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2

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

Morgan
Featured 1 month 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

4

## The mean of 8 numbers is 41 The mean of 2 of the numbers is 29 What is the mean of the other 6 numbers?

Rhys
Featured 1 month ago

Mean = 45

#### Explanation:

Let the 8 numbers be: $a , b , c , d , e , f , g , h$

$\implies \frac{a + b + c + d + e + f + g + h}{8} = 41$

$\implies a + b + c + d + e + f + g + h = 328$

Let the two of those numbers just be $g \mathmr{and} h$

$\implies \frac{g + h}{2} = 29$

$\implies g + h = 58$

So $\implies a + b + c + d + e + f + \textcolor{b l u e}{g + h} = 328$ Becomes...

$\implies a + b + c + d + e + f + \textcolor{b l u e}{58} = 328$

$\implies a + b + c + d + e + f = 270$

Mean of these six numbers...

=> color(red)((a+b+c+d+e+f )/6 = 270/6 = 45

2

## How can we approximate the bell curve?

Rhys
Featured 1 month ago

$y = \frac{1}{\pi} \sech x \text{ For larger } x$

$y = \frac{1}{\sqrt{2 \pi}} \sech x \text{ For smaller } x$

#### Explanation:

A few ways of approximating the bell curve...

The general Bell curve:

color(red)(y = 1/sqrt(2pi) e^(-1/2 x^2 )

There are many reasons to why you may want to approximate the bell curve, one being that it is particulaly difficult to integrate definitely...

So by approximating, and finding a function that is more easy to integrate definitely, and has an antiderivative, this is much eaiser.

The first thing I can think of is utilising a function that already has ${e}^{x}$ within it, then we consider color(blue)(coshx -= 1/2 ( e^x + e^-x )

But by sketching, this doesnt quite have the general shape, so the reciprical could be a good idea:

color(orange)(sech x -=1/coshx -= 2/(e^x+e^(-x) )

Now this looks more like it...

One main property of the bell curve, used in statistics regulaly...

color(blue)(int_-oo ^ oo 1/sqrt(2pi) e^(-1/2 x^2 )  color(blue)(dx=1

This can be proven using polar substitution using the jacobian, a higher level of mathematics....

So one thing we could do, is understand what the total area under $y = \sech x$ is?

Finding: ${\int}_{-} {\infty}^{\infty} \sech x$ $\mathrm{dx}$

We can use the substitution $u = \tanh x$

$\implies \mathrm{du} = {\sech}^{2} x \mathrm{dx}$

$\implies \frac{\mathrm{du}}{\sech x} = \sech x \mathrm{dx}$

So $\int \sech x \mathrm{dx} \implies \int \frac{\mathrm{du}}{\sech} x$

Now use color(red)(1-tanh^2 x -= sech^2 x

$\implies \sech x = \sqrt{1 - {\tanh}^{2} x}$

$\implies \sech x = \sqrt{1 - {u}^{2}}$

$\int \frac{\mathrm{du}}{\sech} x = \int \frac{\mathrm{du}}{\sqrt{1 - {u}^{2}}}$

Now we can use another trig substitution:

$u = \sin \theta$

$\implies \mathrm{du} = \cos \theta d \theta$

$\int \frac{\mathrm{du}}{\sqrt{1 - {u}^{2}}}$

$\implies \int \frac{\cos \theta d \theta}{\sqrt{1 - {\sin}^{2} \theta}}$

$\implies \int \frac{\cos \theta d \theta}{\cos} \theta$

$\implies \int d \theta$

$= \theta + c$

Using our substitutions...

$= {\sin}^{- 1} u + c$

color(blue)(= sin^(-1) (tanhx ) + c

Integrating indefinitely...

