Normal Distribution and Central Limit Theorem

1. Normal Distribution

The normal distribution, also known as the "bell curve" or "Gaussian distribution," is one of the most important and widely used probability distributions in statistics. It describes many continuous random variables that occur in nature and social sciences.

Key characteristics of the normal distribution:

Bell shape: The curve is symmetric around the mean.
Centrality: The mean, median, and mode of the distribution are equal and located at the center of the curve.
Parameters: The distribution is defined by two parameters:
- Mean ($\mu$, mu): Represents the average of the distribution, which is its center of symmetry.
- Standard Deviation ($\sigma$, sigma): Represents the degree of dispersion of the data around the mean. A larger standard deviation indicates a wider spread. The variance is $\sigma^2$.
68-95-99.7 Rule (Empirical Rule):
- Approximately 68% of the values lie within one standard deviation of the mean ($\mu \pm \sigma$).
- Approximately 95% of the values lie within two standard deviations of the mean ($\mu \pm 2\sigma$).
- Approximately 99.7% of the values lie within three standard deviations of the mean ($\mu \pm 3\sigma$).
The total area under the curve is equal to 1 (or 100%), representing the sum of all probabilities.

2. Computational Questions and Calculator

To better understand the concepts, let's try to solve some computational questions. You can use the embedded normal distribution calculator below to check your answers or perform complex calculations.

Question 1: IQ Scores

IQ scores in a certain population are normally distributed with a mean ($\mu$) of 100 and a standard deviation ($\sigma$) of 15.

What percentage of people in the population have an IQ score lower than 85?
What percentage of people have an IQ score higher than 120?
What percentage of people have an IQ score between 90 and 110?
What is the IQ score below which 90% of the population falls (the 90th percentile)?

Use the calculator below to compute the values. Enter the Mean (100) and Standard Deviation (15).

Online Normal Distribution Calculator

The calculator is embedded from statistical-calculators.site.

3. Theoretical Foundations

a. Probability Density Function (PDF)

The probability density function of a random variable $X$ that is normally distributed with mean $\mu$ and standard deviation $\sigma$ is given by the formula:

PDF Formula:

\[ f(x | \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]

Where: $x$ is the value of the random variable, $\mu$ is the mean, and $\sigma$ is the standard deviation.

b. Cumulative Distribution Function (CDF)

The cumulative distribution function ($\Phi$ or $F(x)$) gives the probability that a normally distributed random variable $X$ will take a value less than or equal to a specific $x$:

CDF Formula:

\[ F(x | \mu, \sigma) = P(X \le x) = \int_{-\infty}^{x} f(t | \mu, \sigma) dt \]

c. Z-score and Standard Normal Distribution

A Z-score measures how many standard deviations a value $x$ is from the mean $\mu$:

Z-score Formula:

\[ Z = \frac{X - \mu}{\sigma} \]

A variable $Z$ follows a standard normal distribution with $\mu_Z = 0$ and $\sigma_Z = 1$.

4. Central Limit Theorem (CLT)

The Central Limit Theorem states that the sampling distribution of the sample means ($\Xbar$) approaches a normal distribution as the sample size ($n$) increases, even if the original population is not normally distributed.

If $X_1, \ldots, X_n$ is a random sample from a population with mean $\mu$ and variance $\sigma^2$, then for a sufficiently large $n$:

\[ \Xbar \sim N\left(\mu, \frac{\sigma^2}{n}\right) \]

Where the mean of the sample means is $\mu_{\Xbar} = \mu$ and the standard error of the mean is $\sigma_{\Xbar} = \frac{\sigma}{\sqrt{n}}$.

Importance of the Central Limit Theorem:

It allows for statistical inference about the population mean even without knowing its distribution.
It explains why many natural phenomena appear to be normally distributed.

5. Additional Questions and Solutions

Question 2: Product Weights

The weight of a certain product is normally distributed with a mean of 500 grams and a standard deviation of 10 grams.

What is the probability that a randomly selected product will weigh less than 485 grams?
What is the probability that a product will weigh between 495 and 510 grams?
What is the weight above which only 5% of the products weigh?

You can use the calculator above.

Question 3: Central Limit Theorem Application

The waiting time in line at a supermarket checkout is distributed in some way (not necessarily normal) with a mean of 5 minutes and a variance of 4 minutes² (i.e., a standard deviation of 2 minutes).

