# Probability to z-score and vice versa

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### Question :

Probability to z-score and vice versa

How do I calculate the `z score` of a `p-value` and vice versa?

For example if I have a p-value of `0.95` I should get `1.96` in return.

I saw some functions in scipy but they only run a z-test on an array.

I have access to numpy, statsmodel, pandas, and scipy (I think).

``````>>> import scipy.stats as st
>>> st.norm.ppf(.95)
1.6448536269514722
>>> st.norm.cdf(1.64)
0.94949741652589625
``````

As other users noted, Python calculates left/lower-tail probabilities by default. If you want to determine the density points where 95% of the distribution is included, you have to take another approach:

``````>>>st.norm.ppf(.975)
1.959963984540054
>>>st.norm.ppf(.025)
-1.960063984540054
``````

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

It can be used to get the `zscore` for which x% of the area under a normal curve lies (ignoring both tails).

We can obtain one from the other and vice versa using the `inv_cdf` (inverse cumulative distribution function) and the `cdf` (cumulative distribution function) on the standard normal distribution:

``````from statistics import NormalDist

NormalDist().inv_cdf((1 + 0.95) / 2.)
# 1.9599639845400536
NormalDist().cdf(1.9599639845400536) * 2 - 1
# 0.95
``````

An explanation for the ‘(1 + 0.95) / 2.’ formula can be found in this wikipedia section.