Fred rolls a die 200 times and rolls a "5" 28 times. How unusual is freds experiment? Do you believe his die is fair?... extended ... what if fred had rolled the die 200 times and came up with a "5" only 20 times?

To determine how unusual Fred's experiment is and whether his die is fair, we can use probability theory. Let's break down the steps to calculate this:

Step 1: Calculate the expected number of "5" rolls
Assuming the die is fair, each side (including "5") should have an equal probability of appearing. Since there are 6 sides on a die, the expected number of "5" rolls in 200 rolls would be (1/6) * 200 = 33.33 (approximately).

Step 2: Calculate the standard deviation
To determine the range within which a fair die's rolls should fall, we need to calculate the standard deviation. The formula for the standard deviation of a binomial distribution is √(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.

In our case, n = 200 (number of rolls), p = 1/6 (probability of getting a "5"), and q = 1 - p = 5/6 (probability of not getting a "5").
√(200 * 1/6 * 5/6) ≈ √16.67 ≈ 4.08 (approximately).

Step 3: Calculate z-score
The z-score measures how many standard deviations away from the expected value Fred's result falls. It is calculated by subtracting the expected value from the observed value and dividing it by the standard deviation:
z = (observed value - expected value) / standard deviation.

For Fred's experiment where he rolled a "5" 28 times:
z = (28 - 33.33) / 4.08 ≈ -1.30 (approximately).

For Fred's second scenario where he rolled a "5" 20 times:
z = (20 - 33.33) / 4.08 ≈ -3.27 (approximately).

Step 4: Interpretation of z-score
The z-score allows us to determine how unusual Fred's results are. In a standard normal distribution (with mean 0 and standard deviation 1), a z-score of 2 or greater (or -2 or smaller) is considered statistically significant.

For Fred's experiment with 28 "5" rolls, with a z-score of -1.30, the result is not statistically significant and falls within the expected range.

For Fred's second scenario with only 20 "5" rolls, the z-score of -3.27 indicates a more significant deviation from the expected value, suggesting the result is unusual.

Based on these calculations, we cannot definitively conclude that Fred's die is unfair based solely on these experiments. However, the second scenario of 20 "5" rolls would raise more suspicion compared to the first experiment.