8.58904109589041 years ago

Find the probability of someone winning for deals 1 through 6 ...

Let **winning** deal

Also, note that **Deal #7** MUST be a win if it gets that far.

[Note: **multiply by 2** because **either player** could win]

Now, moving on to the the second deal, this assumes the first deal neither player won.

Continuing for

Finally, since **MUST** be a win for one of the players, this **concludes the game** and the probability of getting to **complement** of the sum of the probabilities for

The table below summarizes the probability distribution and the expected value for D which is equal to approximately 2.5 deals

Hope that helped!

8.586301369863014 years ago

A Pearson's chi-square test can refer to a test of independence or a goodness of fit test.

When we refer to a "Pearson's chi-square test," we may be referring to one of two tests: the Pearson's chi-square test of independence or the Pearson's chi-square goodness-of-fit test.

Goodness of fit tests determine whether a data set's distribution differs significantly from a theoretical distribution. The data must be unpaired.

Tests of independence determine if unpaired observations of two variables are independent of one another.

**Observed values**

**Expected values**

Using the chi-square formula, you determine your chi-square statistic, your degrees of freedom, and your level of significance, and compare your results to a chi-square distribution table. For the data presented above, we could use the chi-square test to determine if males and females differ in the amount of time (more or less than fifteen hours per week) spent on homework.

Both tests analyze unpaired, categorical data and are used when data is nonparametric. Note: by unpaired, we mean that your categories are independent of one another. These tests also can't be used with very small cell counts, such as expected values lower than five.

The results of your chi-square test will only tell you whether or not your observed values fit your expected values (whether those values are to fit an expected distribution or if your two variables are independent of one another). These tests will *not* tell you *how* your observed values differ.

There's a very good tutorial here that walks you through an example in detail.

8.517808219178082 years ago

The probability is

Let's first take a look at the probability for one of those sums.

There are

#(1,1), (1,2), ..., (1, 6)#

#(2, 1), (2, 2), ..., (2,6)#

#...#

#(6,1), (6,2), ..., (6,6)#

The probability of each one of those is

- How many possible combinations of two dice will give you a sum of
#7# ? There are#6# combinations:#(1,6)# ,#(6,1)# ,#(2,5)# ,#(5,2)# ,# (3,4)# and#(4,3)# .

#=> P("sum"=7) = 6 * 1/36 = 6/36 = 1/6#

- For a sum of
#11# , there are#2# combinations:#(5,6)# and#(6,5)# .

#=> P("sum"=11) = 2 * 1/36 = 2/36 = 1/18#

- For a sum of
#12# , there is just#1# combinations:#(6,6)# .

#=> P("sum"=12) = 1/36#

Now, how do you combine those three probabilities?

The events "

For independent events

#P(A " or " B) = P(A) + P(B)#

Thus, our probability is

#P = P("sum"=7) + P("sum"=11) + P("sum"=12)#

#= 6/36 + 2/36 + 1/36 = 9/36#

#= 1/4#

#= 25%#

8.506849315068493 years ago

Draw a **Venn Diagram** (see below)

If you sum the dog and cat owners ...

**intersection** of the two categories (see Venn diagram below) was **counted twice** . The value of the intersection (both cat and dog owners) is

**own both a cat and a dog** .

The count of **owners of a dog and no cat** is

hope that helped

8.484931506849316 years ago

Population variance:

Sample variance:

To calculate the variance:

- Calculate the arithmetic average (the
**mean**) - For each data value
**square the difference**between that data value and the mean - Calculate the
**sum of the squared differences**

If your data represents the entire population:

4. Divide the sum of the squared differences by the number of data values to get the **population variance**

If your data represents only a sample taken from a larger population

4. Divide the sum of the squared differences by 1 less than the number of data values to get the **sample variance**

8.479452054794521 years ago

A 2 person team can be chosen in one of fifteen ways.

The question is not precise because if you treated it literally the answer would be 3. First you choose one team, 4 people are left in the group, the second team takes another 2 people and the remaining create the third team.

But I assume that the question is like "In how many ways a 2 people team can be chosen from 6 people?"

Such question has an answer

There is also another way of calculating the number. A team of 2 chosen from six people is (in mathematics) **a 2 element combination of a six element set**.

The number of such combinations can be calculated as:

8.476712328767123 years ago

Variance (population):

Standard Deviation (population):

The Sum of the data values is

The Mean (

For each of the data values we can calculate the difference between the data value and the mean and then square that difference.

The sum of the squared differences divided by the number of data values gives the population variance (

The square root of of the population variance gives the population standard deviation (

**Note:** I've assumed the data values represent the *entire population*.

If the data values are only a *sample* from a larger population then you should calculate the **sample variance**, **sample standard deviation**,

**Note 2:** Normal statistical analysis is done with the aid of computers (e.g. using Excel) with built-in functions to provide these values.

8.476712328767123 years ago

Choose the closest value in the table.

When the degrees of freedom is very high, the value of the inverse function changes very slowly. This is the case for Student's t-distributions and Chi-Squared where we are normally using the table to look up a value which corresponds to some cumulative probability being met.

These tests are most sensitive when the number of degrees of freedom is low - that's where all the action is. What they are telling us is that if we gather more data (more degrees of freedom) then our answer will get better. But there is a diminishing return as we gather more data, to the point that another point really doesn't change the test by a significant amount. This is reflected in our tables, where at some point they start skipping values and taking larger steps. This can be seen in the following graph, showing a student's t test for an

As the degrees of freedom gets very large the change becomes insignificant, so most tables jump from some high number, like 30, to

So the rule of thumb is to choose the table row closest to the degrees of freedom that you have. The error in doing so will be small, but you can, if you like, interpolate between the values.

If your degrees of freedom is larger than the largest integer entry in the table, use the value for

8.468493150684932 years ago

You form a new statistic which is the difference between the two means which allows you to ask significance questions about it.

In this question we want to know about the difference of two means. This is a function of random variables, i.e.

or, in our case the function is of the sample means:

There are many assumptions that go into the next steps (see this link for details Stat Trek: difference between means ) but for now lets assume that the two distributions we are sampling are approximately normal, and that we only have relatively few sampling points for each (otherwise we would be certain of the values of the means and therefore the difference).

Given this, we can calculate the sample variance of the difference of means from the sample variances of the two samples:

Note that what goes into this calculation is the sample variances divided by the number of points, which is the variance of the calculated means and follows the expected form of the central limit theorem (Variance of sample mean ).

Before we ask questions about the new distribution using t-statistics, we need to know the degrees of freedom which can be approximated from (Welch–Satterthwaite equation ):

This equation allows for a different significance of each point from the two distributions based on their variance and a different number of samples from each. If the distributions have the same variance and we take the same number of samples,

Given all of these, we can use the students-t distribution to ask questions about the probability of the statistic

Where d is the proposed distance between the two means.

8.457534246575342 years ago

Probability density of random variable

The probability density

(a) random variable

(b) random variable

These two above factors are two independent random variable, so the probabilities of combined events must be multiplied.

Now the probability density of

Just to check, integral of this probability density from

Graphically, the probability density of random variable

graph{-ln(x) [-.1, 1, -5, 5]}

To learn more about probability, visit my site Unizor dedicated to presentation of advanced mathematics for high school students.