Final answer:
The probability that the total number of dots after rolling a fair die 20 times falls between 60 and 80 can be bounded using Chebyshev's inequality. After calculation, it is found that the bound on the probability is at least 99.854%.
Step-by-step explanation:
The question is asking us to use the Chebyshev's inequality to provide a bound on the probability that the total number of dots after rolling a fair die 20 times falls between 60 and 80. Chebyshev's inequality is given by the formula P[|M₍ - m| < ∅] ≥ 1 - σ²/n∅² where M₍ is the sample mean, m is the population mean, σ is the population standard deviation, n is the sample size, and ∅ is the margin of error. The population mean (m) of rolling a die is 3.5, and the population standard deviation (σ) is approximately 1.71. Using a sample size (n) of 20 rolls, we want to find the probability that the sum of the rolls (M₍) is between 60 and 80.
Firstly, we calculate the expected sum after 20 rolls. The expected sum is n times m which equals to 20 * 3.5 = 70. Now, let's use this as the center of our range for the calculation of ∅. Since we want the range between 60 and 80, the margin ∅ should be half of the range width, which means ∅ = (80 - 60)/2 = 10. Thus, we will have P[|M₍ - 70| < 10] ≥ 1 - σ²/n∅². Plugging in the numbers, we get P[|M₍ - 70| < 10] ≥ 1 - (1.71²/20*10²) = 1 - (2.92/2000) = 1 - 0.00146 = 0.99854. Therefore, the bound on the probability is at least 99.854%.