use the normal distribution to approximate the desired probability.
find the probability that in 200 tosses we will obtain at least 40 fives.
Probability = N!/(r![N-r]!) * p^r * q^(n-r)
Probability = 200!/(40![200-40]!) * (1/6)^40 * (5/6)^(200-40)
Probability = 0.032865
To find the probability of obtaining at least 40 fives in 200 tosses, we can use the normal distribution approximation.
First, we need to calculate the mean (μ) and the standard deviation (σ) for the binomial distribution. In this case, the binomial distribution represents the probability of getting a five in a single toss, which is 1/6 since there are six possible outcomes (1, 2, 3, 4, 5, 6) and only one of them is a five. Thus, the mean (μ) for the binomial distribution is given by μ = n * p, where n is the number of trials (200 tosses) and p is the probability of success (1/6). The standard deviation (σ) is given by σ = sqrt(n * p * (1 - p)).
μ = 200 * (1/6) = 33.33
σ = sqrt(200 * (1/6) * (1 - 1/6)) = 5.57 (rounded to two decimal places)
Next, we can standardize the desired number of fives by converting it into a z-score. The z-score is calculated as (X - μ) / σ, where X is the desired number of fives.
For at least 40 fives, we need to find the probability P(X ≥ 40). Since the normal distribution is a continuous distribution, we can approximate the binomial distribution with the normal distribution using the Central Limit Theorem.
Using the z-score formula, the z-score is (40 - μ) / σ. Plugging in the values we calculated earlier, we have (40 - 33.33) / 5.57 = 1.2 (rounded to one decimal place).
To find the probability using the z-score, we can refer to a standard normal distribution table or use a statistical calculator. For a z-score of 1.2, the probability is approximately 0.8849 (rounded to four decimal places).
Therefore, the probability of obtaining at least 40 fives in 200 tosses is approximately 0.8849.