1. Concepts & Definitions
1.1. Continous random distribution of probability
1.2. Normal distribution of probability
1.3. Standard normal distribution of probability
1.4. Inverse standard normal distribution
1.6. Inverse Student's T distribution
2. Problem & Solution
2.1. Weight, dimension, and value per HS6
2.2. How to fit a distribution
2.3. Employing standard deviation
2.4. Total time spent in a system
The next figure gives an example of how to find inverse standard normal distribution. This means how to find interval limits given the total percentage P(a ≤ z ≤ b) of the population that should be covered by it.
Examples of finding the following quantities:
• Find interval limits [a, b] with that holds 90% of the values that follow a normal distribution with parameters N(0, 1).
Another way to see this question is that the b upper limit of the interval [a, b] is chosen in a manner that P(z ≤ b) = 1 − α/2 = 0.95 holds as shown in the next figure.
According to previous deductions for a standard normal distribution, the following holds: P(−1.6449 ≤ z ≤ 1.6449) = 0.90.
Find the following quantities:
• Find interval limits [a, b] with that holds 90% of the values that follow a normal distribution with parameters N(40, 8).
• According to previous deductions for a standard normal distribution the following holds: P(−1.6449 ≤ z ≤ 1.6449) = 0.90. To find the interval that encompasses 90% of the values for the normal distribution with parameters N(40, 8):
– For z = −1.6449: x = μ + zσ = 40 − 1.6449·8 = 26.84.
– For z = 1.6449: x = μ + zσ = 40 + 1.6449·8 = 53.16.
The interval must be [26.84, 53.16].
• Find interval limits with 90% of the values for the normal distribution with parameters N(40, 1).
• To find the interval that encompasses 90% of the values for the normal distribution with parameters N(40, 1):
– For z = −1.6449: x = μ + zσ = 40 − 1.6449 · 1 = 38.36.
– For z = 1.6449: x = μ + zσ = 40 + 1.6449 · 1 = 41.64.
The new interval will be significantly smaller than the previous one: [38.36, 41.64].
The following code shows the probability of the numerical example employing PPF as inverse of Normal distribution using norm.cdf command.
from scipy.stats import norm
prob = 0.90
half_alpha = (1 - prob)/2
mean = 0
std = 1
b = norm.ppf(1-half_alpha, loc=mean, scale=std)
a = -b
print('[',a,', ',b,']')
[ -1.6448536269514722 , 1.6448536269514722 ]
Now, it is possible to estimate the interval of values considering normal distribution as N(40, 8).
mu = 40
std = 8
xb = mu + std*b
xa = mu + std*a
print('[',xa,', ',xb,']')
[ 26.84117098438822 , 53.15882901561178 ]
Now, it is possible to estimate the interval of values considering normal distribution as N(40, 1).
mu = 40
std = 1
xb = mu + std*b
xa = mu + std*a
print('[',xa,', ',xb,']')
[ 38.35514637304853 , 41.64485362695147 ]
The previous complete code is available in the following link:
https://colab.research.google.com/drive/1pFXIM0oCkKwDZmRFED2T8SuzMXEYSxFa?usp=sharing