Poisson distribution in SciPy - Step-by-Step Execution

Concept Flow - Poisson distribution

Start with lambda (rate)

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Calculate PMF for k=0

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Calculate PMF for k=1

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Calculate PMF for k=n

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Output probabilities for each k

The Poisson distribution calculates the probability of a number of events (k) happening in a fixed interval, given the average rate (lambda).

Execution Sample

SciPy

from scipy.stats import poisson
lambda_ = 3
k_values = range(6)
pmf_values = [poisson.pmf(k, lambda_) for k in k_values]
print(pmf_values)

This code calculates the Poisson probabilities for k=0 to 5 events when the average rate is 3.

Execution Table

Step	k (events)	lambda (rate)	PMF Calculation	PMF Value
1	0	3	poisson.pmf(0, 3) = e^-3 * 3^0 / 0!	0.0498
2	1	3	poisson.pmf(1, 3) = e^-3 * 3^1 / 1!	0.1494
3	2	3	poisson.pmf(2, 3) = e^-3 * 3^2 / 2!	0.2240
4	3	3	poisson.pmf(3, 3) = e^-3 * 3^3 / 3!	0.2240
5	4	3	poisson.pmf(4, 3) = e^-3 * 3^4 / 4!	0.1680
6	5	3	poisson.pmf(5, 3) = e^-3 * 3^5 / 5!	0.1008
7	End	-	-	Finished calculating PMF for k=0 to 5

💡 All PMF values for k=0 to 5 calculated using lambda=3

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
k	N/A	0	1	2	3	4	5
pmf_values	[]	[0.0498]	[0.0498, 0.1494]	[0.0498, 0.1494, 0.2240]	[0.0498, 0.1494, 0.2240, 0.2240]	[0.0498, 0.1494, 0.2240, 0.2240, 0.1680]	[0.0498, 0.1494, 0.2240, 0.2240, 0.1680, 0.1008]

Key Moments - 2 Insights

Why does the PMF value for k=3 equal the value for k=2?

Because the Poisson formula depends on k and lambda, and for lambda=3, the probabilities for k=2 and k=3 happen to be very close, as shown in execution_table rows 3 and 4.

What does the lambda parameter represent in the Poisson distribution?

Lambda is the average number of events expected in the interval. It is fixed at 3 in this example, as shown in every row of the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the PMF value when k=4?

A 0.2240

B 0.1680

C 0.1008

D 0.0498

💡 Hint

Check the row where k=4 in the execution_table under PMF Value column.

At which step does the PMF calculation for k=1 happen?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look at the Step column in execution_table for k=1.

If lambda was increased to 5, how would the PMF values for k=0 to 5 change?

A Values for higher k would increase, lower k decrease

B They would all decrease

C They would all increase

D Values for lower k would increase, higher k decrease

💡 Hint

Recall that lambda controls the average rate; higher lambda shifts probabilities toward higher k.

Concept Snapshot

Poisson distribution models count of events in fixed time.
Syntax: poisson.pmf(k, lambda)
Outputs probability of k events given average rate lambda.
PMF sums to 1 over all k.
Useful for rare event counts.
Lambda is mean and variance.

Full Transcript

The Poisson distribution calculates the chance of a number of events happening in a fixed time, based on an average rate called lambda. We calculate the probability for each number of events k using the formula poisson.pmf(k, lambda). In the example, lambda is 3, and we find probabilities for k from 0 to 5. The probabilities show how likely each count is. For instance, the chance of 0 events is about 5%, and for 3 events about 22%. Lambda controls the shape: higher lambda shifts probabilities to higher counts. This helps us understand event counts like calls per hour or arrivals per minute.

Practice

1. What does the Poisson distribution model in data science?

easy

A. The probability of a variable being normally distributed

B. The average value of a continuous variable

C. The relationship between two categorical variables

D. The number of events happening in a fixed interval of time or space

Solution

1. Step 1: Understand the Poisson distribution concept

   The Poisson distribution is used to model counts of events that happen independently in a fixed interval of time or space.

