How to calculate probability in a normal distribution given mean & standard deviation?

Question

How to calculate probability in normal distribution given mean, std in Python? I can always explicitly code my own function according to the definition like the OP in this question did: Calculating Probability of a Random Variable in a Distribution in Python

Just wondering if there is a library function call will allow you to do this. In my imagine it would like this:

nd = NormalDistribution(mu=100, std=12)
p = nd.prob(98)

There is a similar question in Perl: How can I compute the probability at a point given a normal distribution in Perl?. But I didn't see one in Python.

Numpy has a random.normal function, but it's like sampling, not exactly what I want.

Kongo

Answer 1

There's one in scipy.stats:

>>> import scipy.stats
>>> scipy.stats.norm(0, 1)
<scipy.stats.distributions.rv_frozen object at 0x928352c>
>>> scipy.stats.norm(0, 1).pdf(0)
0.3989422804014327
>>> scipy.stats.norm(0, 1).cdf(0)
0.5
>>> scipy.stats.norm(100, 12)
<scipy.stats.distributions.rv_frozen object at 0x928352c>
>>> scipy.stats.norm(100, 12).pdf(98)
0.032786643008494994
>>> scipy.stats.norm(100, 12).cdf(98)
0.43381616738909634
>>> scipy.stats.norm(100, 12).cdf(100)
0.5

[One thing to beware of -- just a tip -- is that the parameter passing is a little broad. Because of the way the code is set up, if you accidentally write scipy.stats.norm(mean=100, std=12) instead of scipy.stats.norm(100, 12) or scipy.stats.norm(loc=100, scale=12), then it'll accept it, but silently discard those extra keyword arguments and give you the default (0,1).]

Bathymetry

How would you get probabilities from ranges? Say from 98 - 102? – Automatize

@DSM: In your above example, when you say scipy.stats.norm(100, 12).pdf(98), does that mean the probability of getting 98 in a distribution with mean 100 and stddev 12 is 0.032 ? – Yamamoto

@ThePredator: no, the probability of getting 98 in a normal distribution with mean 100 and stddev 12 is zero. :-) The probability density is 0.032. – Bathymetry

Probability density in that case means the y-value, given the x-value 1.42 for the normal distribution. cdf means what we refer to as the area under the curve. – Strange

@Leon, that's rv.cdf(102) - rv.cdf(98) where rv = scipy.stats.norm(100, 12). – Evangel

@Bathymetry Hi. Can we also call this Likelihood of (mean=100, stddev=12|x=98)? – Cholon

Answer 2

Scipy.stats is a great module. Just to offer another approach, you can calculate it directly using

import math
def normpdf(x, mean, sd):
    var = float(sd)**2
    denom = (2*math.pi*var)**.5
    num = math.exp(-(float(x)-float(mean))**2/(2*var))
    return num/denom

This uses the formula found here: http://en.wikipedia.org/wiki/Normal_distribution#Probability_density_function

to test:

>>> normpdf(7,5,5)  
0.07365402806066466
>>> norm(5,5).pdf(7)
0.073654028060664664

Centralize

Hey, this is a really nice answer. Would you mind providing a step-by step explanation, perhaps? – Harp

This method needs less computation time than scipy – Apogeotropism

But scipy can handle arrays of means, stdevs and samples: mean = [ 5, 10, 20] stddev = [20, 30, 40] for x in ( [ 5, 10, 20], [10, 20, 40], [15, 30, 50], ): prob = scipy.stats.norm(mean, stddev).cdf(x) print(f'prob = {prob}') outputs: prob = [0.5 0.5 0.5] prob = [0.59870633 0.63055866 0.69146246] prob = [0.69146246 0.74750746 0.77337265] – Amphibrach

Answer 3

Here is more info. First you are dealing with a frozen distribution (frozen in this case means its parameters are set to specific values). To create a frozen distribution:

import scipy.stats
scipy.stats.norm(loc=100, scale=12)
#where loc is the mean and scale is the std dev
#if you wish to pull out a random number from your distribution
scipy.stats.norm.rvs(loc=100, scale=12)

#To find the probability that the variable has a value LESS than or equal
#let's say 113, you'd use CDF cumulative Density Function
scipy.stats.norm.cdf(113,100,12)
Output: 0.86066975255037792
#or 86.07% probability

#To find the probability that the variable has a value GREATER than or
#equal to let's say 125, you'd use SF Survival Function 
scipy.stats.norm.sf(125,100,12)
Output: 0.018610425189886332
#or 1.86%

#To find the variate for which the probability is given, let's say the 
#value which needed to provide a 98% probability, you'd use the 
#PPF Percent Point Function
scipy.stats.norm.ppf(.98,100,12)
Output: 124.64498692758187

Incorporeity

I can't thank enough whoever wrote this answer. I was looking everywhere to solve this but couldn't able to find it. And adding the comments with the code really helped me understand what is happening. Thanks a lot. – Submaxillary

Just want to ask one question, how to calculate these probabilities when the data is not normally distributed? What I have to do in this case? – Submaxillary

Answer 4

Starting Python 3.8, the standard library provides the NormalDist object as part of the statistics module.

