Use the normal dist to approx the desired prob. Find the prob of getting at most 30 fives in 200 tosses of a fair 6 sided die
Mean = np
SD = √npq
n = 200
p = 1/6
q = 1 - p = 5/6
Therefore:
Mean = 200(1/6) = 33.33
SD = √[200(1/6)(5/6)] = 5.27
Use z-scores:
z = (x - mean)/SD
z = (30 - 33.33)/5.27 = -0.63
Check a z-table for the probability.
(Remember the problem is asking "at most 30" when looking at the table.)
To find the probability of getting at most 30 fives in 200 tosses of a fair 6-sided die, we can approximate the desired probability using the normal distribution.
To start, we need to calculate the mean and standard deviation of the binomial distribution associated with the problem.
The mean (μ) of a binomial distribution is calculated as: n * p, where n is the number of trials (200 tosses) and p is the probability of success (getting a five on a fair 6-sided die). Since the die is fair, the probability of getting a five is 1/6, so p = 1/6.
μ = n * p = 200 * (1/6) = 33.33
The standard deviation (σ) of a binomial distribution is calculated as: √(n * p * (1 - p))
σ = √(n * p * (1 - p)) = √(200 * (1/6) * (1 - 1/6)) = √(200 * (1/6) * (5/6)) = 5.82
Now that we have the mean (μ) = 33.33 and the standard deviation (σ) = 5.82, we can use the normal distribution to approximate the probability.
To find the probability of getting at most 30 fives, we can convert it into a standard normal distribution by using the continuity correction, which involves subtracting 0.5 from the value we're interested in and adding 0.5 to the next highest integer.
Using the formula for the z-score:
z = (x - μ) / σ
For x = 30:
z = (30 - 33.33) / 5.82 = -0.5685
We want to find the probability of getting at most 30 fives, which is equivalent to finding P(X ≤ 30). For a standard normal distribution, we can use a z-table or a calculator to find the corresponding cumulative probability.
Using a z-table or calculator, we find that P(Z ≤ -0.5685) ≈ 0.2838.
Therefore, the probability of getting at most 30 fives in 200 tosses of a fair 6-sided die is approximately 0.2838.