On a fair 6-sided die, each number has an equal probability p of being rolled. when a fair die is rolled n times, the most likely outcome (the mean) is that each number will be rolled NP times, with a standard deviation of sqrt NP(1-P). Brandon rolls a die 200 times. He will conclude that the die is loaded (unfair) if the number of times any number is rolled is outside 1.5 standard deviations of the mean. What are the minimum and maximum number of times a number can be rolled without brandon concluding that the die is loaded.

Not sure if I am on the right track or not, But I concluded that the deviation is sqrt 33.3 (1-1/6)=5.27

With that 5.27 I multiplied by 1.5 and -1.5 to get +-7.9

With that I took 33.3-7.9 for minimum which was 25.4 and maximum I took 33.3+7.9 which gave me 41.2.

Please let me know if there are any errors with my method.

Lincoln rolled a die 40 times and 12 of the 40 rolls came up as a six. He wanted to see how likely a result of 12 sixes in 40 rolls would be with a fair die, so Lincoln used a computer simulation to see the proportion of sixes in 40 rolls, repeated 200 times. Based on the results of the simulation, what inference can Lincoln make regarding the fairness of the die?

Proportion of Sixes Rolled
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35

(Note: don't use percents as a sample proportion.)

Since the simulation shows
with a sample proportion of
or
, we can conclude that

the result of 12 sixes in 40 rolls is not particularly unlikely with a fair die. Therefore, Lincoln cannot confidently conclude that the die is unfair based on this result alone. He would need more evidence or data to make a stronger inference.

Your method is mostly correct, but there is a small error in your calculations. Let's go through the steps again to find the minimum and maximum number of times a number can be rolled without Brandon concluding that the die is loaded.

First, let's calculate the standard deviation with the given formula: sqrt(NP(1-P)). In this case, p = 1/6 and n = 200.

Standard deviation = sqrt(200 * 1/6 * (1 - 1/6))
= sqrt(200 * 1/6 * (5/6))
= sqrt(1000/36)
≈ 5.27 (rounded to two decimal places)

Now, we can calculate the 1.5 standard deviations: 1.5 * 5.27 ≈ 7.91 (rounded to two decimal places).

To find the minimum and maximum number of times a number can be rolled without Brandon concluding that the die is loaded, we subtract and add the 1.5 standard deviations from the expected mean.

Minimum = NP - 1.5 * sqrt(NP(1-P))
= 200 * 1/6 - 1.5 * 5.27
≈ 33.3 - 7.91
≈ 25.39

Maximum = NP + 1.5 * sqrt(NP(1-P))
= 200 * 1/6 + 1.5 * 5.27
≈ 33.3 + 7.91
≈ 41.21

So, the minimum number of times a number can be rolled without Brandon concluding that the die is loaded is approximately 25.39, and the maximum number is approximately 41.21.