It can be used to get the probability density function (pdf - likelihood that a random sample X will be near the given value x) for a given mean (mu) and standard deviation (sigma):

from statistics import NormalDist

NormalDist(mu=100, sigma=12).pdf(98)
# 0.032786643008494994

Also note that the NormalDist object also provides the cumulative distribution function (cdf - probability that a random sample X will be less than or equal to x):

NormalDist(mu=100, sigma=12).cdf(98)
# 0.43381616738909634

Philosophism

This is the best answer because it uses the native library. Not everyone wants to use scipy – Charleencharlemagne

Answer 5

In case you would like to find the area between 2 values of x mean = 1; standard deviation = 2; the probability of x between [0.5,2]

import scipy.stats
scipy.stats.norm(1, 2).cdf(2) - scipy.stats.norm(1,2).cdf(0.5)

Wed

Thanks - this formula is very hard to find online, but very useful. – Columbite

Answer 6

Note that probability is different than probability density pdf(), which some of the previous answers refer to. Probability is the chance that the variable has a specific value, whereas the probability density is the chance that the variable will be near a specific value, meaning probability over a range. So to obtain the probability you need to compute the integral of the probability density function over a given interval. As an approximation, you can simply multiply the probability density by the interval you're interested in and that will give you the actual probability.

import numpy as np
from scipy.stats import norm

data_start = -10
data_end = 10
data_points = 21
data = np.linspace(data_start, data_end, data_points)

point_of_interest = 5
mu = np.mean(data)
sigma = np.std(data)                                   
interval = (data_end - data_start) / (data_points - 1)
probability = norm.pdf(point_of_interest, loc=mu, scale=sigma) * interval

The code above will give you the probability that the variable will have an exact value of 5 in a normal distribution between -10 and 10 with 21 data points (meaning interval is 1). You can play around with a fixed interval value, depending on the results you want to achieve.

Mosemoseley

I think the questioner is referring to "likelihood" instead of real "probability". – Allier

Answer 7

The formula cited from wikipedia mentioned in the answers cannot be used to calculate normal probabilites. You would have to write a numerical integration approximation function using that formula in order to calculate the probability.

That formula computes the value for the probability density function. Since the normal distribution is continuous, you have to compute an integral to get probabilities. The wikipedia site mentions the CDF, which does not have a closed form for the normal distribution.

Medrano

Thank you for your contribution, although it would fit better as a comment to the answer you are referring at: if I understand well, you aren't really answering to the original question. This way, everyone will see at a first glance what you are talking about. – Ha

Answer 8

I would like to say: the questioner is asking "How to calculate the likelihood of a given data point in a normal distribution given mean & standard deviation?" instead of "How to calculate probability in a normal distribution given mean & standard deviation?".

For "probability", it must be between 0 and 1, but for "likelihood", it must be non-negative (not necessarily between 0 and 1).

You could use multivariate_normal.pdf(x, mean= mean_vec, cov=cov_matrix) in scipy.stats.multivariate_normal to calculate it.

Allier

Answer 9

I wrote this program to do the math for you. Just enter in the summary statistics. No need to provide an array:

One-Sample Z-Test for a Population Proportion:

To do this for mean rather than proportion, change the formula for z accordingly

EDIT:
Here is the content from the link:

import scipy.stats as stats
import math

def one_sample_ztest_pop_proportion(tail, p, pbar, n, alpha):
    #Calculate test stat

    sigma = math.sqrt((p*(1-p))/(n))
    z = round((pbar - p) / sigma, 2)

    if tail == 'lower':
        pval = round(stats.norm(p, sigma).cdf(pbar),4)
        print("Results for a lower tailed z-test: ")


    elif tail == 'upper':
        pval = round(1 - stats.norm(p, sigma).cdf(pbar),4)
        print("Results for an upper tailed z-test: ")


    elif tail == 'two':
        pval = round(stats.norm(p, sigma).cdf(pbar)*2,4)
        print("Results for a two tailed z-test: ")


    #Print test results
    print("Test statistic = {}".format(z))   
    print("P-value = {}".format(pval))
    print("Confidence = {}".format(alpha))

    #Compare p-value to confidence level
    if pval <= alpha:
        print("{} <=  {}. Reject the null hypothesis.".format(pval, alpha))
    else:
        print("{} > {}. Do not reject the null hypothesis.".format(pval, alpha))


#one_sample_ztest_pop_proportion('upper', .20, .25, 400, .05)

#one_sample_ztest_pop_proportion('two', .64, .52, 100, .05)

Homologate

While the link might provide a valuable answer, SO asks users to post their code here on SO Links are useful as a reference, but they tend to break after a while, making solutions inaccessible for future visitors. – Aurify

Answer 10

You can just use the error function that's built in to the math library, as stated on their website.

Rudnick

Nope @user2340146, also in Python 2: docs.python.org/2/library/math.html#math.erf – Scolex