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The binomial distribution is a statistical method used to model the number of successful outcomes in a series of independent and identical experiments. These experiments, known as trials, each have two possible outcomes: success or failure. An example is flipping a coin where a head can represent success and a tail represents failure.
Binomial distributions are characterized by two parameters:

• The number of trials, denoted as \( n \).
• The probability of success in each trial, denoted as \( p \).

In the context of the credit card contract example, we look at whether a person reads every word or not in a contract as `success`, with 44% of people representing the probability \( p = 0.44 \). With a total sample size \( n = 500 \), we utilize the formula for the expected number of successes:
\[ \text{Expected number} = n \times p \]
Calculating for 500 people gives \( 500 \times 0.44 = 220 \). This means, on average, 220 people in the study are expected to read every word of the contract.

When working with large sample sizes in binomial distributions, it is often more practical to use what's known as a normal approximation. This method involves approximating the binomial distribution with a normal distribution, simplifying the calculation of probabilities.
For a normal approximation to be appropriate, the distribution should satisfy the conditions:

• \( np \geq 5 \)
• \( n(1 - p) \geq 5 \)

This is because the distribution becomes bell-shaped when both these conditions are met. Given our example, \( np = 500 \times 0.44 = 220 \) and \( n(1-p) = 280 \), both of which satisfy these conditions. Thus, we can approximate our binomial distribution using normal distribution.
The normal distribution then uses a mean (\( \mu \)) and a standard deviation (\( \sigma \)) calculated by:
\[ \mu = np \] \[ \sigma = \sqrt{np(1-p)} \]
This makes it easier to employ normal distribution tables or tools for probability calculations.

The Z-score is a statistical measure that tells us where a specific value falls in the context of a distribution. It expresses this position in terms of standard deviations from the mean. The formula for a Z-score when using a normal distribution is:
\[ Z = \frac{X - \mu}{\sigma} \]
Here, \( X \) is the value whose position we want to locate, \( \mu \) is the mean and \( \sigma \) is the standard deviation.

• A Z-score of 0 indicates that the value is exactly at the mean.
• A positive Z-score indicates that the value is above the mean.
• A negative Z-score indicates that the value is below the mean.

For example, in our problem, we calculated a Z-score of approximately -1.81 for part (b), assessing the probability that 200 or fewer people read every word of a contract. This Z-score tells us that 200 is 1.81 standard deviations below the mean, which helps to identify the probability using a standard Z-table.

Probability calculations are central to understanding outcomes in statistics. They give us a numerical value ranging from 0 to 1, quantifying the likelihood of an event. For binomial distributions approximated by normal distributions, these probabilities are found using Z-scores and standard normal distribution tables.
In calculating probabilities, consider continuity correction when dealing with discrete distributions. This correction makes the transition between discrete and continuous sampling more accurate by adjusting the specific value slightly, commonly by 0.5, in the context of Z-score calculations.
For example, in part (c) of the exercise, we determine the probability that 15 or more people do not read the contract. We compute:

• \( Z = \frac{14.5 - 20}{4.37} \approx -1.26 \)

Here we adjust 15 to 14.5 for the continuity correction. We then use the Z-score to find the probability from the standard normal table, obtaining \( P(X \geq 15) = 1 - P(X < 15) \approx 0.8953 \), indicating there's nearly an 89.53% chance that at least 15 people won't read their contracts. This step highlights the interplay between probability calculation and statistical methods.