Math Problem Statement

The graph of a uniform probability distribution is given below. If necessary, round results accurate to at least 3 decimal places.

12345678910111213f(x)[Graphs generated by this script: setBorder(50,30,0,0);initPicture(0,13.9,0,1); axes(1,100,1,0,0,1,1);path([[0,0],[4,0],[4,.5],[13,.5],[13,0],[13.9,0]]);fill='red';rect([4,0],[11,.5]); text([-.5,.5],f(x));]

a. What is the probability density function (e.g., what is the height of the line that is the graph of the distribution)?

f(x)=   

b. Find the mean of the distribution.

μ=   

c. Find the standard deviation of the distribution.

σ=   

d. Find the probability represented by the shaded region.    

Solution

Let's break down the components of the problem based on the given graph and information. It appears to be a uniform probability distribution defined over a certain interval, and we're tasked with calculating several values from it.

a. Probability Density Function (PDF)

The height of the line for a uniform distribution is the value of the probability density function f(x)f(x). For a uniform distribution over an interval [a,b][a, b], the PDF is defined as:

f(x)=1baf(x) = \frac{1}{b - a}

From the graph:

  • The distribution starts at x=4x = 4 and ends at x=13x = 13.

Thus, the height f(x)f(x) is:

f(x)=1134=190.111f(x) = \frac{1}{13 - 4} = \frac{1}{9} \approx 0.111

b. Mean of the Distribution (μ)

For a uniform distribution over an interval [a,b][a, b], the mean μ\mu is the midpoint of the interval:

μ=a+b2\mu = \frac{a + b}{2}

Using a=4a = 4 and b=13b = 13:

μ=4+132=172=8.5\mu = \frac{4 + 13}{2} = \frac{17}{2} = 8.5

c. Standard Deviation of the Distribution (σ)

The standard deviation σ\sigma for a uniform distribution over [a,b][a, b] is given by the formula:

σ=ba12\sigma = \frac{b - a}{\sqrt{12}}

Substituting a=4a = 4 and b=13b = 13:

σ=13412=91293.4642.597\sigma = \frac{13 - 4}{\sqrt{12}} = \frac{9}{\sqrt{12}} \approx \frac{9}{3.464} \approx 2.597

d. Probability of the Shaded Region

In a uniform distribution, the probability of a region between two points cc and dd, where cdc \leq d, is given by:

P(cxd)=dcbaP(c \leq x \leq d) = \frac{d - c}{b - a}

To find the exact probability, we'd need to know the boundaries of the shaded region on the graph. If you have that information, I can calculate this probability more specifically.

Would you like further clarification or details?

Related Questions:

  1. How is the height of the PDF determined for a uniform distribution?
  2. What does the standard deviation represent in a uniform distribution?
  3. Why is the mean simply the midpoint of the interval in a uniform distribution?
  4. How can we visually interpret the probability of a region in a uniform distribution graph?
  5. What would happen to the standard deviation if the interval [a,b][a, b] were wider?

Tip:

For a uniform distribution, the area under the curve between two points gives the probability for that interval. The total area under the curve over the entire distribution is always 1.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Statistics
Uniform Distribution

Formulas

f(x) = 1 / (b - a)
μ = (a + b) / 2
σ = (b - a) / sqrt(12)
P(c ≤ x ≤ d) = (d - c) / (b - a)

Theorems

Uniform Distribution Properties
Standard Deviation Formula
Mean Formula

Suitable Grade Level

College Level (Probability and Statistics)