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# Binompdf and binomcdf functions

Using a TI-84 (very similar for TI-85 or TI-89) calculator for making calculations regarding binomial random variables.

## Want to join the conversation?

• I'm curious if there is a hand written formula for cumulative binomial equations?
• Well, yes and no. The binomcdf formula is just the sum of all the binompdf up to that point (unfortunately no other mathematical shortcut to it, from what I've gathered on the internet). So you can't just calculate on paper for large values.

I looked into this specifically because I don't have a graphing calculator and I'll have to write a program into my Casio-FX3650pII which will include a for loop that adds up all the binompdfs up to X. I'm hoping that the 350 byte storage my calculator has will be sufficient for that program! :D

Good luck.
• problem is that I have no idea how to play baseball..... :-(
• You can also use scipy library to do so:
scipy.stats import binom
binom.pmf(4,7,0.35)
binom.cdf(4,7,0.35)
• True...but you would need to know Python to utilize the library. SciPy is a powerful library, among others, used in Machine Learning.
(1 vote)
• What function would I use for something like P(X>8)?
(1 vote)
• Note that the event that X <= 8 is the complement of the event that X > 8.

Therefore, P(X > 8) = 1 - P(X <= 8) = 1 - binomcdf(n, p, 8), where n is the number of trials and p is the probability of success on each trial.

Have a blessed, wonderful day!
• The app Calc84 is a pretty good substitute that seems to work out 'of the box'. I like having it offline rather than relying on internet access.
(1 vote)
• In case of BinomialCDF, I tried solving this problem manually. Because P(X<=4), therefore, I calculated P(X=1)+ P(X=2)+P(X=3)+P(X=4) = 0.89537 which does not match with your result.

Then I calculated P(X=1)+ P(X=2)+P(X=3)+P(X=4)+P(X=5) = 0.94197 which also does not match with your result.

Am I doing it wrong? Please guide.
(1 vote)
• Don't forget to include 𝑝(0), the probability of missing all seven free throws.

– – –

The probability of making exactly 𝑛 free throws is
𝑝(𝑛) = 7!∕(𝑛!(7 − 𝑛)!) ∙ 0.35^𝑛 ∙ (1 − 0.35)^(7 − 𝑛)

The probability of making at most four free throws is
𝑝(0) + 𝑝(1) + 𝑝(2) + 𝑝(3) + 𝑝(4)
• Is there a binomcdf calculator trick or something to do this for questions that want us to find the probability that is greater or more than x without having to find the binompdf of each number and adding them up?
• what can u do if you want to find at most? instead of less than 5 could you do more than 5?
• from a graphical standpoint, what does p(x=4) represent versus p(x<=4)? (assuming p = 0.35 for each trial and we still have 7 trials)