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The binomial distribution is a discrete probability distribution that models the number of successful outcomes in a fixed number of independent trials. Each trial has only two possible results: success or failure. In the context of the exercise, a "success" means an invoice with a first digit of 1 or 2.
Given that we analyze 90 invoices, we define our parameters as follows:

• Number of trials, \( n = 90 \)
• Probability of success for each trial, \( p = 0.477 \)

These parameters indicate that we are using a binomial distribution to find the likelihood of observing different counts of invoices with 1s or 2s. If the actual count deviates significantly from expectations, it might signal irregularities in the data.
The binomial distribution plays a crucial role in validating whether the data aligns with Benford's Law.

When working with large sample sizes, the binomial distribution can be approximated with a normal distribution. This is particularly helpful because it simplifies calculations, such as estimating probabilities. We apply this method to find the chance of seeing 29 or fewer invoices starting with a digit of 1 or 2.
We start by computing the mean \( \mu \) and standard deviation \( \sigma \) using the binomial distribution's characteristics:

• Mean: \( \mu = n \times p = 90 \times 0.477 = 42.93 \)
• Standard deviation: \( \sigma = \sqrt{n \times p \times (1-p)} \approx 4.737 \)

Then, we convert this into a standard normal distribution by determining the z-score, which lets us assess how extreme the observed result is relative to our expectations.

Hypothesis testing is a statistical method that allows us to make decisions using data. In our exercise, hypothesis testing helps us decide whether the invoices follow Benford's Law.
We set up two hypotheses:

• Null Hypothesis \( H_0 \): The invoices follow Benford's Law, with a digit probability of \( p = 0.477 \).
• Alternative Hypothesis \( H_a \): The invoices do not follow Benford's Law, indicating a different digit distribution.

Using the normal approximation, we calculated the z-score for the observed data (29 invoices) and found it to be \( z = -2.94 \). Referring to the z-table, we determined the probability (p-value) of observing such an extreme result if the null hypothesis were true, which was 0.0016.
With such a low p-value, it suggests that the observed pattern is highly unlikely under the null hypothesis, leading us to suspect possible non-genuine invoice data.