${\left[{\sin}^{- 1} \left(\tanh x\right)\right]}_{-} {\infty}^{\infty}$

$\implies {\sin}^{- 1} \left(\tanh \left(\infty\right)\right) - {\sin}^{- 1} \left(\tanh \left(- \infty\right)\right)$

$\implies {\sin}^{- 1} \left(1\right) - {\sin}^{- 1} \left(- 1\right)$

$\implies \frac{\pi}{2} - \left(- \frac{\pi}{2}\right)$

=> color(red)(int_-oo ^oo sechx dx = pi

=> color(red)(int_-oo ^oo 1/pi sechx dx = 1

So color(blue)(y = 1/pi sechx  is a half decent approximation for the bell curve...

But as from looking from the graph we see that for small values of $x$ this may not be the most accurate approximation

So color(blue)(y = 1/sqrt(2pi) sechx  is a good aprpimation for small $x$:

As it has same value for $x = 0$ And for $x$ being small

1

## 4 shoppers are asked to state their preference for 2 brands Coke or Pepsi Let A denote the event that exactly two of the four individuals will prefer Coke. List the elements in A and find P(A)=?

Alan P.
Featured 1 month ago

$A = \left\{< C , C , P , P > , < C , P , C , P > , < C , P , P , C > , < P , C , C , P > , < P , C , P , C > , < P , P , C , C >\right\}$
$P \left(A\right) = \frac{6}{16} = \frac{3}{8}$

#### Explanation:

Explanation of notation:
The ordered sequence: $< \textcolor{red}{C} , \textcolor{b l u e}{P} , \textcolor{g r e e n}{P} , \textcolor{m a \ge n t a}{C} >$
Is intended to indicate that
$\textcolor{w h i t e}{\text{XXX}}$the first person chose $\textcolor{red}{C}$oke;
$\textcolor{w h i t e}{\text{XXX}}$the second person chose $\textcolor{b l u e}{P}$epsi;
$\textcolor{w h i t e}{\text{XXX}}$the third person chose $\textcolor{g r e e n}{P}$epsi; and
$\textcolor{w h i t e}{\text{XXX}}$the fourth person chose $\textcolor{m a \ge n t a}{C}$oke.

With 4 shopper each of whom can choose 1 of 2 drinks,
there are ${2}^{4} = 16$ possible outcome sequences.

Either from the listed set of elements of $A$ (above) or by recognizing that there are ""_4C_2=(4!)/(2!2!)=6 outcomes with 2 out of the 4 selections being Pepsi,
we see that the probability of exactly 2 shopper selecting Pepsi is
$\textcolor{w h i t e}{\text{XXX}} P \left(A\right) = \frac{6}{16} = \frac{3}{8}$

Warning
An assumption was required that a randomly selected shopper would be equally likely to chose Coke or Pepsi. If the population was biased in favor of one or the other this analysis would be invalid. (As an extreme example, for demonstration purposes: suppose everyone loved Pepsi and hated Coke; the the only outcome that would ever occur would be $< P , P , P , P >$ and $P \left(A\right) = 0$.)

2

## Solving Permutations and Combinations.Solving for variable.Find the value of r in 6Pr = 30?

Parabola
Featured 1 month ago

$r = 2$

#### Explanation:

For this problem, I will use the simplest way possible to solve this.
Remember that: nPr=(n!)/((n-r)!)

For $6 P r = 30$, we have: (6!)/((6-r)!)=30=>720/((6-r)!)=30/1
Using the fact that $\frac{a \cdot b}{c \cdot b} = \frac{a}{c}$, we can see that:
720/((6-r)!)=30/1
(720*x)/((6-r)!*x)=30/1
Where $x = \frac{30}{720} \implies \frac{1}{24}$

Therefore,
(720*1/24)/((6-r)!*1/24)=30/1
Which means that (6-r)!*1/24=1. We try to simplify this equation.
(6-r)! =24
Now, we try to think of our basic factorial numbers.
1! =1
2! =2
3! =6
4! =24
Oh! $6 - r$ must equal 4...!

