# Central Limit Theorem

The *central limit theorem* (CLT) states that, for a large enough sample ($$n$$), the distribution of the sample mean will approach normal distribution. This holds for a sample of independent random variables from any distribution with a finite standard deviation. 

Let $${X\_1, X\_2, X\_3,...,X\_n}$$ be a random data set of size $$n$$, that is, a sequence of independent and identically distributed random variables drawn from distributions of expected values given by $$\mu$$ and finite variances given by $$\sigma^2$$. The sample average is:

$$s\_n:=\frac{\sum\_i X\_i}{N}$$

For large $$n$$, the distribution of sample sums $$S\_n$$ is close to normal distribution $$N(\mu^\prime,\sigma^\prime)$$ where: 

* $$\mu^\prime=n \times \mu$$
* $$\sigma^\prime=\sqrt{n} \times \sigma$$

**Task** \
A large elevator can transport a maximum of $$9800$$ pounds. Suppose a load of cargo containing $$49$$ boxes must be transported via the elevator. The box weight of this type of cargo follows a distribution with a mean of $$\mu=205$$ pounds and a standard deviation of $$\sigma=15$$ pounds. Based on this information, what is the probability that all $$49$$ boxes can be safely loaded into the freight elevator and transported?

```
import math

def less_than_boundary_cdf(x, mean, std):
    return round(0.5 * (1 + math.erf((x - mean)/ (std * math.sqrt(2)))), 4)

m = int(input())
n = int(input())
mean = int(input())
devi = int(input())

print(less_than_boundary_cdf(m, n * mean, math.sqrt(n) * devi))
```

**Task** \
The number of tickets purchased by each student for the *University X vs. University Y* football game follows a distribution that has a mean of $$\mu = 2.4$$ and a standard deviation of $$\sigma = 2.0$$.

A few hours before the game starts, $$100$$ eager students line up to purchase last-minute tickets. If there are only $$250$$ tickets left, what is the probability that all $$100$$ students will be able to purchase tickets?

```
import math
def less_than_boundary_cdf(x, mean, std):    
    return round(0.5 * (1 + math.erf((x - mean)/ (std * math.sqrt(2)))), 4)
    
m = int(input())
n = int(input())
mean = float(input())
devi = float(input())

print(less_than_boundary_cdf(m, n * mean, math.sqrt(n) * devi))
```

**Task** \
You have a sample of $$100$$ values from a population with mean $$\mu=500$$ and with standard deviation $$\sigma=80$$. Compute the interval that covers the middle $$95%$$ of the distribution of the sample mean; in other words, compute $$A$$ and $$B$$ such that $$P(A\<x\<B)$$. Use the value of $$z=1.96$$. Note that $$z$$ is the [z-score](https://en.wikipedia.org/wiki/Standard_score).

```
import math

zScore = 1.96
std = 80
n = 100
mean = 500

marginOfError = zScore * (std / math.sqrt(n));
print(mean - marginOfError)
print(mean + marginOfError)
```

The **marginOfError** formula can be found [here](https://www.thoughtco.com/margin-of-error-formula-3126275).

$$\huge E =z\_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$


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