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The Normal Distribution
Diana Mindrila, Ph.D. Phoebe Baletnyne, M.Ed.
Based on Chapter 3 of The Basic Practice of Statistics (6 th^ ed.)
Concepts: Density Curves Normal Distributions The 68-95-99.7 Rule The Standard Normal Distribution Finding Normal Proportions Using the Standard Normal Table Finding a Value When Given a Proportion
Objectives: Define and describe density curves Measure position using percentiles Measure position using z-scores Describe Normal distributions Describe and apply the 68-95-99.7 Rule Describe the standard Normal distribution Perform Normal calculations
References: Moore, D. S., Notz, W. I, & Flinger, M. A. (2013). The basic practice of statistics (6 th ed.). New York, NY: W. H. Freeman and Company.
Density Curves Exploring Quantitative Data
When describing data, always start with a graphical representation. Graphs help identify the overall distribution pattern. Looking at a graph makes it visually clear how spread a variable is, which values occur most frequently, and whether or not the distribution is skewed. Next, obtain more precise information by providing a numerical summary of the data using the mean, median, range, five-number summary, and any other appropriate information. Some distributions are so regular that they can be described by a smooth curve. Real data are represented in a histogram. Curves represent a symbol, or an abstract version of a distribution.
Density curves are lines that show the location of the individuals along the horizontal axis and within the range of possible values. They help researchers to investigate the distribution of a variable. Some density curves have certain properties that help researchers draw conclusions about the entire population.
1. Always plot data first: make a graph.
2. Look for the overall pattern (shape, center, and spread) and
for striking departures such as outliers.
3. Calculate a numerical summary to briefly describe center and
spread.
4. Sometimes the overall pattern of a large number of
observations is so regular that it can be described by a smooth
curve.
A density curve is a curve that:
- is always on or above the horizontal axis
- has an area of exactly 1 underneath it A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values on the horizontal axis is the proportion of all observations that fall in that range.
Density Curves The mean and standard deviation computed from actual observations (data) are denoted by 𝑥̅ and s , respectively The mean and standard deviation of the actual distribution represented by the density curve are denoted by 𝜇 (“mu”) and 𝜎 (“sigma”), respectively. The mean and standard deviation (𝑥̅ and s ) are called statistics , and they can be computed based on observations in the sample. The mean and standard deviation of the density curves (𝜇 and 𝜎) are called parameters. They describe the entire population and are only estimated. With very few exceptions, the real value of the population is unknown and the values must be estimated, with a certain degree of confidence, based on observations from the sample.
Normal Distributions One particularly important class of density curves are the Normal curves, which describe Normal distributions. All Normal curves are symmetric, single-peaked, and bell-shaped. A Specific Normal curve is described by giving its mean 𝜇 and standard deviation 𝜎.
Density curves are used to illustrate many types of distributions. The Normal distribution, or the bell-shaped distribution, is of special interest. This distribution describes many human traits. All Normal curves have symmetry, but not all symmetric distributions are Normal. Normal distributions are typically described by reporting the mean, which shows where the center is located, and the standard deviation, which shows the spread of the curve, or the distance from the mean. When the standard deviation is large, the curve is wider like the example on the left. When the standard deviation is small, the curve is narrower like the example on the right. One example of a variable that has a Normal distribution is IQ. In the population, the mean IQ is 100 and it standard deviation, depending on the test, is 15 or 16. If a large enough random sample is selected, the IQ distribution of the sample will resemble the Normal curve. The large the sample, the more clear the pattern will be.
Normal Curve
Example: IQ score distribution based on the Standford-Binet Intelligence Scale
The smooth curve drawn over the histogram is a mathematical model for the distribution.
The histogram in this image represents a distribution of real IQ scores as measured by the Standford-Binet Intelligence Scale. The blue bars represent the number of individuals who recorded IQ scores within a certain 5-point range. The main purpose of a histogram is to illustrate the general distribution of a set of data. This variable has a mean of 100 and a standard deviation of 15. The curve that is drawn over the histogram is the Normal curve, and it summarized the distribution of the recorded scores.
Normal Curve
The areas of the shaded bars in this histogram represent the proportion of scores in the observed data that are less than or equal to 90. Total: N = 1015 IQ<90: N = 256 (25.22%)
Now the area under the smooth curve to the left of 90 is shaded. If the scale is adjusted so the total area under the curve is exactly 1, then this curve is called a density curve. Total Area = 1 Shaded Area = 0.
