Normal Distribution

Normal Distribution

The probability density of normal distribution is:

$\huge N(\mu, \sigma^2) = \frac{1}{\sigma \sqrt{2\pi}}e^{- \frac{(x-\mu)^2}{2\sigma^2}}$

Here,

• $\mu$ is the mean (or expectation) of the distribution. It is also equal to median and mode of the distribution.

• $\sigma^2$ is the variance.

• $\sigma$ is the standard deviation.

Standard Normal Distribution

If $\mu = 0$ and $\sigma =1$, then the normal distribution is known as standard normal distribution: $\large \phi(x) = \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$

Every normal distribution can be represented as standard normal distribution:

$\large N(\mu, \sigma^2) = \frac{1}{\sigma}\phi(\frac{x-\mu}{\sigma})$

Cumulative Probability

Consider a real-valued random variable, $X$. The cumulative distribution function of $X$ (or just the distribution function of $X$) evaluated at $x$ is the probability that $X$ will take a value less than or equal to $x$:

$F_X(x) = P(X\leq x)$

Also,

$P(a \leq X \leq b) = P(a < X < b) = F_X(b) - F_X(a)$

The cumulative distribution function for a function with normal distribution is:

$\huge \Phi(x) = \frac{1}{2}(1+\text{erf}(\frac{x-\mu}{\sigma\sqrt2}))$

phi = lambda x: 0.5 * (1 + math.erf((x - mean)/(std * (2 ** 0.5))))

Where $\text{erf}$ is the function:

$\large \text{erf}(z) = \frac{2}{\sqrt\pi}f_0^2 e^{-x^2} dx$

Task In a certain plant, the time taken to assemble a car is a random variable, $X$, having a normal distribution with a mean of hours and a standard deviation of $2$ hours. What is the probability that a car can be assembled at this plant in:

1. Less than $19.5$ hours?
2. Between $20$ and $22$ hours?

import math
mean, std = 20, 2
cdf = lambda x: 0.5 * (1 + math.erf((x - mean) / (std * (2 ** 0.5))))

# Less than 19.5
print('{:.3f}'.format(cdf(19.5)))
# Between 20 and 22
print('{:.3f}'.format(cdf(22) - cdf(20)))

Task The final grades for a Physics exam taken by a large group of students have a mean of $\mu = 70$ and a standard deviation of $\sigma = 10$. If we can approximate the distribution of these grades by a normal distribution, what percentage of the students:

1. Scored higher than $80$ (i.e., have a $\text{grade}>80$)?
2. Passed the test (i.e., have a $\text{grade}\geq60$)?
3. Failed the test (i.e., have a $\text{grade}<60$)?

import math

def arg_erf(x, u, o):
    return (x-u)/(o * math.sqrt(2))

def F(x, arg_x):
    return 0.5 * ( 1 + math.erf(arg_x) )

u, o = input().split() # u = 70, o = 10
u, o = float(u), float(o)

a = float(input())   # a = 80
b = float(input())   # b = 60

arg_a = arg_erf(a, u, o)
arg_b = arg_erf(b, u, o)

first_ans = (1 - F(a, arg_a))
second_ans = (1 - F(b, arg_b))
third_ans = F(b, arg_b)

print(round(first_ans*100.0, 2))
print(round(second_ans*100.0, 2))
print(round(third_ans*100.0, 2))