6.1: Introduction to the Normal Distribution

 

Learning Objectives

• Define the normal distribution as a continuous, bell-shaped, symmetric curve modeling many natural phenomena with values clustering around the mean.
• Describe the normal distribution using its mean and standard deviation.
• Use z-scores to standardize values and determine probabilities or areas under the curve with tables or calculators.

Many real-life problems produce a histogram that is a symmetric, unimodal, and bell-shaped continuous probability distribution. For example: height, blood pressure, and cholesterol level. However, not every bell-shaped curve is a normal curve. In a normal curve, there is a specific relationship between its “height” and its “width.”

Normal curves can be tall and skinny or they can be short and fat. They are all symmetric, unimodal, and centered at \(\mu\), the population mean. Figure \(\PageIndex{1}\) shows two different normal curves drawn on the same scale. Both have \(\mu = 100\), but the one on the left has a standard deviation of 10 and the one on the right has a standard deviation of 5. Notice that the larger standard deviation makes the graph wider (more spread out) and shorter.

Figure \(\PageIndex{1}\): Two Normal Distributions with Different Variances

Every normal curve has common features. These are detailed in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): Key Elements of a Normal Distribution

• The center, or the highest point, is at the population mean, \(\mu\).
• The transition points (inflection points) are the places where the curve changes from a “hill” to a “valley”. The distance from the mean to the transition point is one standard deviation, \(\sigma\).
• The area under the whole curve is exactly 1. Therefore, the area under the half below or above the mean is 0.5.

Before looking at the process for finding the probabilities under the normal curve, it is somewhat useful to look at the Empirical Rule that gives approximate values for these areas. The Empirical Rule is just an approximation, and it will only be used in this section to give you an idea of what the size of the probabilities is for different shadings. A more precise method for finding probabilities for the normal curve will be demonstrated in the next section. Please do not use the empirical rule except for rough estimates.

Definition: Empirical Rule

The Empirical Rule for any normal distribution:

• Approximately 68% of the data is within one standard deviation of the mean.
• Approximately 95% of the data is within two standard deviations of the mean.
• Approximately 99.7% of the data is within three standard deviations of the mean.

Figure \(\PageIndex{3}\): Empirical Rule for Normal Distribution

Be careful, there is still some area left over at each end. Remember, the maximum probability can be 100%, so if you calculate 100%-99.7%=0.3%, you will see that for both ends together, there is 0.3% of the curve. Because of symmetry, you can divide this equally between both ends and find that there is 0.15% in each tail beyond the \(\mu \pm 3 \sigma\).

Read and Interpret the Standard Normal Distribution Table

The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. It is used as a reference to understand how data values compare to the average in terms of standard deviations. This distribution helps us calculate probabilities using three basic rules.

Figure \(\PageIndex{4}\): Graph of the Standard Normal Distribution with Mean = 0 and Standard Deviation = 1.

Definition: Rule #1 Finding the Area to the Left of any Z-value

To find the area to the left of any z-value, look up the z-value in the Standard Normal Distribution Table. The value listed is the area.

Example \(\PageIndex{1}\) Finding the Area to the Left of Any Z-Value

Find the area under the Standard Normal Distribution Curve to the left of the given z-values.

1. To the left of z = 2.43.
2. To the left of z = –1.57

Solution

1. Locate the row for the first part of the z-value (2.4). Move across the columns to the one labeled 0.03. Find the value where the row and column meet. This gives the cumulative area to the left of z = 2.43, which is 0.9925.
2. Locate the row for the first part of the z-value (–1.5). Move across the columns to the one labeled 0.07. Find the value where the row and column meet. This gives the cumulative area to the left of z = –1.57, which is 0.0582.

Definition: Rule #2 Finding the Area to the Right of any Z-value

To find the area to the right of any z-value, look up the z-value in the Standard Normal Distribution Table, then subtract the listed area from one.

Example \(\PageIndex{2}\ Finding the Area to the Right of Any Z-Value

Find the area under the Standard Normal Distribution Curve to the right of the given z-values.

