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## What is the difference between discrete probability distribution and continuous probability distribution?

Morgan
Featured 7 months ago

See below.

#### Explanation:

A random variable is a real-valued function defined over a sample space. Consequently, a random variable can be used to identify numerical events of interest in an experiment. Random variables may be either continuous or discrete.

A random variable $Y$ is said to be discrete if it can assume only a finite or countably infinite* number of distinct values.

A set of elements is said to be countably infinite* if the elements in the set can be put into one-to-one correspondence with the positive integers.

Because certain types of random variables occur so frequently in practice, it is useful to have at hand the probability of each value of ra random variable. This collection of probabilities is called the probability distribution of the discrete random variable.

• Distribution functions for discrete random variables are always step functions

Example: Binomial distribution function, $n = 2 , \text{ } p = 1 / 2$

On the other hand, a random variable $Y$ is said to be continuous if it can take on any value in an interval. More precisely, a random variable $Y$ with distribution function $F \left(y\right)$ is said to be continuous if $F \left(y\right)$ is continuous for $- \infty < y < \infty$.

Unfortunately, the probability distribution for a continuous random variable cannot be specified in the same way as outlined above for a discrete random variable; it is mathematically impossible to assign nonzero probabilities to all points on a line interval while satisfying the requirement that the probabilities of the distinct possible values sum to one.

Rather, we define a probability density function for the random variable:

Let $F \left(y\right)$ be the distribution function for a continuous random variable $Y$. Then $f \left(y\right)$, given by

$f \left(y\right) = \frac{\mathrm{dF} \left(y\right)}{\mathrm{dy}} = F ' \left(y\right)$

wherever the derivative exists, is called the probability density function for the random variable $Y$.

Example: A distribution function $F \left(y\right)$ for a continuous random variable

## May I know how to solve it?? Thank you

Bill K.
Featured 7 months ago

Recognize that $M \left(t\right) = \frac{{e}^{6 t} - {e}^{2 t}}{4 t}$ is the moment generating function of a uniform distribution over the interval $\left[2 , 6\right]$. Find the probability by integrating the p.d.f. Calculate $\mu = E \left[X\right]$, $E \left[{X}^{2}\right]$, and ${\sigma}^{2} = E \left[{X}^{2}\right] - {\mu}^{2}$. (You could also try, with computer help, to calculate $M ' \left(0\right)$ and $M ' ' \left(0\right)$ (as limits) to help you confirm the mean and variance.)

#### Explanation:

As you stated, for a continuous random variable $X$ uniform distribution over $\left[a , b\right]$ (for $a \le x \le b$), the p.d.f. is $f \left(x\right) = \frac{1}{b - a}$.

The moment generating function is

$M \left(t\right) = E \left[{e}^{t X}\right] = {\int}_{- \infty}^{\infty} {e}^{t x} f \left(x\right) \mathrm{dx} = \frac{1}{b - a} {\int}_{a}^{b} {e}^{t x} \mathrm{dx}$

$= \frac{1}{\left(b - a\right) t} {e}^{t x} {|}_{x = a}^{x = b} = \frac{{e}^{b t} - {e}^{a t}}{\left(b - a\right) t}$.

(a) Therefore, $M \left(t\right) = \frac{{e}^{6 t} - {e}^{2 t}}{4 t}$ is the moment generating function of a uniform distribution over the interval $\left[2 , 6\right]$ (i.e., $f \left(x\right) = \frac{1}{4}$ for $2 \le x \le 6$).

(b) The probability is $P \left(2 \le X \le 3\right) = {\int}_{2}^{3} f \left(x\right) \mathrm{dx} = \frac{1}{4}$.

(c) In the general case, the mean is

$\mu = E \left[X\right] = {\int}_{- \infty}^{\infty} x f \left(x\right) \mathrm{dx} = \frac{1}{b - a} {\int}_{a}^{b} x \mathrm{dx} = \frac{1}{2 \left(b - a\right)} {x}^{2} {|}_{a}^{b} = \frac{{b}^{2} - {a}^{2}}{2 \left(b - a\right)} = \frac{a + b}{2}$

The mean is also the "first moment" $\mu = E \left[X\right] = {\lim}_{t \to 0} M ' \left(t\right)$ (see https://en.wikipedia.org/wiki/Moment-generating_function). Use a computer-algebra system to help you calculate this if you are interested.

