Featured 1 month ago

See below.

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 **discrete** if it can assume only a finite or countably infinite* number of distinct values.

A set of elements is said to becountably 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,

On the other hand, a random variable **continuous** if it can take on any value in an interval. More precisely, a random variable *continuous* if

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(y)# be the distribution function for a continuous random variable#Y# . Then#f(y)# , given by

#f(y)=(dF(y))/(dy)=F'(y)# wherever the derivative exists, is called the

probability density functionfor the random variable#Y# .

**Example:** A distribution function

Featured 1 month ago

Mean = 45

Let the 8 numbers be:

Let the two of those numbers just be

So

Mean of these six numbers...

Featured 1 month ago

A few ways of approximating the bell curve...

The general Bell curve:

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

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

Now this looks more like it...

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

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

Finding:

We can use the substitution

So

Now use

Now we can use another trig substitution:

Using our substitutions...

Integrating indefinitely...

So

But as from looking from the graph we see that for small values of

So

As it has same value for

Featured 1 month ago

Explanation of notation:

The ordered sequence:

Is intended to indicate that

With 4 shopper each of whom can choose 1 of 2 drinks,

there are

Either from the listed set of elements of

we see that the probability of exactly 2 shopper selecting Pepsi is

**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

Featured 1 month ago

For this problem, I will use the simplest way possible to solve this.

Remember that:

For

Using the fact that

Where

Therefore,

Which means that

Now, we try to think of our basic factorial numbers.

Oh!

We can now solve the equation:

Featured 3 weeks ago

The probability is 30.8%

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) =

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.

Featured 3 weeks ago

Either: 23

Neither: 11

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

Featured 3 weeks ago

See below.

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.

Featured 1 week ago

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

For HHH and for TTT, that is

For TTH and HHT, it's also

When I sum up these results, I get

Notice that if I set

and so in this example, we get:

Now we can do the problem.

We're given the number of rolls as 8, so

Out of 36 possibilities, 15 rolls give a sum greater than 36, giving a probability of

With

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:

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

Featured 3 hours ago

The probability that four out of the ten people have that blood type is

The probability that the other six do not have that blood type is

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

Thus, the probability is

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,

We will use *exactly* four successes.

When using this function on your calculator, enter

#"binompdf"(10, 0.3, 4) ~~ 0.200#