1. To the right of z = 0.95.
2. To the right of z = –3.36

Solution

1. Locate the row for the first part of the z-value (0.9). Move across the columns to the one labeled 0.05. Find the value where the row and column meet. This gives the cumulative area to the left of z = 0.95, which is 0.8289. Now, subtract this area from 1 to get 1 – 0.8289 = 0.1711.
2. Locate the row for the first part of the z-value (–3.3). Move across the columns to the one labeled 0.06. Find the value where the row and column meet. This gives the cumulative area to the left of z = –3.36, which is 0.0004. Now, subtract this area from 1 to get 1 – 0.0004 = 0.9996.

Definition: Rule #3 Finding the Area Between Any Two Z-values

To find the area between any two z-values, look up both z-values in the Standard Normal Distribution Table, then subtract the smaller area from the larger area.

Example \(\PageIndex{3}\) Finding the Area Between Any Two Z-Values

Find the area under the Standard Normal Distribution Curve between the two given z-values.

1. Between z = 1.42 and z = 1.80
2. Between z = –0.47 and z = 2.82
3. Between z = –4.02 and –1.67.

Solution

1. Start by finding the area to the left of z = 1.80. According to the table, the area to the left of z = 1.80 is 0.9641. Next, find the area to the left of z = 1.42, which is 0.9222. To get the area between the two z-values, subtract the smaller area from the larger one: 0.9641 minus 0.9222 equals 0.0419.
2. Start by finding the area to the left of z = 2.82. According to the table, the area to the left of z = 2.82 is 0.9976. Next, find the area to the left of z = –0.47, which is 0.3192. To get the area between the two z-values, subtract the smaller area from the larger one: 0.9976 minus 0.3192 equals 0.6784.
3. Start by finding the area to the left of z = –1.67. According to the table, the area to the left of z = –1.67 is 0.0475. Next, find the area to the left of z = –4.02, which is 0.0001. Since this value is less than z = –3.49, we use the area of 0.0001 as noted below the table. To get the area between the two z-values, subtract the smaller area from the larger one: 0.0475 minus 0.0001 equals 0.0474.

Z-Score Computation

Z-score is a great way to compare different individuals and determine who was ranked relatively higher among his or her group.

The z-score formula: z = \(\dfrac{\text{Individual}-\text{Mean}}{\text{Standard Deviation}}\)

Example \(\PageIndex{4}\) Calculating the Z-scores

Kelly scored 50 points on Exam #1 for her statistics class. The mean of the class was 60, and the standard deviation was 4. John scored a 75 on Exam #1 for his geography class. The mean of the class was 90 with a standard deviation of 8. Who did relatively better compared to their class? Assume the distribution for both classes is normally distributed (bell-shaped).

Solution

• In this case, we can’t compare their exam scores directly because they are two different exams. In addition, the total points for each exam could be different.
• Since we can’t compare their exam scores directly, we need to convert each exam score to a z-score. Then, we will be able to compare their z-scores to know who scored relatively higher among their group.
• We need to use the Z formula to find the relative position of Kelly and John compared to the students taking the same exam.

Z-scores

Calculating Kelly's z-score	Calculating John's z-score
Individual = 50 (Kelly's Score) Mean = 60 Standard Deviation = 4 z = \(\dfrac{\text{Individual}-\text{Mean}}{\text{Standard Deviation}}\) z = \(\dfrac{50-60}{4}\) z = \(\dfrac{-5}{4}\) z = -1.25	Individual = 75 (John's Score) Mean = 90 Standard Deviation = 8 z = \(\dfrac{\text{Individual}-\text{Mean}}{\text{Standard Deviation}}\) z = \(\dfrac{75-90}{8}\) z = \(\dfrac{-15}{8}\) z = -1.875 \(\approx\) -1.88

Table \(\PageIndex{5}\): Computing Z-scores

Since Kelly’s z-score was higher than John’s (-1.25) is farther to the right on the number line than (-1.88), we say that Kelly scored relatively higher among her classmates than John among his classmates.

