# The cdf a certain college library checkout duration X is as follows:

$F(x) \space = \space \begin{Bmatrix} 0 & x<0\\ \frac{4x^2}{49} & 0\le x< 3.5 \\1 & 3.5 \le x \end{Bmatrix}$

Using the above function to compute the following.

– $P(x\le 1)$

– $P(0.5 \le x \le 1)$

– $P(X>0.5)$

– $S = F(\mu)$

– $F'(x)$

– $E(X)$

– $V(X)$

– Charge expected, $E[(h)]$

The main objective of this question is to find the probabilities, mean, and variance for the given expressions when the cumulative distribution function is given.

This question uses the concept of Cumulative distribution function. Another way to explain the distribution of random variables is to use the CDF of a random variable.

Given that:

$F(x) \space = \space \begin{Bmatrix} 0 & x<0\\ \frac{4x^2}{49} & 0\le x< 3.5 \\1 & 3.5 \le x \end{Bmatrix}$

We are given that:

$F (x) \space = \space P(x \space \le \space x)$

a) $P(x \space \le \space 1) = F(1)$

By putting values, we get:

$= \space \frac{4(1)^2}{49}$

$= \frac{4}{49}$

b) $P(0.5 \space \le \space x \space 1)$

$P(x \space \le \space 1) \space – \space P(x \space \le \space 0.5)$

By putting values and simplifying, we get:

$\frac{3}{49}$

c) $P(x \space > \space 0.5)$

$= \space 1 \space – \space P(x \space \le \space 0.5$

$1 \space – \space \frac{4x(0.5)^2}{49}$

$= \space \frac{48}{49}$

d) The CDF at mean is $0.5$, so:

$\int_{0}^{x} \frac{4x^2}{49}\, = \space 0.5$

$\frac{4x^2}{3×49} \space = \space 0.5$

$x \space = \space 2.6388$

e) $F'(x)$, as we already know that:

$f(x) \space = \space \frac{d F(x)}{dx}$

$f(x) \space = \space \frac{8x}{49}$

f)  The mean $E(x)$ is given as:

$\int_{-\infty}^{\infty} x \frac{8x}{49}\,dx$

$= \space 2.33$

g) Variance is calculated as:

$V(X) \space = \space \int_{-\infty}^{\infty} x^2 f(x)\,dx \space – \space \left [ \int_{-\infty}^{\infty} x f(x)\,dx \right ]^2$

By putting the values and simplifying, we get:

$= \space 6.125 \space – \space 5.442$

$= \space 0.683$

Thus the standard deviation is:

$0.8264$

h) The expectation is:

$E(h(x)) \space = \space E(X^2)$

By putting values, we get the final answer:

$6$

Using the given CDF, the probability, mean, and variance are as follows:

• $P(x \space \le \space 1) \space = \space \frac{4}{49}$.
• $P(0.5 \space \le \space x \space 1) \space = \space \frac{3}{49}$.
• $P(x \space > \space 0.5) \space = \space \frac{48}{49}$.
•  The CDF at mean is $0.5$, so x \space = \space 2.6388  $. • F'(x), so$ f(x) \space = \space \frac{8x}{49}$. • The mean$ E(x) is $2.33$.
•  The variance is $0.8264$.
•  The expectation is  $6$.

## Example

Calculate the  probability of $P(x\le 1)$ of  when the CFD of the function is:

$F(x) \space = \space \begin{Bmatrix} 0 & x<0\\ \frac{4x^3}{49} & 0\le x< 3.5 \\1 & 3.5 \le x \end{Bmatrix}$

Given that:

$F(x) \space = \space \begin{Bmatrix} 0 & x<0\\ \frac{4x^3}{49} & 0\le x< 3.5 \\1 & 3.5 \le x \end{Bmatrix}$

$P(x \space \le \space 1) = F(1)$

By putting values, we get:

$= \space \frac{4(1)^3}{99}$

$= \frac{4}{99}$

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