a fair dice is tossed 100 times. use equation 5.20 to bound the probability that the total number of spots is between 300 and 400

p[Mn-mui<abselone]>=
1- sigma square/n(abselone)square

To use equation 5.20 to bound the probability, we need to determine the mean and variance of the total number of spots.

1. Mean (μ):
For a fair die, the mean number of spots on a single toss is ((1+2+3+4+5+6)/6) = 3.5. Since we are tossing the dice 100 times, the mean of the total number of spots (μ) can be calculated as 100 * 3.5 = 350.

2. Variance (σ^2):
The variance of a single toss is calculated as [(1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2]/6 = 2.92. Since we are tossing the dice 100 times, the variance of the total number of spots (σ^2) can be calculated as 100 * 2.92 = 292.

Now, substituting the mean and variance values into equation 5.20, we get:

P[Mn - μ < |ε|] >= 1 - (σ^2 / n) * (|ε|^2)

Given that we want to bound the probability that the total number of spots is between 300 and 400, we can rephrase it as:

P[300 ≤ Mn ≤ 400] = P[Mn - μ < 50] - P[Mn - μ < -50]

Plugging in the values, we get:

P[Mn - 350 < 50] - P[Mn - 350 < -50] >= 1 - (292 / 100)*(50^2) - 1 - (292 / 100)*(50^2)

Simplifying the equation further, we get:

P[Mn - 350 < 50] - P[Mn - 350 < -50] >= 1 - (292 / 100)*(50^2) - 1 - (292 / 100)*(50^2)

Finally, calculate the probability by subtracting the two values:

P[Mn - 350 < 50] - P[Mn - 350 < -50] >= 1 - [ (292 / 100)*(50^2) + (292 / 100)*(50^2) ]