The normal distribution, also known as the Gaussian distribution, is one of the most important probability distributions in statistics. It is characterized by its bell-shaped curve and is defined by two parameters: the mean (μ) and the standard deviation (σ). The normal distribution is widely used because many natural phenomena and measurement errors tend to follow this distribution.

1. Symmetry: The normal distribution is symmetric around its mean.
2. Mean, Median, and Mode: In a normal distribution, the mean, median, and mode are all equal.
3. Bell Shape: The distribution has a bell-shaped curve.
4. Asymptotic: The tails of the distribution approach the horizontal axis but never touch it.
5. Empirical Rule: Approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It is denoted as \( Z \sim N(0, 1) \).

A Z-score indicates how many standard deviations an element is from the mean. It is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where:

• \( X \) is the value,
• \( \mu \) is the mean,
• \( \sigma \) is the standard deviation.

Suppose the heights of adult men are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What is the probability that a randomly selected man is taller than 73 inches?

1. Calculate the Z-score: \[ Z = \frac{73 - 70}{3} = 1 \]

2. Find the probability: Using the Z-table, the probability of \( Z \leq 1 \) is 0.8413. Therefore, the probability of \( Z > 1 \) is: \[ P(Z > 1) = 1 - 0.8413 = 0.1587 \]

So, there is a 15.87% chance that a randomly selected man is taller than 73 inches.

Suppose we want to find the height that separates the tallest 10% of men from the rest. Given the same mean and standard deviation, we need to find the value of \( X \) such that \( P(X > x) = 0.10 \).

1. Find the Z-score: From the Z-table, the Z-score corresponding to the 90th percentile (since \( P(X > x) = 0.10 \)) is approximately 1.28.

2. Calculate the value: \[ X = \mu + Z \cdot \sigma \] \[ X = 70 + 1.28 \cdot 3 = 73.84 \]

So, the height that separates the tallest 10% of men is approximately 73.84 inches.

Given a normal distribution with a mean of 50 and a standard deviation of 5, calculate the Z-score for the value 60.

Solution: \[ Z = \frac{60 - 50}{5} = 2 \]

Given a normal distribution with a mean of 100 and a standard deviation of 15, what is the probability that a randomly selected value is less than 85?

Solution:

1. Calculate the Z-score: \[ Z = \frac{85 - 100}{15} = -1 \]

2. Find the probability: Using the Z-table, the probability of \( Z \leq -1 \) is 0.1587.

So, there is a 15.87% chance that a randomly selected value is less than 85.

Given a normal distribution with a mean of 200 and a standard deviation of 20, find the value that separates the lowest 5% of the distribution.

Solution:

1. Find the Z-score: From the Z-table, the Z-score corresponding to the 5th percentile is approximately -1.645.

2. Calculate the value: \[ X = \mu + Z \cdot \sigma \] \[ X = 200 + (-1.645) \cdot 20 = 167.1 \]

So, the value that separates the lowest 5% of the distribution is approximately 167.1.

In this section, we explored the normal distribution, its properties, and practical applications. We learned how to calculate Z-scores, find probabilities, and determine values from given probabilities. Understanding the normal distribution is crucial for many statistical analyses and real-world applications. In the next section, we will delve into other important distributions, further expanding our statistical toolkit.