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The concept of a probability distribution is fundamental in statistics as it describes the likelihood of various outcomes in an experiment or process. In the case of the multiple-choice exam in the exercise, the number of correct answers can be described by a specific probability distribution called the binomial distribution.

A binomial distribution is applicable when the experiment meets certain criteria: there are a fixed number of trials (in this case, 100 questions), each trial has only two possible outcomes (correct or incorrect), the probability of success is the same for each trial (here, a 1/5 chance of guessing correctly), and trials are independent of each other. The distribution then shows us how likely it is to achieve any number of successes (in this case, correct answers) across those trials.

The shape of the binomial distribution depends on the number of trials and the probability of success. It allows us to answer questions like, 'What is the likelihood of guessing exactly 20 questions correctly?' or 'What is the probability of guessing more than 50 questions correctly?' Understanding this helps in predicting outcomes and making informed decisions based on probabilities.

The expected value, also known as the mean, of a probability distribution is a critical measure that represents the average outcome if an experiment were repeated many times. For a binomial distribution, the expected value is calculated as the product of the number of trials and the probability of success, which is denoted by the formula \( \mu = np \).

In the context of the given exercise, the expected score on the exam, where a student guesses on all questions, would be \( \mu = 100 \times \frac{1}{5} = 20 \). This means that if a student took many such exams, each time randomly guessing every answer, they should average about 20 correct responses over time. While it's possible to score higher or lower on a single test, the expected value provides a benchmark for the long-term average performance.

Moving deeper into the characteristics of the binomial distribution, we encounter variance and standard deviation. Variance \( \sigma^2 \) measures the spread of the distribution, and standard deviation \( \sigma \) is the square root of variance, providing a gauge of how much we can expect outcomes to vary from the expected value.

The exercise shows that variance is found by multiplying the number of trials by the probability of success and the probability of failure \( \sigma^2 = np(1-p) \). The resulting variance for the test is 16, and the standard deviation, being the square root of variance, comes out to 4. This standard deviation signifies that the scores on this type of exam by guessing would typically deviate from the mean by about 4 questions in either direction.

Understanding the standard deviation is crucial as it helps interpret the spread of data around the mean. If we were to look at the likelihood of the student scoring over 50, we would see that it falls many standard deviations away from the expected value—indicating that achieving such a score by guessing alone would be extremely improbable.