Answered: 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),…

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 Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

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,

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

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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.

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