Read/review the following resources for this activity:

Textbook: Chapter 4 (All Sections)
Minimum of 1 scholarly source
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Topic: Poisson Probability Distribution

The Poisson Distribution is a discrete probability distribution where the number of occurrences in one interval (time or area) is independent of the number of occurrences in other intervals.

April Showers bring May Flowers!! Research the “Average Amount of Days of Precipitation in April” for a city of your choice.

In your initial post,


Introduce the City and State. Let us know a fun fact!
Tell us the average number of days of precipitation in that city for the month of April.
Cite your Source

What is the probability of having exactly 10 days of precipitation in the month of April?
What is the probability of having less than three days of precipitation in the month of April?
What is the probability of having more than 15 days of precipitation in the month of April?


Write a sentence for each of the probabilities explaining what those probabilities mean in context of days of precipitation in your chosen city.

Would any of the situations be considered unusual? Why or Why Not?

NOTE: this is the link for chapter 4, textbook we are using:


Discrete Probability Distributions
Discrete probability distributions are a unique type of probability experiment where the outcomes are countable. That is, we can count the number of outcomes that are possible. After introducing discrete probability distributions, we will work with two major named distributions this week. The first is the binomial and the second is Poisson. The geometric probability will also be described.

Probability Distribution
Suppose we want to study the number of doctors seeing patients in an office at any given time. Let x represent the number of doctors seeing patients at any given time in a doctor’s office. There are four doctors, so the possible values of x are 0, 1, 2, 3, and 4. That is, the number of doctors seeing patients must be one of those values. We can count the number of doctors seeing patients, so this is a discrete variable.

Suppose further that, after several weeks of analyzing the number of doctors seeing patients, the following probability list was obtained.

Number of Calls On Hold, x
Probability of x, denoted P(x)
0 0.25
1 0.38
2 0.21
3 0.11
4 0.05
That is, the probability that one doctor is seeing a patient at any given time is 0.38 = 38%; the probability that two doctors are seeing patients at any given time is 0.21 = 21%; and so on.

Is this a probability distribution? The answer is yes if the following two conditions are met.

Each probability is between 0 and 1 (inclusive).
The sum of the probabilities = 1.
In this example, each probability is between 0 and 1. The sum of the probabilities = 0.25 + 0.38 + 0.21 + 0.11 + 0.05 = 1.

Thus, the above table is a probability distribution.

Mean and Expected Value
For a given discrete probability distribution, we can calculate the mean or expected value of x. For these variables, the mean and the expected value are the same thing.

The expected value of x = mean of x = the sum of the x value multiplied by its corresponding probability.

Example: Using the discrete probability distribution above, find the expected number of doctor’s seeing patients at any given time.

Expected value = (0 * 0.25) + (1 * 0.38) + (2 * 0.21) + (3 * 0.11) + (4 * 0.05) = 0 + 0.38 + 0.42 + 0.33 + 0.20 = 1.33.

This means that the average number of doctors seeing patients at any given time is 1.33.

To understand this value, suppose the number of doctors seeing patients was recorded every minute on the minute. Suppose this was done every minute for two weeks. We would have a lot of data. The mean of these data would be approximately 1.33, meaning that on average about one and one-third of all doctors are seeing patients. Now, as a manager of the practice, you can decide if that is a large enough patient load. If your practice were struggling financially, then you might decide that you only have enough patients for three doctors rather than four.

Assumptions for the Binomial

The experiment is repeated a number (n) of tries where each try is independent of the other attempts.
There are only two possible outcomes for each try. These outcomes are called success or failure. A success means that the criteria were met, not necessarily that a good thing happened.
The probability of a success P(S) must remain constant on each attempt.
The random variable x counts the number of successful outcomes throughout the n tries.
Example: Assume that 8% of the population is left-handed. Find the probability of selecting three left-handed students in a class of 20 students.

Solution: Identify the values for the distribution: n = 20 because we have 20 students; x = 3 because we want three successes; and p = 0.08 (the probability of having a left-handed student).

We can also determine the mean, variance, and standard deviation for the binomial distribution.

m = n * p

s2 = n * p * q

s = √(n * p * q)

So in the example above with the left-handed students, on average we would expect to find 20 * 0.08 = 1.6 or about two left-handed students in that class. This is using the formula for the mean above.

Let’s do more examples with details on how to do this in Excel.