The second moment is $E \left[{X}^{2}\right]$ (which also equals ${\lim}_{t \to 0} M ' ' \left(t\right)$). This is

$E \left[{X}^{2}\right] = {\int}_{- \infty}^{\infty} {x}^{2} f \left(x\right) \mathrm{dx} = \frac{1}{b - a} {\int}_{a}^{b} {x}^{2} \mathrm{dx} = \frac{1}{3 \left(b - a\right)} {x}^{3} {|}_{a}^{b} = \frac{{b}^{3} - {a}^{3}}{3 \left(b - a\right)} = \frac{{a}^{2} + a b + {b}^{2}}{3}$

The variance ${\sigma}^{2}$ (or ${\theta}^{2}$) is

${\sigma}^{2} = E \left[{X}^{2}\right] - {\mu}^{2} = \frac{{a}^{2} + a b + {b}^{2}}{3} - \frac{{a}^{2} + 2 a b + {b}^{2}}{4}$

$= \frac{4 {a}^{2} + 4 a b + 4 {b}^{2} - 3 {a}^{2} - 6 a b - 3 {b}^{2}}{12} = \frac{{a}^{2} - 2 a b + {b}^{2}}{12}$

$= \frac{{\left(a - b\right)}^{2}}{12} = \frac{{\left(b - a\right)}^{2}}{12}$.

## A standard number cube is tossed. What is the probability that the number will be less than or equal to 3?

EZ as pi
Featured 5 months ago

$P \left(3 \text{ or less on a die}\right) = \frac{3}{6} = \frac{1}{2}$

#### Explanation:

To determine the probability of an event happening, there are two values you need to determine first.

1. How many possible outcomes are there altogether?
In this case with a toss of the die, there are $6$ outcomes, being the numbers from $1$ to $6$.

2. Of the possible outcomes how many of them will meet our conditions? IN this case we want a number equal to or less than $3$, ... the numbers $1 , 2 , 3$. There are $3$ desired outcomes.

$\text{Probability" = "number of desirable outcomes"/"total number of possible outcomes}$

$P \left(3 \text{ or less on a die}\right) = \frac{3}{6} = \frac{1}{2}$

Write a probability as a fraction and give in its simplest form.

## Help me about mathematical statistics please? I really need help for preparing my midterm in mathematical statistics. Thank you very much for your help.

Geoff K.
Featured 4 months ago

Given ${f}_{X , Y} \left(x , y\right) = \left\{\left(4 {x}^{2} y + 2 {y}^{5} \text{,", 0 <= x <= 1", " 0 <= y <= 1), (0",", "otherwise}\right)\right]$

Let $A$ be the set of all possible values of $X$. Then $A = \left[0 , 1\right] .$
Let $B$ be the set of all possible values of $Y$. Then $B = \left[0 , 1\right] .$
The support is the 2-D region $\left\{\left(x , y\right) | 0 \le x \le 1 , 0 \le y \le 1\right\}$.

a) $E \left(X\right)$

When $X$ and $Y$ are jointly distributed, we have the general formula for the expected value of a function of $X$ and $Y$:

$E \left[g \left(X , Y\right)\right] = {\int}_{A} {\int}_{B} g \left(x , y\right) {f}_{X , Y} \left(x , y\right) \text{ "dy" } \mathrm{dx}$

Of course, the function $g \left(X , Y\right)$ can be whatever we choose. If we choose $g \left(X , Y\right) = X$, we get

$E \left[X\right] = {\int}_{A} {\int}_{B} x \text{ "f_(X,Y)(x,y)" "dy" } \mathrm{dx}$

For this question:

$E \left(X\right) = {\int}_{x = 0}^{1} {\int}_{y = 0}^{1} x \left(4 {x}^{2} y + 2 {y}^{5}\right) \text{ "dy" } \mathrm{dx}$

$\textcolor{w h i t e}{E \left(X\right)} = {\int}_{0}^{1} x {\int}_{0}^{1} \left(4 {x}^{2} y + 2 {y}^{5}\right) \text{ "dy" } \mathrm{dx}$

$\textcolor{w h i t e}{E \left(X\right)} = {\int}_{0}^{1} x {\left[2 {x}^{2} {y}^{2} + \frac{1}{3} {y}^{6}\right]}_{y = 0}^{1} \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left(X\right)} = {\int}_{0}^{1} x \left(2 {x}^{2} + \frac{1}{3}\right) \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left(X\right)} = {\int}_{0}^{1} \left(2 {x}^{3} + \frac{1}{3} x\right) \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left(X\right)} = {\left[\frac{1}{2} {x}^{4} + \frac{1}{6} {x}^{2}\right]}_{x = 0}^{1}$