2. Step 2: Match the concept to the options

   The number of events happening in a fixed interval of time or space correctly describes this as the number of events in a fixed interval, while others describe unrelated concepts.

3. Final Answer:

   The number of events happening in a fixed interval of time or space -> Option D

4. Quick Check:

   Poisson models event counts in intervals = A [OK]

Hint: Poisson counts events in fixed intervals, not averages or categories [OK]

Common Mistakes:

• Confusing Poisson with normal distribution
• Thinking Poisson models averages instead of counts
• Mixing Poisson with categorical data analysis

2. Which of the following is the correct way to import the Poisson distribution from scipy.stats?

easy

A. from scipy.stats import poisson

B. import poisson from scipy.stats

C. from scipy import poisson.stats

D. import scipy.poisson

Solution

1. Step 1: Recall correct Python import syntax

   Python imports use 'from module import object' format. The Poisson distribution is in scipy.stats as 'poisson'.

2. Step 2: Check each option's syntax

   from scipy.stats import poisson uses correct syntax. Options A, B, and C have incorrect import statements.

3. Final Answer:

   from scipy.stats import poisson -> Option A

4. Quick Check:

   Correct import syntax = D [OK]

Hint: Use 'from scipy.stats import poisson' to import Poisson [OK]

Common Mistakes:

• Using wrong import order or keywords
• Trying to import from wrong submodules
• Using 'import' instead of 'from ... import ...'

3. What is the output of this code?

from scipy.stats import poisson
rv = poisson(mu=3)
prob = rv.pmf(2)
print(round(prob, 3))

medium

A. 0.2240

B. 0.224

C. 0.149

D. 0.050

Solution

1. Step 1: Understand the pmf function with mu=3 and k=2

   The pmf gives the probability of exactly k=2 events when the average rate mu=3.

2. Step 2: Calculate the Poisson pmf for k=2, mu=3

   Using formula: P(k=2) = (3^2 * e^-3) / 2! ≈ 0.224

3. Final Answer:

   0.224 -> Option B

4. Quick Check:

   Poisson pmf(2,3) ≈ 0.224 [OK]

Hint: Use pmf(k) with mu to get exact event probability [OK]

Common Mistakes:

• Confusing pmf with cdf
• Rounding incorrectly
• Using wrong mu or k values

4. Identify the error in this code snippet:

from scipy.stats import poisson
rv = poisson(4)
prob = rv.pmf(3)
print(prob)

medium

A. poisson is not imported correctly

B. pmf cannot be called on rv object

C. No error, code runs correctly

D. The parameter should be named mu=4, not just 4

Solution

1. Step 1: Check the poisson object creation

   poisson(4) creates a frozen distribution with mean 4. The parameter can be passed positionally.

2. Step 2: Verify pmf method usage

   Calling pmf(3) on the frozen distribution is correct and prints the probability of 3 events.

3. Final Answer:

   No error, code runs correctly -> Option C

4. Quick Check:

   Frozen distribution with positional mu works = A [OK]

Hint: Frozen distribution accepts positional mu parameter [OK]

Common Mistakes:

• Thinking mu must be named explicitly
• Assuming pmf is not a method of frozen distribution
• Believing import is incorrect without error

5. You want to find the probability of observing at most 2 events when the average rate is 1.5. Which scipy code correctly calculates this?

hard

A. poisson.cdf(2, mu=1.5)

B. poisson.pmf(2, mu=1.5)

C. poisson.sf(2, mu=1.5)

D. poisson.cdf(1.5, mu=2)

Solution

1. Step 1: Understand the problem requirement

   We want the probability of at most 2 events, which means P(X ≤ 2).

2. Step 2: Identify the correct scipy function

   poisson.cdf(k, mu) gives P(X ≤ k). So poisson.cdf(2, mu=1.5) is correct.

3. Final Answer:

   poisson.cdf(2, mu=1.5) -> Option A

4. Quick Check:

   At most k events = cdf(k) [OK]

Hint: Use cdf(k) for probability of up to k events [OK]

Common Mistakes:

• Using pmf instead of cdf for cumulative probability
• Confusing survival function (sf) with cdf
• Swapping mu and k parameters