Question

The following is the p.d.f (Probability Density Function) of a continuous random variable X:
 \[F(x) = \dfrac{x}{{32}},0 < x < 8; = 0,\] otherwise
Also find its value at \[x = 0.5\] and \[9\] .

Answer 1

Hint: First find the c.d.f (Cumulative Distribution function) for the continuous random variable given. Use of basic integration formula. Then just put the given values in the function and get the results.
Using the range given like any value of \[x \geqslant 8\] will give 1 as value.

Complete step-by-step answer:
The value of function given is \[F(x) = \dfrac{x}{{32}}\] for the range of \[0 < x < 8; = 0\] .
First find the c.d.f (Cumulative Distribution function) for the continuous random variable given.
Since, we know that c.d.f (Cumulative Distribution function) of a continuous variable \[X\] is given by:
 \[F(x) = \int\limits_{ - \infty }^x {f(y)dy} \]
But in the given density function \[F(x)\] ,the range of \[X\] starts at \['0'\] .
So now the function becomes: \[F(x) = \int\limits_0^x {f(y)dy} \] . ………..(i)
Just put the values and find the function:
Since it was given that \[F(x) = \dfrac{x}{{32}}\] so like that only \[F(y) = \dfrac{y}{{32}}\] .Put this value in (i) and we get:
 \[
  F(x) = \int\limits_0^x {f(y)dy} \\
   = \int\limits_0^x {\dfrac{y}{{32}}dy} \;
 \]
From the integration formulas \[F(x) = \int\limits_0^2 {ydy} = {\left[ {{y^2}} \right] ^2}_0\] , solve the function and we get:
 \[
  F(x) = \int\limits_0^x {\dfrac{y}{{32}}dy} \\
  F(x) = {\left[ {\dfrac{{{y^2}}}{{64}}} \right] _0}^x \;
 \]
Now put the upper and lower limits in the function like \[{\left[ {{y^2}} \right] ^2}_0 = \left[ {{2^2} - {0^2}} \right] = 2\] and we get:
 \[
  F(x) = {\left[ {\dfrac{{{y^2}}}{{64}}} \right] _0}^x = \left[ {\dfrac{{{x^2}}}{{64}} - \dfrac{{{0^2}}}{{64}}} \right] \\
   = \left[ {\dfrac{{{x^2}}}{{64}} - 0} \right] = \dfrac{{{x^2}}}{{64}} \\
 \]
Therefore, the c.d.f(Cumulative Distribution Function) for a given continuous range is:
 \[
  F(x) = \dfrac{{{x^2}}}{{64}}\forall x \in R \;
 \] .
Now, for \[F(0.5)\] just put the value in the c.d.f obtained and we get:
 \[F(x) = \dfrac{{{x^2}}}{{64}}\]
 \[F(0.5) = \dfrac{{{{(0.5)}^2}}}{{64}} = \dfrac{{0.25}}{{64}} = 0.00390625\] , for \[x\] in range \[0 < x < 8\] .
Now, for \[x \geqslant 8\] will give 1 as value:
 \[F(9) = 1\]
Therefore, the p.d.f (Probability Density Function) of a continuous random variable X:
 \[F(x) = \dfrac{x}{{32}},0 < x < 8; = 0,\] otherwise
 \[F(0.5) = 0.00390625\] and \[F(9) = 1\] .
So, the correct answer is “ \[F(0.5) = 0.00390625\] and \[F(9) = 1\] ”.

Note: Always check for the range before finding the values.
Use the correct integration formula for ease and put the appropriate value in the upper and lower limit’s part.
There can be a chance of error in solving the integration part.