16.2k views
3 votes
1200 dice are rolled. What is the approximate probability that the sum of dice is at most 4000? Explain your answer. Express your answer in terms of Φ(x)=∫−[infinity]x​2π​1​e−t2/2dt

1 Answer

3 votes

Answer:

To calculate the approximate probability that the sum of 1200 dice is at most 4000, you can use the Central Limit Theorem (CLT) since the sum of multiple dice rolls follows a binomial distribution. The CLT states that the distribution of the sum (or average) of a large number of independent, identically distributed random variables becomes approximately normal.

In this case, rolling a fair six-sided die follows a discrete uniform distribution, which can be approximated by a continuous normal distribution when the number of dice rolls is large. The mean (μ) of a single die roll is 3.5, and the variance (σ²) is approximately 2.917.

For the sum of 1200 dice rolls:

- Mean (μ_sum) = 1200 * μ = 1200 * 3.5 = 4200

- Variance (σ²_sum) = 1200 * σ² = 1200 * 2.917 ≈ 3500.4

Now, you want to find the probability that the sum is at most 4000, which can be expressed as P(X ≤ 4000), where X is the sum of 1200 dice rolls.

To use the normal distribution, you need to standardize the value using the Z-score formula:

Z = (X - μ_sum) / √(σ²_sum)

Z = (4000 - 4200) / √(3500.4)

Z ≈ -1.1304

Now, you can find the probability using the standard normal distribution function Φ(x):

P(X ≤ 4000) ≈ Φ(-1.1304)

You can look up the Z-score -1.1304 in a standard normal distribution table or use a calculator to find the cumulative probability.

P(X ≤ 4000) ≈ 0.1282

So, the approximate probability that the sum of 1200 dice rolls is at most 4000 is approximately 0.1282, or about 12.82%.

User Ramnath
by
7.9k points