# Normal Distribution

#### [Normal Distribution](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Normal_distribution#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](https://en.wikipedia.org/wiki/Normal_distribution#Cumulative_distribution_function)

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:

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

```python
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?

```python
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 $$\text{grade}<60$$)?

```python
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))
```


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