We can now solve the equation:
$6 - r = 4$
$- r = - 2$
$r = 2$

2

## A card is drawn from a well shuffled pack of cards, what is the probability that is either spade or ace?

Dwight
Featured 3 weeks ago

The probability is 30.8%

#### Explanation:

There are 52 cards in all, and 13 are spades. In addition, 4 are aces, but one of these aces is also a spade! We won't count it twice.

So, in total, we have 13 + 3 = 16 cards that are either spade or ace (including one that is both spade and ace, which I assume is to be included in the sample of "successful choices").

Probability = (successful events) / (all events) = $16 \div 52 = 0.308$ or 30.8 %

Note: If "either a spade or an ace" is not meant to include the card that is both spade and ace, then reduce the number of successful events to 15.

1

## In a group of 50 persons,30 like tea,25 like coffee and 16 like both.How many like 1)either tea or coffee 2)neither tea or coffee? please represent in cardinal form of set and a venn diagram

tara t.
Featured 3 weeks ago

Either: 23
Neither: 11

#### Explanation:

Using a Venn Diagram - which to me is just a diagrammatic method of representing the information. I just go step by step with whatever information the question gives me, draw it out and then proceed from there.

So from the question, I can see that there are to be two groups of people (those who like tea and those who like coffee). Also, I can gather that there is an overlap in these two groups (i.e. those who like both). Which means that the Venn Diagram looks like this:

Now, the number of people belonging to the red section , the overlap, is information that is given to me in clear words in the question: 16 . The number of people who like both tea and coffee.

This means that now I can, through subtraction, find the number of people who like ONLY tea or ONLY coffee.

ONLY tea: 30 - 16 = 14
ONLY coffee: 25 - 16 = 9

These numbers I would fill in the either of the white sections of the circles.

To find out how many people like neither tea nor coffee, I would add up the three numbers that I have written within the circle and subtract from 50.

9 + 14 + 16 = 39
50 - 39 = 11.

This number I would write outside both circles (as seen in diagram).

The number of people who like EITHER tea or coffee would be the number of people not in the overlap and not outside either circle.

Hence it would be:
14 + 9 = 23

1

## How does probability differ from statistics?

Nimo N.
Featured 3 weeks ago

See below.

#### Explanation:

The answer is fairly complex; there are a lot of attempts to answer the question. The development of Statistics relied on Probability theory.

The practice of a Statistician, involves collecting data and examining its various aspects.

The information gained about a population is used to make statements about large scale behaviors and trends. Trying to make such predictions about a given individual is next to folly, although a lot of people try to do it.

Scientific research frequently uses statistical methods to make predictions and develop theories about how things actually work.

Probability is centered around looking at the number of ways some event can happen and calculating the likelihood of a particular outcome. For example, taking a look at the number of ways a coin might land if tossed into the air and allowed to land, then asking what is the likelihood that it would fall with "heads" showing, or that it might actually land on its edge (not too likely, but a possible outcome).

One can look to Statistics to produce a probabilistic view of possible events, and that sort of thing is done when the weather is discussed on the evening news. In fact, it is quite common to hear the commentator say things like, "it is 60% likely to rain on Wednesday." Such statements are probabilistic in nature, though based on statistical studies.

I found a good discussion regarding this topic on (which is not necessarily "the end of the story"):
https://www3.cs.stonybrook.edu/~skiena/jaialai/excerpts/node12.html

One very good statement from that source is:

"Probability is primarily a theoretical branch of mathematics, which studies the consequences of mathematical definitions. Statistics is primarily an applied branch of mathematics, which tries to make sense of observations in the real world."

A search on "probability v.s. statistics" in your browser will pull-up a lot of other sources to fill-in the story.

2

## A pair of fair six-sided dice is thrown eight times. Find the probability that a score greater than 7 is scored no more than five times?

Parzival S.
Featured 1 week ago

$\cong 0.9391$

#### Explanation:

Before we get into the question itself, let's talk about the method for solving it.

Let's say, for instance, that I want to account for all the possible results from flipping a fair coin three times. I can get HHH, TTT, TTH, and HHT.

The probability of H is $\frac{1}{2}$ and the probability for T is also $\frac{1}{2}$.

