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The binomial distribution is a probability distribution. It helps you find the probability of a given number of successes from a fixed number of trials. Each trial has two possible outcomes: success or failure. The trials must be independent, meaning the outcome of one trial does not affect the others. For instance, consider flipping a coin. Getting heads might be considered a success, and tails a failure. If you flip the coin 10 times, the binomial distribution can tell you the likelihood of getting exactly 6 heads.

When working with large sample sizes in a binomial distribution, calculations can be tedious. To simplify, we can use the normal approximation. This method transforms the binomial distribution into a normal distribution, which is easier to handle. But, before using this method, certain conditions must be met: the product of the number of trials (n) and probability of success (p) should both be greater than 5. This ensures that the distribution is roughly symmetric and bell-shaped—for these exercises, both criteria were satisfied. In our problem, with n=80 and p=0.15, calculating np and n(1-p) gives 12 and 68. Both are greater than 5, hence the normal approximation can be applied.

The binomial coefficient in the binomial probability formula helps in finding the number of ways to choose k successes out of n trials. It is denoted as \( \binom{n}{k} \) and calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In our problem, \( n = 80 \) and \( X = 18 \). Hence, the binomial coefficient becomes \( \binom{80}{18} \). It calculates as: \( \frac{80!}{18!62!} \). factorials can lead to large numbers and are best computed using a calculator or software to avoid manual errors.

The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. This distribution is useful in converting values from any normal distribution into a common scale, called the Z-score. The Z-score tells you how many standard deviations a data point is from the mean. For example, in our exercise, we converted X = 18 to the Z-score using the formula: \( Z = \frac{X - \text{µ}}{\text{σ}} \). With mean \( \text{µ} = 12 \) and standard deviation \( \text{σ} \approx 3.182 \), Z became \( \frac{18 - 12}{3.182} \approx 1.886 \). We then used standard normal distribution tables to find probabilities corresponding to this Z-score.