The entire area under the curve represents all the individuals in the sample. If only part of the area is shaded, this represents the proportion of individuals who scored below a certain point. In this above example, the area under the curve represents all the individuals in the sample. In this case, they add up to 1,015. This number represents 100% of the sample. The shaded area in the above example represents the individuals who had an IQ score below 90. This group consists of 256 individuals. To find the percentage, divide the number in the group by the total number, and then multiply by 100. In this case, 256 divided by 1015 times 100 results in a percentage of 25.22. This means that 25.22% of the individuals in this sample had an IQ score below 90. The Normal curve is used to find proportions from the entire population, rather than just from the sample. The values for the entire population are often unknown, but if the variable has a Normal distribution, the proportion can be found using only the population mean and standard deviation for that variable. Rather than using percentages, statisticians use decimals. Therefore, the entire area under the curve is 1. Using the properties of the Normal curve, the shaded are in the above example is 0.2546. This will be explained in greater detail later.
Technically, the two tails of the Normal curve extent to positive or negative infinity, but these numbers would be limited for certain variables like IQ, which cannot be smaller than zero. The proportion of individuals who are located more than three standard deviations above or below the mean is extremely small: only 0.3%.
The 68-95-99.7 Rule Example
Figure 1 illustrates how to apply the 68-95-99.7 Rule to the distribution of IQ scores. In this example, the population mean is 100 and the standard deviation is 15. Based on the 68-95-99.7 Rule, approximately 68% of the individuals in the population have an IQ between 85 and 115. Values in this particular interval are the most frequent. Approximately 95% of the population has IQ scores between 70 and 130. Approximately 99.7% of the population has IQ scores between 55 and 145. Only approximately 0.3% of the population has IQ scores outside of this interval (less than 55 or higher than 145).
Normal Distributions Example
Example:
Joe: IQ = 111 Sigma = 15 Pop. Mean = 100
Joe’s IQ on the z distribution:
z = (111-100)/ z = 11/ z = 0.
In this example, an individual score on a Normal distribution is given. The top image shows the IQ score distribution. The bottom image shows the curve from this distribution transformed into z scores. To find the z score for Joe’s IQ:
- Subtract the mean from the score (score – mean) = (111 – 100) = 11
- Divide the difference by the standard deviation 11/15 = 0. Now that a z score has been obtained, it would be helpful to find out the proportion of individuals who have an IQ below 111, or a z score below 0.73. In other words, what is the area under the curve on the left side of this specific score?
Mean = 0
Joe’s z score = 0.
The Standard Normal Table All Normal distributions are the same when they have been turned into z scores. Therefore, areas under any Normal curve can be found using a single table.
To find the proportion of observations from the standard Normal distribution that are less than 0.73, use table A:
For every z score, areas on the left side of the curve have already been computed and are listed in a probability table. Statistics textbooks generally store these tables in the appendices. This table lists the first two digits of the z score vertically and the last digit horizontally. In this example, to find the area under the curve for a z score of 0.73, start by finding 0.7 on the left. Then find 0.03 at the top. Finally, find the cell where this row and column meet. The value in this cell (0.7673) is the area under the curve for a z score of 0.73. This value means the probability of a z score being lower than this one is 0.7673. Simply put, 76.73% of the population has a z score at or below 0.73. In this case, 76.73% of the population has an IQ equal to or lower than 111.
Z .00 .01 .02.
The Standard Normal Table
Table A is a table of areas under the standard Normal curve.
The table entry for each value z is the area under the curve to
the left of z.
P (z < 0.73) =.
Normal Distributions The Normal distribution is very useful for comparing variables that are measured on different scales. Example: a graduate student has a score of 25 on a quiz and a score of 56 on the final exam. On which assessment did the student perform better? This depends on the distribution of the two variables. ♦ If the quiz is out of 30 points and the exam is out of 100 points, it may seem clear that the quiz performance was better. ♦ However, using the Normal curve, the most important information is the mean and the standard deviation. ♦ Standardized scores (e.g. z scores) can help to compare scores measured on different scales.
Quiz Exam x 1 = 25 x 2 = 56 Mean = 20 Mean = 68 St. Dev. = 5 St. Dev. = 12 Z 1 = (25-20)/5 Z 2 = (56-68)/ = 5/5 = – 12/ = 1 = – 1
After calculating the z scores for both assessments, it can be concluded that this student performed better on the quiz, even though the raw score of 25 was lower than the raw score of 56. The student’s performance on the quiz was one standard deviation above the mean and the student’s performance on the exam was one standard deviation below the mean, resulting in a higher performance on the quiz. This example makes it clear that if it is necessary to compare scores that are on different scales, the scores must be standardized or put on the same scale. In this case, the standardized scores are z scores, but there are many other kinds of standardized scores. If raw scores only are compared, the results can be misleading, as this example demonstrated. It is important to note that these comparisons are based on the assumption that the two variables have a Normal distribution in the population.
Normal Calculations
How to Solve Problems Involving Normal Distributions
State: Express the problem in terms of the observed variable x.
Plan: Draw a picture of the distribution and shade the area of interest under the curve.
Do: Perform calculations.
- Standardize x to restate the problem in terms of a standard Normal variable z.
- Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve.
Conclude: Write the conclusion in the context of the problem.
