FAQ: Introduction to Probability Distributions - Using the Cumulative Distribution Function in Python

Frequently Asked Questions Python FAQ
1

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This community-built FAQ covers the “Using the Cumulative Distribution Function in Python” exercise from the lesson “Introduction to Probability Distributions”.

Paths and Courses
This exercise can be found in the following Codecademy content:

Master Statistics with Python

Probability

FAQs on the exercise Using the Cumulative Distribution Function in Python

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2

Why does the cdf show a slightly different output compared to the summed up pmf function while calculating the same probabilities? Shouldn’t they be the same?

prob_3 = stats.binom.cdf(5, 10, 0.5) - stats.binom.cdf(1, 10, 0.5)
print(prob_3)

## Output: 0.6123046874999999

print(stats.binom.pmf(2, n=10, p=.5) + stats.binom.pmf(3, n=10, p=.5) + stats.binom.pmf(4, n=10, p=.5) + stats.binom.pmf(5, n=10, p=.5))

## Output: 0.6123046875000006
1 Like
3

I think that’s just a limitation of floating point numbers and operations on them won’t give you perfect accuracy to an unlimited number of decimal places because there’s only a finite amount of space used to store the numbers.
(Calculating the cdf may also use different code than just the sum of the pmf internally.)

Although they’re both about 0.6123046875; that seems close enough.

1 Like
4

I just figured I should add this since it’s bugging me and perhaps anyone who reads this may find it useful. The exercise makes out like it is much harder to use the pmf for calculating the probability x falls within some range, but actually, I find it easier because you can simplify it by using the range() function and the .sum() method:

Suppose we want to calculate the probability that we observe between 2 and 7 heads. Using the cdf, we would write:

import scipy.stats as stats

prob_cdf = stats.binom.cdf(7, 10, 0.5) - stats.binom.cdf(1, 10, 0.5)
print(prob_cdf)
# output: 0.9345703125

Which is not too tedious. But using the pmf, we can just write:

import scipy.stats as stats

prob_pmf = stats.binom.pmf(range(2, 8), 10, 0.5).sum()
print(prob_pmf)
# output: 0.9345703125

This kind of seems more intuitive to me so I just thought I would share in case others see this and were thinking the same thing.

5

I think checkpoint 3 on this one is wrong. If you want to calculate the probability of flipping 5 or more heads. You’d want to subtract from 1, the probability of flipping 0 to 4 heads which would be stats.binom.cdf(4, 10, 0.5). The answer that they claimed is correct in 1 - stats.binom.cdf(5, 10, 0.5) is actually calculating the probability of flipping 6 or more heads. Am I crazy?

6

Step 3 states:

… the probability of observing more than 5 heads from 10 fair coin flips.

It is not asking for the probability of 5 or more heads.
Instead, it is asking for more than 5 heads i.e. 6 or more heads.

7

Ahh, gotcha! That makes more sense than.

Thank you.