Lesson 3.3: The Normal Distribution

Introduction

In this lesson, students will learn about the normal distribution, a fundamental concept in statistics that describes how data points are distributed in a symmetrical, bell-shaped curve. Understanding the normal distribution is crucial, as it allows us to model real-world scenarios where many outcomes are random and can be approximated by this distribution.

Learning Objectives

By the end of this lesson, you should be able to:

• Understand the properties of the normal distribution and recognize the bell-shaped density curve.
• Apply the empirical rule, which states that approximately 95% of observations lie within two standard deviations and 99.8% within three.
• Determine probabilities and unknown parameters using a calculator.
• Analyze the distribution of the sample mean of a normal variable.
• Model real-world situations using the normal distribution.

Properties of the Normal Distribution

The normal distribution is a continuous probability distribution that is symmetric about its mean. Here are the key properties:

1. Bell-shaped Curve: The graph of the normal distribution takes the form of a bell-shaped curve. The highest point on the curve corresponds to the mean, median, and mode of the distribution, which are all equal in a perfectly normal distribution.

1. Mean and Standard Deviation: The mean ($\mu$) determines the center of the distribution, while the standard deviation ($\sigma$) controls the spread. A smaller $\sigma$ results in a steeper curve, indicating that more data points are concentrated around the mean.

1. Empirical Rule: This rule helps us understand how data is dispersed in a normal distribution:

• About 68% of data falls within one standard deviation ($\mu \pm \sigma$).
• Approximately 95% of data falls within two standard deviations ($\mu \pm 2\sigma$).
• About 99.8% of data falls within three standard deviations ($\mu \pm 3\sigma$).

Worked Example

Suppose the heights of 1,000 adult males follow a normal distribution with a mean height of 70 inches and a standard deviation of 3 inches. We can use this information to answer the following questions:

Question 1: What percentage of adult males are taller than 76 inches?

Solution:

1. Calculate the Z-score:

The Z-score formula is:

$$Z = \frac{X - \mu}{\sigma}$$

where:

• $X$ is the value we are comparing (76 inches)
• $\mu$ is the mean (70 inches)
• $\sigma$ is the standard deviation (3 inches)

Plugging in the values:

$$Z = \frac{76 - 70}{3} = \frac{6}{3} = 2$$

1. Find the area to the right of Z = 2:

Using a standard normal distribution table, we find that the area to the left of $Z = 2$ is approximately 0.9772.

Thus, the area to the right (which corresponds to heights greater than 76 inches) is:

$$P(X > 76) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228$$

This means approximately 2.28% of adult males are taller than 76 inches.

Determining Probabilities Using a Calculator

Calculators and statistical software can provide efficient tools for calculating probabilities and Z-scores associated with the normal distribution. It is essential to be familiar with how to use these tools.

Z-Score Calculation

To calculate the Z-score using a calculator, you typically input the mean and standard deviation alongside the value you want to evaluate.

Finding Probabilities

To find probabilities, you will often use the normal distribution function available in calculators under the statistics menu. You will input the mean, standard deviation, and the desired value to find the probability associated with that value.

Worked Example

Consider a normal distribution with $\mu = 50$ and $\sigma = 5$.

Question 2: What is the probability of a value being less than 53?

Solution:

1. Calculate the Z-score:

$$Z = \frac{53 - 50}{5} = 0.6$$

1. Use the calculator: Input this Z-score into your calculator’s normal distribution function (or lookup table).

You will find that:

$$P(X < 53) \approx 0.7257$$

This means there is approximately a 72.57% chance that a value is less than 53.

The Distribution of the Sample Mean

When dealing with sample sizes, the Central Limit Theorem (CLT) states that the sampling distribution of the sample mean will approach a normal distribution, regardless of the original distribution, as long as the sample size is large enough (usually $n \geq 30$).

Key Points about the Sample Mean

• The mean of the sampling distribution ($\mu_{\bar{x}}$) is equal to the mean of the population ($\mu$).
• The standard deviation of the sampling distribution, called the standard error ($SE$), is given by:

$$SE = \frac{\sigma}{\sqrt{n}}$$

• Distributions become more concentrated around the mean as sample size increases.

Worked Example

Suppose you take a random sample of 36 students' test scores from a school where the average score is 75 with a standard deviation of 10. What is the probability that the sample mean is greater than 78?

Solution:

1. Calculate the standard error:

$$SE = \frac{10}{\sqrt{36}} = \frac{10}{6} = 1.67$$

1. Calculate the Z-score for the sample mean:

$$Z = \frac{78 - 75}{1.67} = \frac{3}{1.67} \approx 1.79$$

1. Find the probability:

Using the calculator or Z-table, find $P(Z > 1.79)$:

$$P(Z > 1.79) \approx 0.0367$$

This means there is approximately a 3.67% chance that the sample mean will be greater than 78.

Modelling Real World Situations

The normal distribution can be applied to various real-world situations, such as:

• Heights, weights, and test scores often follow a normal distribution.
• Measurement errors in manufacturing processes.
• Natural phenomena like blood pressures, IQ scores, and many others.

When modeling these situations, assess the data to determine if it fits or approximately fits a normal distribution. Once established, you can apply the properties of the normal distribution to make predictions or infer conclusions about the population.

Example Application

Suppose a company claims that their light bulbs have an average lifespan of 800 hours with a standard deviation of 100 hours. To assess quality control, they want to see what percentage of bulbs last between 750 hours and 850 hours.

1. Calculate the Z-scores:

For 750 hours:

$$Z_1 = \frac{750 - 800}{100} = -0.5$$

For 850 hours:

$$Z_2 = \frac{850 - 800}{100} = 0.5$$

1. Find probabilities using the calculator:

$$P(750 < X < 850) = P(Z < 0.5) - P(Z < -0.5) \approx 0.6915 - 0.3085 = 0.383$$

Thus, approximately 38.3% of bulbs will last between 750 and 850 hours.

Conclusion

In this lesson, students explored the normal distribution, its properties, and its applications. You have learned to determine probabilities and model real situations using the normal distribution, which is an essential technique in statistics.

Understanding the normal distribution not only aids in evaluating data but also helps inform decision-making in real-life contexts. Familiarity with using calculators for these calculations will enhance your statistical skills as you advance in your studies.

Study Notes

• The normal distribution is defined by its mean ($\mu$) and standard deviation ($\sigma$).
• The empirical rule states:
• 68% within $\mu \pm 1\sigma$
• 95% within $\mu \pm 2\sigma$
• 99.8% within $\mu \pm 3\sigma$
• To find probabilities, compute Z-scores and use standard normal distribution tables or calculators.
• The Central Limit Theorem states that the sample mean will be approximately normally distributed for sufficiently large sample sizes.
• Standard error (SE) is $\frac{\sigma}{\sqrt{n}}$ for sample distributions.
• Real-world applications of the normal distribution include heights, test scores, and quality control in manufacturing.