# Day 6 - The Central Limit Theorem - Blog
Data Science for Political and Social Phenomena
Author: Guillaume Redoulès

# Day 6 - The Central Limit Theorem

## Problem 1

A large elevator can transport a maximum of \(9800\) kg. 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 \(\mu=205\) kg and a standard deviation of \(\sigma=15\) kg. Based on this information, what is the probability that all boxes can be safely loaded into the freight elevator and transported?

### Mathematical explanation

This problem can be tackled with the central limit theorem. Since the number of boxes is "large", the sum of the weight approaches normal distribution with : * \(\mu' = n\mu\) * \(\sigma'=\sigma\sqrt{n}\)

If we want to know the probability of the sum of the mass of all boxes to be under a certain weight, we can compute the cumulative density function :

$$P(x<9800) = F_X(9800)$$

    max_load = 9800
    n = 49
    mu = 205
    st_dev = 15

    import math

    def cumulative(x,mean,sd):
        return 0.5*(1+math.erf((x-mean)/(sd*math.sqrt(2))))

    mu_group = n*mu
    st_dev_group = st_dev*math.sqrt(n)

    print(f"Probability that all the boxes can be lifted by the elevator : {cumulative(max_load, mu_group, st_dev_group)}")

    Probability that all the boxes can be lifted by the elevator : 0.009815328628645315

## Problem 2

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 students will be able to purchase tickets?

### Mathematical explanation

We want to know if the sum of all the purchases will exceed the total supply of tickets. Since each student buy follows a Normal distribution and that the number of students is relatively high, the probability that all the students will be able to buy a ticket can be computed by applying the central limit theorem. The total number of tickets bought follows a normal distribution of mean \(\mu'=n*\mu\) and of standard deviation \(\sigma'=\sigma\sqrt{n}\)

    ticket_supply = 250
    n_students = 100
    mu = 2.4
    st_dev = 2

    mu_group = n_students*mu
    st_dev_group = st_dev*math.sqrt(n_students)

    print(f"Probability that all the students can purchase tickets : {cumulative(ticket_supply, mu_group, st_dev_group)}")

    Probability that all the students can purchase tickets : 0.691462461274013

## Problem 3

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 \(P(A<x<B)=0.95\). Use the value of \(z=1.96\). Note that is the z-score.

### Mathematical explanation

The margin of error can be computed with :

$$MoE = \frac{z-\sigma}{\sqrt{n}}$$

Knowing this, we can figure out what values for x paint the exact middle of the distribution for 0.95 probability, that means theres a 0.0025 leftover on both sides to make the total of 1.

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

    marginOfError = zScore * std / math.sqrt(n);
    print("A =",mean- marginOfError)
    print("B =",mean + marginOfError)

    A = 484.32
    B = 515.68

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