$\textcolor{w h i t e}{E \left(X\right)} = \frac{1}{2} + \frac{1}{6}$

$\textcolor{w h i t e}{E \left(X\right)} = \frac{2}{3}$

b) $E \left(Y\right)$

I'll leave the calculation of $E \left(Y\right)$ as an exercise. Just choose $g \left(X , Y\right) = Y .$ You should get $E \left(Y\right) = \frac{46}{63.}$

c) $E \left(3 X + 7 Y\right)$

Using the rules of expectation, we get

$E \left(3 X + 7 Y\right) = E \left(3 X\right) + E \left(7 Y\right)$

$\textcolor{w h i t e}{E \left(3 X + 7 Y\right)} = 3 E \left(X\right) + 7 E \left(Y\right)$

$\textcolor{w h i t e}{E \left(3 X + 7 Y\right)} = 3 \left(\frac{2}{3}\right) + 7 \left(\frac{46}{63}\right)$

$\textcolor{w h i t e}{E \left(3 X + 7 Y\right)} = 2 + \frac{46}{9}$

$\textcolor{w h i t e}{E \left(3 X + 7 Y\right)} = \frac{64}{9}$

d) $\text{Var} \left(X\right)$

Recall by the definition of variance:

$\text{Var} \left(X\right) = E \left\{{\left[X - E \left(X\right)\right]}^{2}\right\} = E \left({X}^{2}\right) - {\left[E \left(X\right)\right]}^{2}$

We already know $E \left(X\right)$, so we just need $E \left({X}^{2}\right) .$ This is found by letting $g \left(X , Y\right) = {X}^{2} :$

$E \left({X}^{2}\right) = {\int}_{x = 0}^{1} {\int}_{y = 0}^{1} {x}^{2} \left(4 {x}^{2} y + 2 {y}^{5}\right) \text{ "dy" } \mathrm{dx}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = {\int}_{0}^{1} {x}^{2} {\int}_{0}^{1} \left(4 {x}^{2} y + 2 {y}^{5}\right) \text{ "dy" } \mathrm{dx}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = {\int}_{0}^{1} {x}^{2} {\left[2 {x}^{2} {y}^{2} + \frac{1}{3} {y}^{6}\right]}_{y = 0}^{1} \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = {\int}_{0}^{1} {x}^{2} \left(2 {x}^{2} + \frac{1}{3}\right) \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = {\int}_{0}^{1} \left(2 {x}^{4} + \frac{1}{3} {x}^{2}\right) \text{ } \mathrm{dx}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = {\left[\frac{2}{5} {x}^{5} + \frac{1}{9} {x}^{3}\right]}_{x = 0}^{1}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = \frac{2}{5} + \frac{1}{9}$

$\textcolor{w h i t e}{E \left({X}^{2}\right)} = \frac{23}{45}$

Thus,

$\text{Var} \left(X\right) = E \left({X}^{2}\right) - {\left[E \left(X\right)\right]}^{2}$

$\textcolor{w h i t e}{\text{Var} \left(X\right)} = \frac{23}{45} - {\left(\frac{2}{3}\right)}^{2}$

$\textcolor{w h i t e}{\text{Var} \left(X\right)} = \frac{23}{45} - \frac{4}{9}$

$\textcolor{w h i t e}{\text{Var} \left(X\right)} = \frac{1}{15}$

e) $\text{Var} \left(Y\right)$

Once again, I'll leave this one as practice. You should get $\text{Var} \left(Y\right) = \frac{797}{15876} \approx 0.0502 .$

f) $E \left(X Y\right)$

Same as before, but with $g \left(X , Y\right) = X Y$. You should get $E \left(X Y\right) = \frac{10}{21.}$

g) $E \left(X Y + 14\right)$

We use the properties of expectation again to get

$E \left(X Y + 14\right) = E \left(X Y\right) + E \left(14\right)$

$\textcolor{w h i t e}{E \left(X Y + 14\right)} = \frac{10}{21} + 14$

$\textcolor{w h i t e}{E \left(X Y + 14\right)} = \frac{304}{21}$

h) $\text{P} \left(X + Y \le 1\right)$

We're looking for the probability of being in the part of the support that satisfies $x + y \le 1$. So we need to integrate ${f}_{X , Y} \left(x , y\right)$ over this partial region. The added restriction is:

$x + y \le 1$
$\implies \text{ } y \le 1 - x$

Combining this with the given restrictions $x \ge 0$ and $y \ge 0 ,$ our integration bounds are the $x$-axis, the $y$-axis, and $y = 1 - x .$

We can picture covering this region as follows: as $x$ goes from $0$ to $1 ,$ $y$ will go from $0$ to $1 - x .$

Thus, our probability is

$\text{P"(X+Y<=1)= int_(x=0)^1int_(y=0)^(1-x)f_(X,Y)(x,y)" "dy" } \mathrm{dx}$

#color(white)("P"(X+Y<=1))= int_(0)^1int_(0)^(1-x)(4x^2y+2y^5)" "dy" "dx#

#color(white)("P"(X+Y<=1))= int_(0)^1 [2x^2y^2]_(y=0)^(1-x)" "dx#

#color(white)("P"(X+Y<=1))= int_(0)^1 2x^2(1-x)^2" "dx#

#color(white)("P"(X+Y<=1))= int_(0)^1 (2x^2-4x^3+2x^4)" "dx#

$\textcolor{w h i t e}{\text{P} \left(X + Y \le 1\right)} = {\left[\frac{2}{3} {x}^{3} - {x}^{4} + \frac{2}{5} {x}^{5}\right]}_{x = 0}^{1}$

$\textcolor{w h i t e}{\text{P} \left(X + Y \le 1\right)} = \frac{2}{3} - 1 + \frac{2}{5}$

$\textcolor{w h i t e}{\text{P} \left(X + Y \le 1\right)} = \frac{1}{15}$

## What is the variance for the following data, 2 4 5 7 ? Please show working.[steps].

Featured 3 months ago

$\textcolor{red}{{\sigma}^{2} = 3.25}$

#### Explanation:

To find the variance, we first need to calculate the mean.

To calculate the mean, simply add all the data points, then divide by the number of data points.

The formula for the mean $\mu$ is

$\mu = \frac{{\sum}_{k = 1}^{n} {x}_{k}}{n} = \frac{{x}_{1} + {x}_{2} + {x}_{3} + \cdots + {x}_{n}}{n}$

Where ${x}_{k}$ is the $k$th data point, and $n$ is the number of data points.

For our data set, we have:

$n = 4$

$\left\{{x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}\right\} = \left\{2 , 4 , 5 , 7\right\}$

So the mean is

$\mu = \frac{2 + 4 + 5 + 7}{4} = \frac{18}{4} = \frac{9}{2} = 4.5$

Now to calculate the variance, we find out how far away each data point is from the mean, then square each of those values, add them up, and divide by the number of data points.

The variance is given the symbol ${\sigma}^{2}$

The formula for the variance is:

${\sigma}^{2} = \frac{{\sum}_{k = 1}^{n} {\left({x}_{k} - \mu\right)}^{2}}{n} = \frac{{\left({x}_{1} - \mu\right)}^{2} + {\left({x}_{2} - \mu\right)}^{2} + \ldots + {\left({x}_{n} - \mu\right)}^{2}}{n}$

So for our data:

${\sigma}^{2} = \frac{{\left(2 - 4.5\right)}^{2} + {\left(4 - 4.5\right)}^{2} + {\left(5 - 4.5\right)}^{2} + {\left(7 - 4.5\right)}^{2}}{4}$

${\sigma}^{2} = \frac{{\left(- 2.5\right)}^{2} + {\left(- 0.5\right)}^{2} + {\left(0.5\right)}^{2} + {\left(2.5\right)}^{2}}{4}$

${\sigma}^{2} = \frac{6.25 + 0.25 + 0.25 + 6.25}{4} = \frac{13}{4} = \textcolor{red}{3.25}$

## From a normal pack of 52 playing cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?

Ratnaker Mehta
Featured 3 months ago

$\frac{7}{13}$.

#### Explanation:

Let, $A \mathmr{and} B$ denote, the events that the card selected is an Ace

and a Black-coloured, resp.

Then, the Reqd. Prob. is, $P \left(A \cup B\right)$.

We know that, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) \ldots \ldots \left(\star\right)$.

For P(A), there are $52$ cards in a normal pack, out which $1$ can be

selected in $52$ ways.

There are $4$ aces in a pack, so, $1$ can be chosen in $4$ ways.

$\therefore P \left(A\right) = \frac{4}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{1}\right)$.

Similarly, $P \left(B\right) = \frac{26}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{2}\right)$.

Note that, $A \cap B$ denotes the event that the selected card is

an ace and black-coloured, i.e., a black-coloured ace.

Out of $52$ cards, there are $2$ black-coloured aces.

Clearly, $P \left(A \cap B\right) = \frac{2}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{3}\right)$.

From $\left(\star\right) , \left({\star}_{1}\right) , \left({\star}_{2}\right) \mathmr{and} \left({\star}_{3}\right)$, we have,

$\text{The Reqd. Prob.} = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$, as already

derived by Respected Andrea S. !

Enjoy Maths.!

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