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 of Mickey Mantle getting exactly 5 hits in 15 times at bat in his best year, we can use both the normal approximation and the binomial distribution.
1. Normal Approximation:
Step 1: Calculate the mean (μ) and the standard deviation (σ).
The mean (μ) is the product of the batting average (0.365) and the number of at-bats (15):
μ = 0.365 * 15 = 5.475
The standard deviation (σ) is the square root of the product of the batting average (0.365), the complementary probability (1 - 0.365), and the number of at-bats (15):
σ = sqrt(0.365 * (1 - 0.365) * 15) ≈ 1.294
Step 2: Convert the discrete number of hits (5) into a continuous z-score.
The z-score is calculated by subtracting the mean (μ) from the desired value (5 hits) and dividing the result by the standard deviation (σ):
z = (5 - μ) / σ = (5 - 5.475) / 1.294 ≈ -0.367
Step 3: Find the probability using a standard normal distribution table.
Using the z-score calculated in Step 2, find the corresponding probability from a standard normal distribution table. In this case, look for the probability of z ≤ -0.367. Let's assume this probability is P_n.
2. Binomial Distribution:
The binomial distribution can be used to calculate the probability of exactly k successes in n independent trials, given a probability p of success in each trial.
In this case, we have n = 15 (number of at-bats) and p = 0.365 (batting average). We want to find the probability of exactly k = 5 hits.
The binomial formula for calculating this probability is:
P(k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
- nCk is the number of combinations of n items taken k at a time.
- p^k is the probability of k successes.
- (1 - p)^(n - k) is the probability of n - k failures.
Using this formula, calculate P(k) where k = 5.
3. Comparison:
To compare the results from the normal approximation with the binomial distribution, compare the probability P_n (from the normal approximation) with P(k) (from the binomial distribution).
If the values are close, the normal approximation can be considered a reasonable estimate of the binomial probability. If they are significantly different, it suggests that the binomial distribution should be used instead.
Hope this helps explain how to calculate the probability and compare the results using both the normal approximation and the binomial distribution!