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

Bill K.
Featured 4 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

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

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

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

## What is moment generating function?

Morgan
Featured 4 months ago

See below.

#### Explanation:

The parameters $\mu$ (mean) and $\sigma$ (standard deviation) locate the center and describe the spread associated with the values of a random variable $Y$. They do not, however, provide a unique characterization of the distribution of $Y$; many different distributions possess the same means and standard deviations.

• The $k$th moment of a random variable $Y$ taken about the origin is defined to be $E \left({Y}^{k}\right)$ and is denoted by ${\mu}_{k}^{'}$

• The moment-generating function $m \left(t\right)$ for a random variable $Y$ is defined to be $m \left(t\right) = E \left({e}^{t Y}\right)$. We say that a moment-generating function for $Y$ exists if there exists a positive constant $b$ such that $m \left(t\right)$ is finite for $\left\mid t \right\mid \le b$

If $m \left(t\right)$ exists, then for any positive integer $k$,

#(d^km(t))/(dt^k)]_(t=0)=m^(k)(0)=mu_k^'#

In other words, if you find the $k$th derivative of $m \left(t\right)$ with respect to $t$ and then set $t = 0$, the result will be ${\mu}_{k}^{'}$.

Then we find that various probability distributions have their own unique moment-generating function.

## Four cards are drawn out from a packet of cards casually. What is the probability to find 2 cards of them to be spade? @probability

Parzival S.
Featured 2 months ago

#(57,798)/(270,725)~~21.35%#

#### Explanation:

Let's first see the number of ways we can pick 4 cards from a pack of 52:

#C_(n,k)=(n!)/((k!)(n-k)!)# with $n = \text{population", k="picks}$

#C_(52,4)=(52!)/((4!)(48!))=(52xx52xx50xx49)/24=270,725#

How many ways can we draw 4 cards and have exactly 2 of them be spades? We can find that by choosing 2 from the population of 13 spades, then choosing 2 cards from the remaining 39 cards:

#C_(13,2)xxC_(39,2)=(13!)/((2!)(11!))xx(39!)/((2!)(37!))=(13xx12)/2xx(39xx38)/2=57,798#

This means the probability of drawing exactly 2 spades on a 4 card draw from a standard deck is:

#(57,798)/(270,725)~~21.35%#

## In a class of 50 students, 18 take choir, 26 take concert band, and 2 students take both. How many students are not in either choir or concert band?

Nimo N.
Featured 1 month ago

Please see below and refer to the diagram included.

#### Explanation:

Problem:
In a class of In a class of 50 students, 18 take choir, 26 take concert band, and 2 students take both. How many students are not in either choir or concert band?, 18 take choir, 26 take concert band, and 2 students take both.
How many students are not in either choir or concert band?

Summary:
The Universe is the entire collection of students in the class under consideration:
Universe: 50 students in a class.
Concert Band: cb = 26 students.
Choir: c = 18 students.

The drawing below is an illustration of a Venn Diagram. Mr. Venn is given credit for the invention of a way to deal with problems like this one.

The rectangle contains the "universe", the total collection of people or items under consideration.

Each circle contains numbers that represent the people or items belonging to a particular subgroup. There are two subgroups:

a circle labelled cb contains the total number of people in the universe who are taking concert band.

a circle labelled c contains the total number of people in the universe who are taking choir.

the overlapping area of the two circles contains the number of people who are taking both subjects c, and cb.

Adding the numbers in each part of the circles in the diagram gives the number of people who are in either one of the two classes or both of them.

number in cb + number in c + number in both + number not in either one of the two classes makes-up the total universe, 50.

People taking the specified classes:
$c + c b + \text{both} = 16 + 24 + 2 = 42$
People not taking either of the specified classes:
$\text{universe"\ - \ "people taking specified classes} = 50 - 42 = \textcolor{p u r p \le}{8}$

Summary:
$c + c b + \text{both" + "neither} = 16 + 24 + 2 + \textcolor{p u r p \le}{8} = \textcolor{b l u e}{50}$

## 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 1 week 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.!

## There are 20 students in a class with 7 boys . (a) In how many different ways can a team of 2 boys and 6 girls formed ? (b) In how many different ways can a team of 6 be formed such that there is a least one boy and at least one girl in the team ?

Parzival S.
Featured 1 week ago

a. 36,036 ways
b. 278,460 ways

#### Explanation:

There are 7 boys and 13 girls in the class.

When picking teams, we're looking at combinations (we don't care in what order the players are picked). The general formula is:

#C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)# with $n = \text{population", k="picks}$

a

We want a team with 2 boys (from a population of 7) and 6 girls (from a population of 13):

$\left(\begin{matrix}7 \\ 2\end{matrix}\right) \left(\begin{matrix}13 \\ 6\end{matrix}\right) = 21 \times 1716 = 36036$ ways

b

We want a team of 6 with at least 1 boy and 1 girl. So let's first have 1 guaranteed boy and 1 guaranteed girl:

$\left(\begin{matrix}7 \\ 1\end{matrix}\right) \left(\begin{matrix}13 \\ 1\end{matrix}\right)$

Now we need to fill in the remaining team. We can pick any of the remaining 18 students to fill in the remaining 4 spots:

$\left(\begin{matrix}7 \\ 1\end{matrix}\right) \left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}18 \\ 4\end{matrix}\right) = 7 \times 13 \times 3060 = 278460$ ways

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