Consider a soccer player who is working on penalty goals. This player has a 20% chance of making a penalty goal and tries 50 times during practice. What is the probability that the player makes 18 goals out of the 50 attempts?

open a worksheet;
remembering that the binomial formula is set up as =BINOM.DIST(number_s, trial, probability_s, cumulative); and
in this case, the formula would be =BINOM.DIST(18, 50, 0.20, false).
The probability of making 18 goals is 0.00375 or less than 1%.

Let’s continue with this same example. What would be the average number of goals that this soccer player would make? This can be done by hand, rather than the through Excel. The mean for binomial distributions is the number times the probability, np. In this case, that would be 50 * 0.2 = 10. On average, we would expect this player to make 10 out of the 50 goals. That is why making 18 has a very low probability.

In another example, we can look at cumulative probabilities. Say we were to plant trees. Based on past gardening experiences, we estimate that about 60% of them will grow. What is the probability that no more than 30 trees will grow out of 85 that were planted? Notice that this is different than the example above. Previously, we were looking at the probability of 18 goals. That is one number value, so it is a straight probability calculation. In this example, we are looking at no more than 30, which is a cumulative probability. That means we are looking at the probability that 0, 1, 2, 3, 4, . . . 28, 29, or 30 trees will grow.

In Excel

remember that the binomial formula is set up as =BINOM.DIST(number_s, trial, probability_s, cumulative); and
in this case, the formula would be =BINOM.DIST(30, 85, 0.60, true).
The probability of having at most 30 trees grow is much less than 1%. We will probably get more than 30 trees.

Now, let’s make this a more difficult problem. If we plant 85 trees, what is the probability that at least 45 will grow? In this example, we are looking for at least 45. This means we want the probability that 45, 46, 47, 48, . . . .84, or 85 trees grew. This is a cumulative probability. Now, there is one more issue here. Cumulative probability calculations go from 0 through a higher number. We want 45 through 85. What we do is use the complement rule. Notice:

0 through 44 is the complement of 45 through 85.

So in our calculations, we will find the cumulative probability of 0 through 44 and then subtract that from 1.

Probability (45 through 85) = 1 – Probability(0 through 44).

To do this in Excel, open a new worksheet and then type =BINOM.DIST(44,85,0.60,true). The probability in the session window is 0.076, so we have about an 8% chance of having 44 or few trees grow. To find the probability of having 45 or more trees grow is

1 – 0.08 = 0.92 or 92%.

So we have a 92% chance of growing at least 45 trees.

Binomial Probabilities

Poisson and Geometric Probabilities
Assumptions for the Poisson

During an interval, the experiment counts the number of successes. The number of successes is x. An interval could be time, area, or volume.
The probability of the event occurring is the same for each interval.
The number of occurrences in one interval is independent of the number of occurrences in other intervals.
Examples of Poisson distributions would be the probability of how many students come to office hours during a week (the week would be the interval), how much rain falls during a month (the month is the interval), or how many insurance claims come in during the year (the year is the interval).

Let’s consider some examples. Going back to the soccer game, the team might score three goals on average during a game. What is the probability that the team scores two goals during a game? This is a Poisson distribution because the game is the interval. Also note that we need to have the number of goals that happen on average in order to calculate this probability.

To do this in Excel,

remember that the poisson formula is set up as
=POISSON.DIST(x, mean, cumulative); and
in this case, the formula would be =POISSON.DIST(2, 3, false).
The probability is 0.22 or 22%.

In another example, imagine that the number of times a child asks for his or her mom’s help in an hour averages 10 (although many moms would consider this average low). What is the probability that the child will ask for help at most six times? This is a cumulative probability because the values of 0, 1, 2, 3, 4, 5, and 6 should all be included. Follow similar steps as above, and type true at the end so that the probabilities are cumulated. The P(x<=6) is 0.13 or about 13%. That is a low probability, so Mom will probably be asked more than six times each hour.

Geometric probabilities are based on finding the probability that the firstsuccess occurs on a specific try. For example, what is the probability of making your first sale on your fourth attempt? In other words, if you had

no sale;
no sale;
no sale; or
That is your first sale on the fourth attempt. This probably needs to be done by hand and the formula is P(x) = pq(x-1). In this example, if there was a 30% chance of a sale on each attempt, then the probability of your first sale on the fourth attempt would be

P(x=4) = (0.3)(0.7)(4-1) = (0.3)(0.7)3 = 0.1029 or 10.29%.

Note: If you search the Internet for geometric probabilities, most of the sites describe probabilities based on shapes. That is not the type of geometric probabilities we use in this course.

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