91Ó°ÊÓ

In statistics, a binomial distribution represents the number of successes in a fixed number of independent experiments. Each experiment has two possible outcomes: success or failure. The probability of success is denoted by \(p\), and the total number of experiments is \(n\). For our exercise involving voting:

• Total number of trials (n): 1002
• Probability of success (p): 0.61 (since 61% of eligible voters actually did vote)

The mean (\( \mu \)) and the standard deviation (\( \sigma \)) for a binomial distribution can be calculated using:

• Mean: \( \mu = np \)
• Standard deviation: \( \sigma = \sqrt{np(1-p)} \)

These formulas help us to understand the central tendency and the variability of the distribution. In our context:
\( \mu = 1002 \times 0.61 = 611.22 \)
\( \sigma = \sqrt{1002 \times 0.61 \times (1-0.61)} = 15.476 \)

Normal approximation is often used when the sample size is large. It helps simplify the calculations. For a large sample size, the binomial distribution can be approximated by a normal distribution with the same mean and standard deviation.
In our exercise, the sample size is 1002, which is large enough to use the normal approximation. The steps are:

• Calculate the mean, \( \mu \), and standard deviation, \( \sigma \), as previously described.
• Use these values to transform the binomial problem into a normal one.
• We then convert the binomial variable into a z-score so we can use the z-table.
  

For our problem, we want to find the probability that at least 701 people voted. First, we compute the z-score of 701:
\( Z = \frac{701 - \mu }{\sigma } = \frac{701 - 611.22 }{15.476} = 5.8 \)
A z-score of 5.8 indicates how many standard deviations the value 701 is from the mean. High z-scores, like in this case, often indicate a very low probability of occurrence.

A z-score measures how many standard deviations an element is from the mean. It can be calculated using the formula:
\( Z = \frac{X - \mu}{\sigma } \)
Here, \(X\) is the value for which we want to find the probability, \( \mu \) is the mean, and \( \sigma \) is the standard deviation of the distribution. In our exercise:
The z-score formula helps to standardize our binomial distribution, enabling us to use the normal distribution (z-table) to find probabilities. For the value 701:
\( Z = \frac{701 - 611.22 }{15.476} = 5.8 \)
Using the z-table, we find the area to the left of the z-score, which provides the cumulative probability. Because a z-score of 5.8 is very high, it means the probability of at least 701 voters is almost zero.
This shows it is very unlikely that randomly selecting 1002 eligible voters would result in 701 or more actual voters, given that 61% is the true voting rate.