What is the approximate distribution of the average waiting time for a sample of 36 customers? (Specify the mean and standard error of the sample mean)
What is the probability that the average waiting time for 36 customers will be between 4.5 and 5.5 minutes?

For part b, you can use the normal calculator after calculating the parameters of the mean's distribution.

Solutions and Guidelines

Solution for Question 1 (IQ Scores)

Given: $\mu = 100$, $\sigma = 15$.

$P(X < 85)$:
- In the calculator: Mean=100, SD=15. Select "P(X<a)" and enter a=85.
- Manual calculation (for understanding): $Z = (85 - 100) / 15 = -15 / 15 = -1$. From Z-table (or calculator with $\mu=0, \sigma=1$), $P(Z < -1) \approx 0.1587$. So, about 15.87%.
$P(X > 120)$:
- In the calculator: Mean=100, SD=15. Select "P(X>b)" and enter b=120.
- Manual calculation: $Z = (120 - 100) / 15 = 20 / 15 \approx 1.333$. $P(Z > 1.333) = 1 - P(Z \le 1.333) \approx 1 - 0.9082 = 0.0918$. So, about 9.18%.
$P(90 < X < 110)$:
- In the calculator: Mean=100, SD=15. Select "P(a<X<b)" and enter a=90, b=110.
- Manual calculation: $Z_1 = (90 - 100) / 15 \approx -0.667$. $Z_2 = (110 - 100) / 15 \approx 0.667$. $P(-0.667 < Z < 0.667) = P(Z < 0.667) - P(Z < -0.667) \approx 0.7486 - 0.2514 = 0.4972$. So, about 49.72%.
Score $x$ such that $P(X < x) = 0.90$:
- In the calculator: Mean=100, SD=15. Select "Find X given P(X<a)" or "Inverse Normal" and enter P=0.90.
- Manual calculation: Find $z$ such that $P(Z < z) = 0.90$. From table, $z \approx 1.28$. $X = \mu + Z\sigma = 100 + 1.28 \times 15 = 100 + 19.2 = 119.2$.

Solution for Question 2 (Product Weights)

Given: $\mu = 500$ grams, $\sigma = 10$ grams.

$P(X < 485)$:
- In the calculator: Mean=500, SD=10. "P(X<a)", a=485.
- $Z = (485 - 500) / 10 = -1.5$. $P(Z < -1.5) \approx 0.0668$. About 6.68%.
$P(495 < X < 510)$:
- In the calculator: Mean=500, SD=10. "P(a<X<b)", a=495, b=510.
- $Z_1 = (495 - 500) / 10 = -0.5$. $Z_2 = (510 - 500) / 10 = 1$. $P(-0.5 < Z < 1) = P(Z < 1) - P(Z < -0.5) \approx 0.8413 - 0.3085 = 0.5328$. About 53.28%.
Weight $x$ such that $P(X > x) = 0.05$ (equivalent to $P(X \le x) = 0.95$):
- In the calculator: Mean=500, SD=10. "Find X given P(X<a)" (P=0.95) or "Find X given P(X>b)" (P=0.05).
- Find $z$ such that $P(Z \le z) = 0.95$. From table, $z \approx 1.645$. $X = \mu + Z\sigma = 500 + 1.645 \times 10 = 500 + 16.45 = 516.45$ grams.

Solution for Question 3 (Central Limit Theorem)

Given: $\mu = 5$ minutes, $\sigma = 2$ minutes, $n = 36$.

Distribution of average waiting time $\Xbar$:
- Mean of the sample mean: $\mu_{\Xbar} = \mu = 5$ minutes.
- Standard error of the mean: $\sigma_{\Xbar} = \sigma / \sqrt{n} = 2 / \sqrt{36} = 2 / 6 = 1/3 \approx 0.333$ minutes.
- Therefore, $\Xbar \sim N(5, (1/3)^2)$ approximately.
$P(4.5 < \Xbar < 5.5)$:
- Now use a normal distribution with $\mu = 5$ and $\sigma = 1/3 \approx 0.333$.
- In the calculator: Mean=5, SD=0.3333. "P(a<X<b)", a=4.5, b=5.5.
- Manual calculation: $Z_1 = (4.5 - 5) / (1/3) = -0.5 / (1/3) = -1.5$. $Z_2 = (5.5 - 5) / (1/3) = 0.5 / (1/3) = 1.5$. $P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5) \approx 0.9332 - 0.0668 = 0.8664$. About 86.64%.