In his best year, Mickey Mantles batting average was o.365. In that year, what was the probability that he would get exactly 5 hits in 15 times at bat? Compare the results from the normal approximation with the results from a calculation using a binomial distribution
To find the probability that Mickey Mantle would get exactly 5 hits in 15 times at bat, we can approach it using both the normal approximation and the binomial distribution.
1. Using the normal approximation:
To use the normal approximation, we need to assume that the number of hits follows a normal distribution, which is a reasonable assumption for large sample sizes. We also need to calculate the mean and standard deviation.
Mean (μ) = n * p
(where n is the number of trials and p is the probability of success)
In this case, n = 15 (times at bat) and p = batting average = 0.365.
μ = 15 * 0.365 = 5.475 (approximately)
Standard Deviation (σ) = √(n * p * (1 - p))
σ = √(15 * 0.365 * (1 - 0.365)) = √(15 * 0.365 * 0.635) ≈ 1.517 (approximately)
Now, we can use the normal distribution to calculate the probability of exactly 5 hits. However, since we want the probability of a discrete event (5 hits) rather than a continuous range, we will use a continuity correction.
P(X = 5) ≈ P(4.5 < X < 5.5)
≈ P((4.5 - μ) / σ < (X - μ) / σ < (5.5 - μ) / σ)
≈ P((4.5 - 5.475) / 1.517 < X < (5.5 - 5.475) / 1.517)
You can then use a standard normal distribution table or a calculator to find the probability associated with these z-scores.
2. Using the binomial distribution:
Alternatively, we can directly calculate the probability using the binomial distribution formula.
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where (n C k) represents the number of combinations of n items taken k at a time.
In this case, n = 15, k = 5, and p = 0.365.
P(X = 5) = (15 C 5) * 0.365^5 * (1 - 0.365)^(15 - 5)
You can calculate this using a calculator or by hand to find the exact probability.
Comparing the results from both approaches will give you an idea of how well the normal approximation works compared to the exact binomial distribution.