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Probability is a measure of how likely an event is to occur. In the context of binomial distribution, which is used to model situations with two possible outcomes, probability helps us determine the likelihood of a specific number of 'successes' in a set of trials.

Let's take the example of commuters where each commuter has a 5% probability (\( p = 0.05 \)) of having a commute over one hour. In a binomial distribution, we calculate the probability of various outcomes using the binomial probability formula:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
  
• \( \binom{n}{k} \) is the binomial coefficient that shows the number of ways to choose \( k \) successes from \( n \) trials.

For example, if we want to find out the probability that 3 out of 20 commuters have a travel time of over an hour, substitute \( n = 20 \), \( k = 3 \), and \( p = 0.05 \) into the formula. This calculation shows the probability of observing exactly 3 commuters out of 20 with a longer commute on that day.

Similarly, to determine the probability that none of the 20 commuters have a commute over an hour, set \( k = 0 \). This approach helps in understanding how the chance of a specific number of successes in a series of independent trials is obtained using probability.

Expected Value, often denoted by \( E[X] \), is a vital concept in probability that tells us the average outcome we can expect from a random event, in the long run. For a binomial distribution, the expected value is calculated as:

• \( E[X] = n \times p \)

where \( n \) is the number of trials and \( p \) is the probability of success on each trial.

In our exercise, we consider a company with 2000 employees, each having a 5% probability of commuting for more than an hour. The expected number of such employees is found by:

• \( E[X] = 2000 \times 0.05 = 100 \)
  

This indicates that, on average, 100 employees out of 2000 are expected to have a commute longer than one hour. The expected value thus provides a useful measure to anticipate the typical number of successes in repeated trials over time.

Variance measures the spread or dispersion of a set of outcomes in a probability distribution. It essentially tells us how much the numbers differ from the expected value. In a binomial distribution, variance is given by:

• \( \text{Var}(X) = n \times p \times (1-p) \)
  
• where \( n \) is the total number of trials and \( p \) is the probability of success.
  

For instance, in our problem concerning the employees, with \( n = 2000 \) and \( p = 0.05 \), the variance calculates to:

• \( \text{Var}(X) = 2000 \times 0.05 \times 0.95 = 95 \)
  

This value of 95 indicates how much variation or fluctuation we can expect in the number of employees whose commute exceeds one hour. Variance gives us an idea of the consistency of the outcomes, representing how clustered or spread out the results are around the expected value.

Standard Deviation is a key concept that quantifies the amount of variation or dispersion in a set of values, providing insight into how spread out the numbers are around the mean (expected value).

In the context of a binomial distribution, standard deviation is the square root of the variance:

• \( \sigma = \sqrt{\text{Var}(X)} \)

Using our example of 2000 employees, where the variance is 95, the standard deviation is calculated as:

• \( \sigma = \sqrt{95} \approx 9.75 \)

This standard deviation value tells us about the average distance of the commute times from the expected value. A smaller standard deviation indicates that the data points are closer to the average, whereas a larger standard deviation suggests a wider spread of results. Thus, it helps in understanding the consistency of the number of commuters with longer commute times.