# Statistics challenge and solution

June 01, 2017 3 minutes reading time Math

This is a statistic challenge that should be solvable for business analysts. For reference, an example solution is provided. The topic is not part of a mathematics advanced course in most federal states of Germany, which is not because it’s difficult - on the contrary. The application of the central limit theorem of statistics (CLT) can be helpful in many decision-making processes.

When I was still using Facebook, I wrote it to mock those who posted math riddles. What I learned: they can’t be mocked - at least not with this.

## Challenge

You work for an international logistics company. Your company should ship urgent cargo from Munich to Hamburg, 36 boxes from one of your regular customers, but he does not know the weight of the boxes. Historical records show that the customer’s boxes weigh 72 kg on average, with a standard deviation of 3 kg. The only aircraft currently available can hold cargo up to a maximum of 2,630 kg.

Your boss wants to know from you: what is the probability that the cargo can be transported?

## Solution

The solution is based on the central limit theorem. It states that if we have a sufficiently large sample $\bar{x}$ of size ${n}$, we can expect the mean $\mu_ {\bar{x}}$ of this sample to be equal to the average value $\mu_ {x}$ of the customer’s historical data ${x}$, and the relation as described under $(2)$ exists between the standard deviation $\sigma_ {x}$ of the customer’s historical data and the standard deviation of the sample $\sigma_ {\bar{x}}$.

In addition, the probability density function of the means $\mu_ {\bar{x}}$ of enough samples $\bar{x}$ of the size ${n}$ approximates the normal distribution $\mathcal{N}(\mu_ {\bar{x}}, \sigma_ {\bar{x}}^{2})$.

This allows us to use the standard normal distribution $\mathcal{N}(0, 1)$ by normalizing. There exist lookup tables for the standard normal distribution, which saves the calculation.

• Since the aircraft takes a maximum of 2,630 kg of cargo, there is a critical mass that must not be exceed for the average of the boxes:
$(1)$ $$x_{crit} = \frac{2,630\ kg}{36} = 73.06\ kg$$

• For the mean values ​​and standard deviations:
$(2)$ $$\mu_ {\bar{x}} = \mu_ {x} = 72\ kg, \qquad \sigma_ {\bar{x}} = \frac{\sigma_ {x}}{\sqrt{n}} = \frac{3\ kg}{\sqrt{36}} = 0.5\ kg$$

• We can get the ${z}$ value for our $x_{crit}$ in $(1)$ from the mean $\mu_ {\bar{x}}$ and the standard deviation $\sigma_ {\bar{x}}$ in $(2)$:
$(3)$ $$z = \frac{x_{crit} - \mu_ {\bar{x}}}{\sigma_ {\bar{x}}} = \frac{73.06\ kg - 72\ kg}{0.5\ kg} = 2.12$$

The ${z}$ value is our ${x_{crit}}$ converted to its equivalent in the standard normal distribution. With this value, we can look up the cumulative probability in the ${z}$ table. We get $p(2.12) = 0.983$.

This corresponds to a probability of 98.3% that we have a sample whose mean is 73.06 kg or less.