Expected value, variance, and standard deviation

Geometric distribution

The geometric distribution is used when you’re waiting for a first success in a series of independent trials. Unlike the binomial distribution, which counts successes in a fixed number of trials, the geometric distribution counts how many trials it takes to get that first success. The key formulas are defined below.

Definitions

Geometric distribution

A probability distribution for a sequence of independent trials with two possible outcomes: success and failure. The probability of success on each trial is $p$, and the probability of failure is $1-p$. This distribution is used to find the probability that the first success happens on a particular trial:

$$P(X=k)=(1-p{)}^{k-1}\cdot p$$

where $X$ is the number of trials until the first success, $p$ is the probability of success on each trial, and $k$ is the trial on which the first success occurs, $k=1,2,3,\dots$ Expected value for the geometric distribution : The average number of trials needed to achieve the first success:

$$E(X)=\frac{1}{p}$$

Variance for the geometric distribution

A measure of how spread out the number of trials is around the mean:

$$\text{Var}(X)=\frac{1-p}{p^2}$$

Standard deviation for the geometric distribution

The typical deviation from the expected number of trials:

$$\sigma =\sqrt{\frac{1-p}{p^2}}$$

Calculator Commands

You can also use software to calculate geometric probabilities. geometpdf $(p,x)$ gives the probability that the first success occurs on the $x$th trial, where $p$ is the probability of success on a single trial. geometcdf $(p,x)$ gives the cumulative probability that the first success occurs on or before the $x$th trial, where $p$ is the probability of success on a single trial.

Conditions for Using the Geometric Distribution

Sidenote

Important to know!

In order to use the geometric distribution, all of the “BITS” conditions must be met:

• Binary: The outcome of each trial can be classified into exactly two categories: “success” or “failure.”
• Independent: The trials must be independent. (If sampling without replacement, the sample size should be less than 10% of the population.)
• Trials: The trials are not fixed; we are counting the number of trials until the first success.
• Same probability: The probability of success ($p$) must be the same for every trial.

Practice problem

Example 1:

Suppose that only $2\%$ of Americans voted third party in the last election. Liam is randomly selecting Americans until he finds one who voted third party in the last election.

Part a.

Define the random variable of interest for Liam and state how the random variable is distributed and with what parameter(s).

Since Liam is counting the number of people needed until the first third party voter is found, this is a geometric setting. The parameter $p=0.02$ represents the probability that any one selected American voted third party.

Solution:

Part b.

What is the expected number of people that Liam will have to randomly select in order to find one who voted third party?

Use the expected value formula for a geometric distribution:

$$E(X)=\frac{1}{p}=\frac{1}{0.02}=50$$

Solution:

Part c.

What is the standard deviation for the number of people that Liam will have to select in order to find one who voted third party?

Use the standard deviation formula for a geometric distribution:

$$\sigma =\sqrt{\frac{1-p}{p^2}}=\sqrt{\frac{1-0.02}{0.02^2}}=\sqrt{2450}=49.4975$$

Solution:

Part d.

What is the probability that the first third party voter that Liam finds is the tenth person selected?

There are two approaches to this problem.

Solution 1 (without technology): Use the geometric probability formula with $p=0.02$ and $k=10$:

$$P(X=10)=(1-0.02{)}^{10-1}\cdot 0.02=0.016675$$

Solution 2 (with technology): geometpdf $(p,x)$ gives the probability that the first success occurs on the $x$th trial. With $p=0.02$ and $x=10$:

$$\text{geometpdf}(0.02,10)=0.016675$$

Solution:

Part e.

What is the probability that the first third party voter that Liam finds is one of the first five people selected?

We want $P(X\le 5)$, meaning the first success happens on trial 1, 2, 3, 4, or 5. You could compute and add

$$P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$$

but it’s more efficient to use cumulative probability. geometcdf $(p,x)$ gives the cumulative probability that the first success occurs on or before the $x$th trial:

$$\text{geometcdf}(0.02,5)=0.0961$$

Solution:

Part f.

What is the probability that the first third party voter that Liam finds is either the 15th person selected or someone selected sometime after the 15th person is selected?

We want $P(X\ge 15)$. Use the complement: the opposite of “first success on trial 15 or later” is “first success on or before trial 14.”

First, compute $P(X\le 14)$ using the cumulative distribution:

$$\text{geometcdf}(0.02,14)=0.246$$

Now subtract from 1:

$$P(X\ge 15)=1-P(X\le 14)=1-0.246=0.754$$

Solution:

Part g.

What is the probability that the first non-third party voter that Liam finds is the third one who is selected?

For this part, the definition of “success” has changed. Now “success” is selecting a person who did not vote third party, so $p=1-0.02=0.98$. We want the first success on the third trial ($x=3$):

$$\text{geometpdf}(0.98,3)=0.000392$$

Solution: