91Ó°ÊÓ

Understanding the concepts of mean and standard deviation is crucial when working with binomial distributions. The mean, often symbolized as \( \mu \), represents the expected number of successes in a set of trials. For a binomial distribution, it is computed using the formula \( \mu = n \times p \). Here, \( n \) is the total number of trials, and \( p \) is the probability of a success on each trial. In our exercise, with \( n = 100 \) and \( p = 0.20 \), the mean comes out to be 20. This means that out of 100 trials, we expect around 20 successful outcomes.

The standard deviation, denoted as \( \sigma \), measures how much the actual number of successes is expected to vary from the mean. The formula for the standard deviation in a binomial distribution is \( \sigma = \sqrt{n \times p \times (1 - p)} \). By substituting \( n = 100 \), \( p = 0.20 \), we find \( \sigma = \sqrt{16} = 4 \). This value indicates a typical deviation of 4 successes from the mean under repeated trials, providing an insight into the distribution's spread.

Normal approximation is a method used to simplify calculations when dealing with a large number of trials in a binomial distribution. This technique is valid when both \( np \) and \( n(1-p) \) exceed 5. This criterion ensures that the distribution of outcomes is sufficiently symmetric to be well-approximated by a normal distribution.

In our problem, we check this condition: \( np = 20 \) and \( n(1-p) = 80 \). Both values are comfortably above 5, which means the normal approximation can be applied. When applicable, using a normal distribution can make it easier to calculate probabilities, particularly when dealing with cumulative probabilities or larger numbers of trials.

Probability calculation in binomial distribution requires understanding and applying specific methods to find the likelihood of different outcomes. For instance, to find the probability of exactly 24 successes out of 100 trials, we can use the binomial probability formula: \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \( \binom{n}{k} \) represents the number of ways we can choose \( k \) successes from \( n \) trials. An example calculation with \( k = 24 \), given our parameters, results in a probability of approximately 0.0404.

For calculating probabilities over a range, like 18 to 22 successes, we sum the individual probabilities for each outcome within this range, which can involve direct computation or approximation methods like the normal approximation. This total comes out to approximately 0.2852. Finally, for cumulative probabilities like 15 or fewer successes, you might utilize binomial tables or software to achieve an efficient result, culminating in approximately 0.1302 for this scenario.