What we saw in the last example was that any sum or difference of independent Normal random variables is also normally distributed. The mean and standard deviation of the resulting normal distribution can be found using the appropriate rules for means and variances.
Some random variables count the number of times the outcome of interest occurs in a fixed number of repetitions. They are called binomial random variables.
Other variables count the number of repetitions of the chance process it takes for the outcome of interest to occur. They are known as geometric random variables.
Binomial Random Variables and Settings
In binomial cases:
- Perform repeated trials of the same chance process
- Number of trials are fixed in advance
- Knowing the outcome of one trial tells us nothing about the outcome of any other trial. That is, the trials are independent
- Our chances of getting a specific event (called a “success”) is the same for each trial.
- When these conditions are met, we have a binomial setting.
A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs.
Remember the four conditions for a binomial setting using the acronym BINS:
Is it Binary? - Can the possible outcomes of each trial be classified as “success” or “failure”? Is it Independent? - Are the trials independent? That is, does knowing the result of one trial tell us anything about the result of any other trial? If yes, then it is not independent. Is the Number fixed?- The number of trials n of the chance process must be fixed in advance. Is Success the same? - Is there the same probability p of success on each trial?
Binomial distribution
The count X of successes in a binomial setting is a binomial random variable.
The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of success on any one trial (which should all be the same). The possible values of X are the whole numbers from 0 to n.
Use this specific pattern of calculation for any binomial probability. P(X = k) = P(exactly k successes in n trials) = number of arrangements * p^k(1-p)^(n-k)
To use this formula, we must count the number of arrangements of k successes in n observations. This number is called the binomial coefficient. We use the following to do the counting without actually listing all the arrangements.
The number of ways of arranging k successes among n observations is given by the binomial coefficient. The formula for the binomial coefficient is:
nCk “said n choose k” = (n!)/(k!(n-k)!)
For k = 0,1,2,..., n where n! = n(n-1)(n-2)...(3)(2)(1) Remember that 0! = 1.
The formula for binomial coefficients uses factorial notation. For any positive whole number n, its factorial n! is n! = n(n-1)(n-2)...(3)(2)(1).
Let’s look at how a binomial coefficient is determined with an example:
Simplify the binomial coefficient 6C4. 6!/(4!2!) = 15
Instead of doing this manually, you can use your calculator (which would be much more efficient when you have larger numbers).
Alpha → Window -> 8 -> type in your values.
Binomial Probability Formula
If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2,..., n. If k is any one of these values,
P(X=k)=nCk*pk(1-p)n-k
Again, this can be done using a calculator: 2nd -> vars -> binomial pdf You use a binomial pdf when you are finding the area under the graph for a single probability. For example, P(X=3) would use binomial pdf. You use a binomial cdf when you are finding the area under the graph for cumulative probabilities. P(X≤3) would use binomial cdf.
Binomial CDF is the same as P(X < k), so if you are dealing with more (>), subtract the answer from 1, the total area.
How to find Binomial Probabilities
- State the distribution and the values of interest. Specify a binomial distribution with the number of trials n, success probability p, and the values of the variable clearly identified.
- Perform calculations––show your work! Do one of the following:
- Use the binomial probability formula to find the desired probability or
- Use the binompdf or binomcdf command and label each of the inputs
- Answer the question
When using cdf, make sure to add or subtract 1 from the x-value if it’s ≥ or ≤ (because cdf doesn’t account for the equal type :/ ).
When describing distributions, the center is the mean (or μx̅) and the spread is the variance and standard deviation.
Mean and Standard Deviation of a Binomial Random Variable
If a count X of successes has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are:
µx = np Ox = √np(1-p)))
Important note: These formulas are solely for binomials. You will learn other formulas for the means and distributions as the type of variables change.
Almost all real-world sampling, such as taking an SRS from a population of interest, is done without replacement. However, sampling without replacement leads to a violation of the independence condition.
We can use binomial distributions in the statistical setting of selecting an SRS. When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution with n equal to the sample size and p equal to the proportion of successes in the population. What counts as “much larger”? In practice, the binomial distribution gives a good approximation as long as we don’t sample more than 10% of the population. We refer to this as the 10% condition.
10% Condition
When taking an SRS of size n from population of size N, we can use a binomial distribution to model the count of successes in the sample as long as n<(1/10)N.
Because all real-world sampling is done without replacement, we will always need to check the 10% condition when working with sample data.
Large Counts condition
Suppose that a count X of successes has the binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately normal with: Mean = μx̅ = np and Standard Deviation Ox = √(np(1-p))
As a rule of thumb, we will use the Normal approximation when n is so large that Np > 10 and n(1-p) > 10. Remember to check both conditions.
Important note: When you are talking about Normal distribution, you have to make sure you specify that the distribution is approximately Normal and not exactly Normal.