|Read/review the following resources for this activity:
Textbook: Chapter 4 (All Sections)
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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!
What is the probability of having exactly 10 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: https://openstax.org/books/introductory-statistics/pages/4-introduction
THIS IS THE LESSON IN CASE YOU NEED IT:
Discrete Probability Distributions
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
Is this a probability distribution? The answer is yes if the following two conditions are met.
Each probability is between 0 and 1 (inclusive).
Thus, the above table is a probability distribution.
Mean and Expected Value
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.
The experiment is repeated a number (n) of tries where each try is independent of the other attempts.
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;
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.
remember that the binomial formula is set up as =BINOM.DIST(number_s, trial, probability_s, cumulative); and
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.
During an interval, the experiment counts the number of successes. The number of successes is x. An interval could be time, area, or volume.
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
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
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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