Poisson Probability Distribution

Solved Examples: Poisson Distribution

Samuel Dominic Chukwuemeka (SamDom For Peace)

Prerequisite Topics
Poisson Distribution Calculators
Poisson Distribution Table
Texas Instruments (TI) calculators for Probability

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.
Use the functions in your TI-84 Plus or TI-Nspire to solve some of the questions in order to save time.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.
You may use Possion Distribution Formulas or Poisson Distribution Tables or both.
However, if you are not asked to round your answers; then it is very necessary that you do not use Poisson Distribution Tables.
In other words, if you are not asked to round your answers; use Poisson Distribution Formulas.

(1.) ACT At a local post office, on average, $3$ customers are in line when the post office closes each day.
The probability, $P$, that exactly $n$ customers are in line when the post office closes can be modeled by the equation: $P = \dfrac{3^n e^{-3}}{n!}$
Given that $e^{-3} \approx 0.05$, which of the following values is closest to the probability that exactly $2$ customers are in line when the post office closes?

$ F.\;\; 0.08 \\[3ex] G.\;\; 0.11 \\[3ex] H.\;\; 0.15 \\[3ex] J.\;\; 0.23 \\[3ex] K.\;\; 0.45 \\[3ex] $

Poisson Distribution: Probability in Combinatorics

$ P = \dfrac{3^n e^{-3}}{n!} \\[5ex] n = 2 \\[3ex] e^{-3} \approx 0.05 \\[3ex] P = \dfrac{3^2 * 0.05}{2!} \\[5ex] P = \dfrac{9 * 0.05}{2 * 1} \\[5ex] P = \dfrac{0.45}{2} \\[5ex] P = 0.225 \\[3ex] P \approx 0.23 $

(5.) 1,000,000 radioactive atoms of a radioactive element decayed to 974,228 atoms in a year. Assume 365 days in a year.
(a.) Calculate the mean number of radioactive atoms that decayed in a day. Round to 3 decimal places as needed.
(b.) Determine the probability that 52 radioactive atoms decayed on a given day. Round to 6 decimal places as needed.

$N_i$ = initial number of atoms present in the radioactive element
$N_r$ = number of atoms remaining in the radioactive element after some time, $t$
$N_{TD}$ = total number of atoms that has decayed after some time, $t$

$ N_i = N_r + N_{TD} \\[3ex] N_{TD} = N_i - N_r \\[3ex] N_{TD} = 1000000 - 974228 \\[3ex] N_{TD} = 25772\;atoms \\[3ex] (a.) \\[3ex] \mu = \dfrac{N_{TD}}{365} \\[5ex] \mu = \dfrac{25772}{365} \\[5ex] \mu = 70.60821918 \\[3ex] \mu \approx 70.608 \\[3ex] (b.) \\[3ex] P(x) = \dfrac{\mu^x * e^{-\mu}}{x!} \\[5ex] x = 52 \\[3ex] \mu = 70.60821918 \\[3ex] P(x) = \dfrac{70.60821918^{52} * e^{-70.60821918}}{52!} \\[5ex] P(x) = 0.003707363 \\[3ex] P(x) \approx 0.003707 $