Answered: Find the constant c so that f (x) is a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

f (x) = c/x^3/4 for 0 < x < 1; Find the constant c so that f (x) is a
pdf of some random variable X, and then find the cdf, F (x) = P(X ≤ x). Sketch graphs
of the pdf f (x) and the cdf F (x), and find the mean μ and variance σ².

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1: we will define the f(x) then find out the constant value c

Given

$f left parenthesis x right parenthesis equals c over x to the power of bevelled 3 over 4 end exponent space f o r space 0 less than x less than 1 space$

we find the constant $c$ such that $f left parenthesis x right parenthesis$ is a pdf of some random variable $X$

we need to ensure that the integral of $f left parenthesis x right parenthesis$ over its entire domain equals to $1$ :

$integral subscript 0 superscript 1 f left parenthesis x right parenthesis d x equals 1 W e space h a v e space g i v e n space f left parenthesis x right parenthesis equals c over x to the power of bevelled 3 over 4 end exponent space f o r space 0 less than x less than 1 integral subscript 0 superscript 1 c over x to the power of bevelled 3 over 4 end exponent d x equals 1 c open parentheses fraction numerator x to the power of negative 3 over 4 plus 1 end exponent over denominator negative 3 over 4 plus 1 end fraction close parentheses subscript 0 superscript 1 equals 1 c open parentheses fraction numerator x to the power of 1 fourth end exponent over denominator 1 fourth end fraction close parentheses subscript 0 superscript 1 equals 1 c open parentheses fraction numerator 1 to the power of 1 fourth end exponent minus 0 to the power of 1 fourth end exponent over denominator 1 fourth end fraction close parentheses equals 1 c equals 1 fourth$

the value $bold italic c bold equals bold 1 over bold 4$

Hence the pdf of $X$ is,

$f left parenthesis x right parenthesis equals fraction numerator 1 over denominator 4 x to the power of bevelled 3 over 4 end exponent end fraction space space f o r space space 0 less than x less than 1$

Now we will find the cdf of $X$

$F left parenthesis x right parenthesis equals P left parenthesis X less or equal than x right parenthesis space space space space space space space space equals integral subscript 0 superscript x f left parenthesis t right parenthesis d t space space space space space space space space equals integral subscript 0 superscript x fraction numerator 1 over denominator 4 t to the power of bevelled 3 over 4 end exponent end fraction d t space space space space space space space space equals 1 fourth open parentheses t to the power of negative 3 over 4 plus 1 end exponent over blank to the power of negative 3 over 4 plus 1 end exponent close parentheses subscript 0 superscript x space space space space space space space space equals 1 fourth open parentheses t to the power of 1 fourth end exponent over blank to the power of 1 fourth end exponent close parentheses subscript 0 superscript x space space space space space space space space equals 1 fourth open parentheses fraction numerator x to the power of 1 fourth end exponent minus 0 to the power of 1 fourth end exponent over denominator blank to the power of 1 fourth end exponent end fraction close parentheses F left parenthesis x right parenthesis equals x to the power of bevelled 1 fourth end exponent$

Hence the cdf of $X$ is,

$F left parenthesis x right parenthesis equals x to the power of bevelled 1 fourth end exponent space space f o r space 0 less than x less than 1$