# Chapter 3

### Example 
Suppose there are 30 resistors, 7 of them do not work. You randomly choose 3 of them. Let $X$ be the number of defective resistors. Find the probability distribution of $X$.

$$ X = [0,3]$$
$$P(X=0) = { {7 \choose 0} {23 \choose 3} \over {30 \choose 3} } = 0.436$$
$$P(X=1) = { {7 \choose 1} {23 \choose 2} \over {30 \choose 3} } = 0.436$$
$$P(X=2) = { {7 \choose 2} {23 \choose 1} \over {30 \choose 3} } = 0.119$$
$$P(X=3) = { {7 \choose 3} {23 \choose 0} \over {30 \choose 3} } = 0.009$$

Probability distribution:
$$P(X = x) =
\begin{cases}
0.436 & x=0 \\
0.436 & x=1 \\
0.119 & x=2 \\
0.009 & x=3
\end{cases}
$$

## The Cumulative Distribution Function
The cumulative distribution function (CDF), $F(x)$, of a discrete random variable, $x$, with probability distribution, $f(x)$, is:
$$F(x) = P(X \le x)$$

Find CDF for the example above:
$$F(0) = P(X \le 0) = P(X = 0) = 0.436$$
$$F(1) = P(X \le 1) = P((X = 0) \cup (X=1)) = 0.872$$
$$F(2) = P(X \le 2) = P((X=0) \cup (X=1) \cup (X=2)) = 0.991$$
Since 3 is the largest possible value for $x$:
$$F(3) = P(X \le 3) = 1$$

As a piecewise function:
$$F(x) = 
\begin{cases}
0 & x < 0 \\
0.436 & 0 \le x < 1 \\
0.872 & 1 \le x < 2 \\
0.991 & 2 \le x < 3 \\
1 & x \ge 3
\end{cases}$$

### Exercise
Suppose that a days production of 850 manufactured parts contains 50 parts that to not conform to customer requirements. 2 parts are selected at random from the batch. Let $X$ be the number of non-conforming parts.

#### a)
Find the probability distribution for $X$
$$P(X = 0) = { {50 \choose 0} {800 \choose 2} \over {850 \choose 2 }} = 0.8857$$
$$P(X = 1) = { {50 \choose 1} {800 \choose 1} \over {850 \choose 2 }} = 0.1109$$
$$P(X = 2) = { {50 \choose 2} {800 \choose 0} \over {850 \choose 2 }} = 0.0034$$

$$P(X = x) =
\begin{cases}
0.8857 & x=0 \\
0.1109 & x=1 \\
0.0034 & x=2
\end{cases}
$$

#### b)
Find the CDF $F(x)$
$$F(x) = 
\begin{cases}
0 & x < 0 \\
0.8857 & 0 \le x < 1 \\
0.9966 & 1 \le x < 2 \\
1 & x \ge 2
\end{cases}$$

#### c)
Plot $F(x)$:

## Continuous Probability Distributions
A continuous random variable is a variable that can take on any value within a range. It takes on infinitely many possible value within the range.
![](NormalDistribution.png)

For a continuous distribution, $f(x)$:
$$P(X = x) = 0$$
$$P(x_0 \le X \le x_1) = \int\limits_{x_0}^{x_1} f(x) dx$$
$$P(X \ge x_0) = \int\limits_{x_0}^{\infty} f(x) dx$$

### Definition
The function, $f(x)$, is a probability density function fo the continuous random variable, $X$, defined over $\Reals$ if:

1.
$$f(x) \ge 0, \forall x \in \Reals$$
2.
$$\int\limits_{-\infty}^{\infty} f(x) dx = 1$$
3.
$$P(x_0 \le X \le x_1) = P(x_0 < X < x_1)$$
$$= P(x_0 \le X < x_1)$$
$$= P(x_0 < X \le x_1)$$

### Example
Suppose that the error in the reaction temperature in $^\circ \text{C}$ for a controlled lab experiment is a continuous random variable, $X$, having PDF:
$$f(x) =
\begin{cases}
{x^2 \over 3} & -1 < x < 2 \\
0 & elsewhere
\end{cases}$$

#### a)
Verify that $f(x)$ is a PDF.
$$\int\limits_{-1}^{2} {x^2 \over 3} dx \stackrel{?}{=} 1$$
$${1 \over 3} \left[{1 \over 3} x^3 \Big\vert_{-1}^{2}\right] = {1\over9}[8- (-1)] = 1$$

#### b)
Find $P(0 < X < 0.5)$:
$$P(0 < X < 0.5) = \int\limits_0^{0.5} {x^2 \over 3}dx$$
$${1\over9}\left[x^3 \Big|_0^{0.5}\right] = {1\over9}[0.125] = 0.01389$$

### Definition
The CDF, $F(x)$ of a continuous random variabl, $X$, with probability density function $f(x)$ is:
$$F(x) = P(X \le x) = \int\limits_{-\infty}^x f(t) dt$$

**Note:**

1.
$$P(a < X < b) = F(b) - F(a)$$
2.
$$f(x) = {d\over dx}F(x)$$

### Example 
Find the CDF of the previous example
$$f(x) =
\begin{cases}
{x^2 \over 3} & -1 < x < 2 \\
0 & elsewhere
\end{cases}$$

$$F(x) = \int\limits_{-1}^x {t^2 \over 3} dt$$
$${1/over 9}\left[t^3\Big|_{-1}^x\right] = {1\over 9}\left[x^3 + 1\right]$$
$$F(x) = \begin{cases}
0 & t < -1 \\
{1\over 9} \left[x^3 + 1\right] & -1 \le x \le 2 \\
1 & elsewhere
\end{cases}$$

### Example
The proportion of the budget for a certain type of industrial copany that is allotted to environmential and pollution control is coming under scrutiny. A data collection project determines that the distribution of these proportions is given by:
$$f(y) = \begin{cases}
k(1-y)^4 & 0 \le y \le 1 \\
0 & elsewhere
\end{cases}$$

Find $k$ that renders $f(y) a valid density function:
$$\int\limits_0^1 k(1-y)^4dy = 1$$
$${k\over5} = 1$$
$$\therefore k = 5$$