,6X = 4, 5, 6:

P(3<X6)=P(X=4)+P(X=5)+P(X=6)P(3 < X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)

Substituting the values in terms of kk:

P(3<X6)=9k+11k+13k=(9+11+13)k=33kP(3 < X \leq 6) = 9k + 11k + 13k = (9 + 11 + 13)k = 33k

Substituting k=149k = \frac{1}{49}:

P(3<X6)=33×149=3349P(3 < X \leq 6) = 33 \times \frac{1}{49} = \frac{33}{49}

Final Answers:

  • k=149k = \frac{1}{49}
  • P(X<4)=1649P(X < 4) = \frac{16}{49}
  • P(3<X6)=3349P(3 < X \leq 6) = \frac{33}{49}

Would you like further details or explanations on any step? Here are some related questions for further practice:

  1. How do you calculate the expected value of XX using this PDF?
  2. What is the variance of XX based on the given probabilities?
  3. How do cumulative distribution functions (CDF) relate to PDFs in discrete variables?
  4. Can you find P(X5)P(X \geq 5) from the given PDF?
  5. What is the probability that XX equals an odd number?

Tip: Always check that the total sum of probabilities equals 1 to confirm that your PDF is valid.

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Math Problem Analysis

Mathematical Concepts

Probability
Discrete Random Variables
Probability Density Function

Formulas

Sum of probabilities must equal 1: P(X = 0) + P(X = 1) + ... + P(X = 6) = 1
Calculation of probability: P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Calculation of probability: P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6)

Theorems

Basic Probability Theorem

Suitable Grade Level

Undergraduate