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When considering situations with two possible outcomes, like a student either attending or not attending university, a binomial distribution can model this scenario effectively. Each admitted student can be seen as a success if they decide to attend, and a failure if they don't.
In our exercise, the number of trials, or the total number of students admitted, is 2,500, and the probability of a student attending, which we consider a 'success', is 70%. This distribution helps us understand the range of likely outcomes for student attendance.
To set up the model:

• The number of trials ( ) is 2,500.
• The probability of success (p) is 0.70.

Considering the large number of trials and a binary outcome, the binomial distribution is perfect for estimating the probability of different possible numbers of students attending.

Given a large sample size, as in our case with 2,500 admitted students, it is often practical to use the normal approximation to a binomial distribution for ease of calculation. This involves treating the binomial distribution as if it were a normal distribution.
The idea behind this simplification is that for large enough numbers, the behavior of the binomial distribution closely resembles that of a normal distribution with the same mean and standard deviation.
The parameters for our normal distribution for this scenario are:

• The mean is calculated as the number of trials multiplied by the probability of success, i.e., 2,500 * 0.70 = 1,750.
• The standard deviation is derived from the binomial formula, which we'll discuss in the next section.

Once you've gathered these parameters, you can use the characteristics of the normal distribution to calculate the likelihood of a given outcome, such as the attendance exceeding dormitory capacity.

In probability theory, the standard deviation of a binomial distribution tells us how much the outcome is likely to vary from the mean.
For our situation, where we have a large number of trials with a set probability of success, the standard deviation is crucial for understanding the typical variation in outcomes. It's calculated using the formula:

• \(\sigma = \sqrt{n \cdot p \cdot (1 - p)}\)

In our particular example, \(n = 2500\) and \(p = 0.70\), leading to the calculation:

• \(\sigma = \sqrt{2500 \cdot 0.70 \cdot 0.30} = \sqrt{525} \approx 22.91\)

This standard deviation of approximately 22.91 gives us an idea of how much fluctuation to expect around the mean attendance of 1,750. Understanding this spread is critical for making informed predictions about the number of students likely to show up.

A Z-score helps in determining how far away a particular value is from the mean of a distribution, measured in terms of standard deviations. This statistic is crucial for assessing the probability of different potential outcomes in a normal distribution approximation.
In our exercise, we calculate the Z-score to find out the likelihood that more than 1,786 students will attend. The formula is:

• \(z = \frac{x - \text{mean}}{\sigma}\)

For the scenario, substituting the values gives:

• \(z = \frac{1786 - 1750}{22.91} \approx 1.57\)

This Z-score of approximately 1.57 suggests that 1,786 is 1.57 standard deviations above the mean of 1,750.
By consulting a standard normal distribution table, we find that the probability of the outcome being less than this Z-score is 0.9429. To discover the probability of exceeding this dorm capacity, we calculate:

• \(1 - 0.9429 = 0.0571\)

Thus, there's roughly a 5.71% chance of exceeding dorm capacity.