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In the context of a binomial distribution, the expected value is a way to predict the average outcome of a random experiment that is repeated many times. It gives us a sense of what to expect in the long run from a series of trials where each trial has a binary outcome, like success or failure.
To find the expected value for a binomial distribution, we use the formula: \( E(X) = n \times p \).

• \( n \) represents the total number of trials.
• \( p \) is the probability of achieving a success on each trial.

In our given problem, we have 20 trials with a success probability of 0.40. Thus, the expected value is \( E(X) = 20 \times 0.40 = 8 \).
This means that, on average, we can expect 8 successes out of 20 trials. Keep in mind that the expected value is a theoretical average and doesn't guarantee that exactly 8 successes will happen—it represents a mean over many repetitions of the experiment.

Standard deviation measures the amount of variability or dispersion in a set of values. In a binomial distribution, it tells us how spread out the numbers of successes are around the expected value. A larger standard deviation means more spread out results.
We calculate the standard deviation using the formula: \( \sigma = \sqrt{n \times p \times (1-p)} \).

• \( n \) is the number of trials.
• \( p \) is the probability of success.
• \( 1-p \) is the probability of failure, which complements the probability of success.

Using our exercise values, where \( n = 20 \) and \( p = 0.40 \), the standard deviation is \( \sigma = \sqrt{20 \times 0.40 \times 0.60} \Approx 2.19 \).
This value shows us that the number of successes typically varies by around 2.19 from the expected value of 8.

Cumulative probability is crucial when determining the likeliness of an event occurring within a specified range in a probability distribution. For binomial distributions, it can be used to calculate the probability of getting a number of successes that is less than, equal to, or more than a specific number.
In this particular exercise, we are interested in the probability of achieving fewer than 3 successes. This means we need to find the sum of probabilities of having 0, 1, and 2 successes.

• \( P(X = 0) \approx 0.0008 \)
• \( P(X = 1) \approx 0.0105 \)
• \( P(X = 2) \approx 0.0489 \)

Hence, the cumulative probability is \( P(X < 3) = 0.0008 + 0.0105 + 0.0489 = 0.0602 \).
This value is the combined probability of having fewer than 3 successes, and is used to assess whether such an outcome is unusual in our experiment's context.

A probability distribution describes how the probabilities of all possible outcomes of a random variable are assigned. In a binomial distribution, we specifically look at the probabilities of obtaining a certain number of successes in a fixed number of trials, where each trial is independent and has the same probability of success.
The formula for finding the probability of getting exactly \( k \) successes in \( n \) trials is:
\( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \).

• \( \binom{n}{k} \) is the binomial coefficient, and it accounts for all the different ways to choose \( k \) successes out of \( n \) trials.
• \( p^k \) denotes the probability of \( k \) successes.
• \( (1-p)^{n-k} \) represents the probability of the remaining trials resulting in failure.

This formula, when used appropriately, allows us to create a full binomial probability distribution table, as seen in the problem's solution.
By listing out the probabilities of all possible numbers of successes, the distribution table helps in visualizing and confirming outcomes, such as determining if particular results are unusual or common.