For HHH and for TTT, that is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each.

For TTH and HHT, it's also $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each, but since there are 3 ways I can get each result, it ends up being $3 \times \frac{1}{8} = \frac{3}{8}$ each.

When I sum up these results, I get $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$ - which means I now have all the possible results of the coin flip accounted for.

Notice that if I set $H$ to be $p$ and therefore have $T$ be ~p, and also notice that we have a line from the Pascal's Triangle $\left(1 , 3 , 3 , 1\right)$, we've set up a form of:

sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))

and so in this example, we get:

$= {C}_{3 , 0} {\left(\frac{1}{2}\right)}^{0} {\left(\frac{1}{2}\right)}^{3} + {C}_{3 , 1} {\left(\frac{1}{2}\right)}^{1} {\left(\frac{1}{2}\right)}^{2} + {C}_{3 , 2} {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2}\right)}^{1} + {C}_{3 , 3} {\left(\frac{1}{2}\right)}^{3} {\left(\frac{1}{2}\right)}^{0}$

$= 1 \left(1\right) \left(\frac{1}{8}\right) + 3 \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) + 3 \left(\frac{1}{4}\right) \left(\frac{1}{2}\right) + 1 \left(\frac{1}{8}\right) \left(1\right)$

$= \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$

Now we can do the problem.

We're given the number of rolls as 8, so $n = 8$.

$p$ is the sum greater than 7. To find the probability of getting a sum greater than 7, let's look at the possible rolls:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

Out of 36 possibilities, 15 rolls give a sum greater than 36, giving a probability of $\frac{15}{36} = \frac{5}{12}$.

With p=5/12, ~p=7/12

We can write out the entire sum of possibilities - from getting all 8 rolls being a sum greater than 7 all the way to getting all 8 rolls being a sum of 7 or less:

$= {C}_{8 , 0} {\left(\frac{5}{12}\right)}^{8} {\left(\frac{7}{12}\right)}^{0} + {C}_{8 , 1} {\left(\frac{5}{12}\right)}^{7} {\left(\frac{7}{12}\right)}^{1} + {C}_{8 , 2} {\left(\frac{5}{12}\right)}^{6} {\left(\frac{7}{12}\right)}^{2} + {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8} = 1$

but we're interested in summing up only those terms that have our greater than 7 sum happening 5 times or less:

$= {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8}$

$\cong 0.9391$

2

## In Bengal, 30% of the population has a certain blood type.What is the probability that exactly four out of a randomly selected group of 10 Bengalis will have that blood type?

Fleur
Featured 3 hours ago

$0.200$

#### Explanation:

The probability that four out of the ten people have that blood type is $0.3 \cdot 0.3 \cdot 0.3 \cdot 0.3 = {\left(0.3\right)}^{4}$.

The probability that the other six do not have that blood type is ${\left(1 - 0.3\right)}^{6} = {\left(0.7\right)}^{6}$.

We multiply these probabilities together, but since these outcomes can happen in any combination (for example, person 1, 2, 3, and 4 have the blood type, or perhaps 1, 2, 3, 5, etc.), we multiply by ${\textcolor{w h i t e}{I}}_{10} {C}_{4}$.

Thus, the probability is ${\left(0.3\right)}^{4} \cdot {\left(0.7\right)}^{6} \cdot {\textcolor{w h i t e}{I}}_{10} {C}_{4} \approx 0.200$.

â€”â€”â€”

This is another way to do it:

Since having this specific blood type is a Bernoulli trial (there are only two outcomes, a success and a failure; the probability of success, $0.3$, is constant; and the trials are independent), we can use a Binomial model.

We will use $\text{binompdf}$ because the "pdf", probability density function, allows us to find the probability of exactly four successes.

When using this function on your calculator, enter $10$ for the number of trials, $0.3$ for $p$ (the probability of success), and $4$ for the $X$ value.

$\text{binompdf} \left(10 , 0.3 , 4\right) \approx 0.200$