to Probability and Statistics

Question

] The density function for a random variable X is given by f(x)={(_0 , otherwise^(kx, if0≤x≤2)) ¦ where k is a constant
Find k.
Find p(1/2<x<1)
Find p(x<1)
Find the mean and standard deviation.
Find the cumulative distribution function.

Answer 1

I'm assuming you mean the function f(x) = kx for 0≤x≤2, 0 otherwise.

 

For f(x) to be a probability density function (pdf) it has to be non-negative and normalized to 1:

∫_-∞^∞ f(x) dx = 1

In this case

∫_-∞^∞ f(x) dx =∫₀² (kx) dx =2k = 1, which implies k=1/2.

Therefore, f(x) = x/2 for 0≤x≤2, 0 otherwise.

 

We have

P(1/2<x<1) = ∫_1/2¹ x/2 dx = 3/16

P(x<1) = ∫₀¹ x/2 dx = 1/4

The mean is defined as

µ = ∫_-∞^∞ x f(x) dx

so that

µ = ∫₀² x²/2 dx = 4/3

The variance is defined as

σ² = ∫_-∞^∞ (x-µ)² f(x) dx

so that

σ² = ∫₀² (x-4/3)² x/2 dx= 2/9,  σ = √2 /3

 

The cumulative distribution function (cdf) is 0 for x≤0,

∫₀ ^x t/2 dt = x²/4 for 0≤x≤2,

and 1 for 2≤x.

 

Answer 2

We need to understand the differences between discrete variables vs continuous variables.  

 

(1) An example of discrete variables is balling scores from 5 times of playing balling: 89, 102, 110, 100, 92.  The variables are discrete integers in this case.  

 

(2) A continuous variable can be any real number in the given region.  For example, a student arrived late in school five times last month.  The amount of time for which the student was late compared to the required entrance time can be any real number, on the condition that we can measure the time as precisely as possible.  Let's suppose they were  2 min 3.456 seconds, 5 min 4.221 seconds, 6 min 0.0011 seconds, 10 min 50.554 seconds, and 4 min 443 seconds.  

 

Discrete variables have probability for each variable, but continuous variables have a density function instead.  The probability of continuous variables is related to the area under the given density function.  

 

Now, let's solve the given problem. 

(i) Find k. 

The probability is the area under the given density function in the given region.  And we know that the sum of probability is 1.  Because the density function is 0 other than 0≤x≤2, we only care about the region between 0 and 2.  The area under the density function has the form of a triangle, where vertices are (0,0), (2,0) and (2, 2k).  So the area of the triangle is

 

(1/2)*(2)*2k = 2k

 

This value should be 1, as I said the total probability is 1. 

 

So, 

2k = 1

and 

k = 1/2 = 0.5

 

(ii) Find p (1/2 <x <1)

This is the area under the density function in the given range. 

Density function is f(x) = 2x

So, the area under the function between 1/2 <x <1 is

3/16

 

answer is  3/16

 

(iii) p (x<1)

The area under the density function in x<1, which is 1/4

Ans: 1/4

 

(iv) Find the mean and St Dev

mean = ∫xf(x) dx = ∫x*0.5x dx = ∫0.5x² dx

When we calculate the integral in 0≤x≤2, mean is 4/3

 

Variance is calculated as below.

 

Var =  ∫x²f(x) dx - mean²                  0≤x≤2

 

When we plug in the density function in the given region and calculate,

 

Var = 2/9

 

St Dev = square root of Variance = sqrt(2)/3

 

(v) Cumulative dist function

 

∫f(x) dx 

which is,

 1/4*x²

 

Hope this helps you to understand.