Math Problem Statement

Let X be a continuous random variable with probability density function f(x) = x^5 / k, ∀x ∈ [0, 4]. (a) Find k so that f(x) is a valid probability density function. (Please round all answers to 4 decimal places.) (b) Find the probability that X = 3. (Please round all answers to 1 decimal place.) (c) Find the probability that X > 3. (Please round all answers to 4 decimal places.) (d) What is the mean of X? (Please round all answers to 4 decimal places.) (e) What is the variance of X? (Please round all answers to 4 decimal places.)

Solution

To solve this problem, let's go step by step for each part:


a) Find kk so that f(x)f(x) is a valid probability density function

For f(x)f(x) to be a valid probability density function (PDF), the integral over its range must equal 1:

04x5kdx=1\int_0^4 \frac{x^5}{k} \, dx = 1

The integral of x5x^5 is:

x5dx=x66\int x^5 \, dx = \frac{x^6}{6}

Evaluate the definite integral:

1k04x5dx=1k[x66]04=1k(4660)=1k40966\frac{1}{k} \int_0^4 x^5 \, dx = \frac{1}{k} \left[ \frac{x^6}{6} \right]_0^4 = \frac{1}{k} \left( \frac{4^6}{6} - 0 \right) = \frac{1}{k} \cdot \frac{4096}{6}

Set the integral equal to 1:

1k40966=1\frac{1}{k} \cdot \frac{4096}{6} = 1

Solve for kk:

k=40966=682.6667k = \frac{4096}{6} = 682.6667


b) Find the probability that X=3X = 3

Since this is a continuous distribution, the probability that X=3X = 3 is:

P(X=3)=f(3)=35kP(X = 3) = f(3) = \frac{3^5}{k}

Substitute k=682.6667k = 682.6667:

f(3)=35682.6667=243682.6667=0.3561f(3) = \frac{3^5}{682.6667} = \frac{243}{682.6667} = 0.3561


c) Find the probability that X>3X > 3

We calculate P(X>3)P(X > 3) by integrating f(x)f(x) from 3 to 4:

P(X>3)=34x5kdxP(X > 3) = \int_3^4 \frac{x^5}{k} \, dx

The integral of x5x^5 is:

34x5kdx=1k[x66]34\int_3^4 \frac{x^5}{k} \, dx = \frac{1}{k} \left[ \frac{x^6}{6} \right]_3^4

Substitute k=682.6667k = 682.6667:

P(X>3)=1682.6667[466366]P(X > 3) = \frac{1}{682.6667} \left[ \frac{4^6}{6} - \frac{3^6}{6} \right]

P(X>3)=1682.6667[409667296]P(X > 3) = \frac{1}{682.6667} \left[ \frac{4096}{6} - \frac{729}{6} \right]

P(X>3)=1682.666733676=33674096=0.8224P(X > 3) = \frac{1}{682.6667} \cdot \frac{3367}{6} = \frac{3367}{4096} = 0.8224


d) What is the mean of XX?

The mean μ\mu is given by:

μ=04xf(x)dx=04x6kdx\mu = \int_0^4 x \cdot f(x) \, dx = \int_0^4 \frac{x^6}{k} \, dx

The integral of x6x^6 is:

x6dx=x77\int x^6 \, dx = \frac{x^7}{7}

Evaluate the definite integral:

μ=1k[x77]04=1682.6667477\mu = \frac{1}{k} \left[ \frac{x^7}{7} \right]_0^4 = \frac{1}{682.6667} \cdot \frac{4^7}{7}

μ=1682.6667163847=163844788.6667=3.4230\mu = \frac{1}{682.6667} \cdot \frac{16384}{7} = \frac{16384}{4788.6667} = 3.4230


e) What is the variance of XX?

The variance σ2\sigma^2 is:

[ \sigma^2 = E[X^2] - (E[X])^2 ]

First, calculate E[X2]E[X^2]:

E[X2]=04x2f(x)dx=04x7kdxE[X^2] = \int_0^4 x^2 \cdot f(x) \, dx = \int_0^4 \frac{x^7}{k} \, dx

The integral of x7x^7 is:

x7dx=x88\int x^7 \, dx = \frac{x^8}{8}

Evaluate the definite integral:

[ E[X^2] = \frac{1}{k} \left[ \frac{x^8}{8} \right]_0^4 = \frac{1}{682.6667} \cdot \frac{4^8}{8} ]

E[X2]=1682.6667655368=655365461.3333=12.0000E[X^2] = \frac{1}{682.6667} \cdot \frac{65536}{8} = \frac{65536}{5461.3333} = 12.0000

Now calculate the variance:

[ \sigma^2 = E[X^2] - (E[X])^2 = 12.0000 - (3.4230)^2 = 12.0000 - 11.7151 = 0.2849 ]


Final Answers:

a) k=682.6667k = 682.6667
b) P(X=3)=0.3561P(X = 3) = 0.3561
c) P(X>3)=0.8224P(X > 3) = 0.8224
d) Mean (μ\mu) = 3.4230
e) Variance (σ2\sigma^2) = 0.2849

Would you like further clarification or any specific part re-explained? Here are 5 follow-up questions:

  1. How is the normalization constant kk calculated for a probability density function?
  2. Why is the probability at a single point zero for a continuous distribution?
  3. How can the cumulative distribution function (CDF) help with these calculations?
  4. What is the relationship between E[X2]E[X^2] and variance in a distribution?
  5. How would the answers change if the range of XX was different?

Tip: Always check that a PDF integrates to 1 over its range to ensure it's valid!

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Math Problem Analysis

Mathematical Concepts

Probability Density Functions
Definite Integration
Mean and Variance of Continuous Random Variables

Formulas

Integral of PDF over range = 1: ∫[0,4] f(x) dx = 1
Mean: μ = ∫[0,4] x * f(x) dx
Variance: σ^2 = E[X^2] - (E[X])^2

Theorems

Normalization condition for a valid PDF
Calculation of expectation and variance for continuous random variables

Suitable Grade Level

University Level - Probability and Statistics