Important Notes

• Kelly and John’s z-scores are both negative. This means their exam scores were below the mean score.
• If the z-score is positive, it means the individual’s value is above the mean.
• If the z-score is zero, it means the individual’s value is exactly the mean.

Finding Probability Less Than a Z-score

Using the standard normal distribution and z table, the z-score can also tell us roughly the percentage of individuals in the group that was below a value (or someone’s value). This can be done if the data is roughly normally distributed.

Example \(\PageIndex{5}\) Finding Probability Less Than z-score

Kelly recently took Exam #1 for her statistics class and scored 55 points. The class average for the exam was 60, with a standard deviation of 4. Kelly is curious about how her score compares to her classmates. Using the standard normal distribution, we determine that her z-score is -1.25.

Assuming the exam scores are normally distributed, approximately what percentage of students scored lower than Kelly? Additionally, what does this tell us about how she performed relative to the rest of the class?

Solution

In this case, we are trying to find \(P(z < -1.25)\)

We can find our answer by using a standard normal distribution and the related table.

We calculated Kelly’s z-score earlier, which was -1.25. Since the value of Kelly’s z-score is negative, it will be to the left of the center. We shade the area to the left side, z = -1.25.

Figure \(\PageIndex{6}\): Standard Normal Distribution, Area Shaded to the Left of z=-1.25

Kelly’s z-score = -1.25

To find the area or probability to the left of z = -1.25, we use the standard normal distribution table provided below. We go down -1.2 and move over to the right 0.05 to get the area of 0.1056. This area (probability) represents the area to the left of z=-1.25.

Figure \(\PageIndex{7}\): Standard Normal Distribution Table. Describes How to Find Probability of \(z < -1.25\)

\(P(z < -1.25)\approx 0.1056 \approx 10.56%\)

Thus, the probability of students who scored less than Kelly on the statistics exam is 0.1056 (10.56%).

Note: Computing the Area to the Left of a Z-value Using the TI-84+ Calculator

Steps to Enable Stat Wizard:

1. Press [MODE] to open the mode settings.
2. Use the down arrow key to scroll down until you see [STAT WIZARDS].
3. Highlight [STAT WIZARDS] and use the right arrow key to toggle between ON and OFF.
4. Select [ON] and press [ENTER] to confirm.
5. Press [2nd] then [QUIT] to exit the mode settings.

Steps to Find the Area/Probability Using a TI-84+ Calculator:

1. Press the [2nd] button.
2. Press [VARS] to open the [DISTR] menu.
3. Select [2: normalcdf(] by either scrolling down and pressing [ENTER] or simply pressing 2.
4. Enter the lower bound, which for a standard normal distribution extends to negative infinity. Since the calculator cannot input infinity, use -E99 (which represents a very large negative number). To enter -E99, press (-) (the negative sign is not the same as the subtraction sign), then [2nd], then [EE], and finally type 99.
5. Enter the upper bound as -1.25 (the given z-score).
6. Enter the mean as 0 and the standard deviation as 1 for the standard normal distribution.
7. Press [ENTER] to compute the probability.

The calculator will display 0.1056.

Finding Probability Greater Than a z-score

Using the standard normal distribution and z table, the z-score can also tell us roughly the percentage of individuals in the group that was above a value (or someone’s value). This can be done as long as the data is roughly normally distributed.

Example \(\PageIndex{6}\) Finding Probability Greater Than z-score

Kelly scored 55 points on Exam #1 for her statistics class. The mean of the class was 60, and the standard deviation was 4. Determine approximately the percentage (or proportion) of students who scored above Kelly’s result (z = - 1.25). Assume the exam scores for the class are normally distributed.

Solution

In this case, we are trying to find \(P(z>-1.25)\).

The inequality (>) is pointing to the right, so we shade on the right side of our z-score. This shaded area will represent all the z-values greater than z = -1.25.

We can find our answer by using a standard normal distribution curve and a z-table. However, the standard normal distribution table gives the area to the left of -1.25.

Shade on the right side of -1.25 because we want to find the probability of the students who scored greater than Kelly’s exam score.

Figure \(\PageIndex{8}\): Standard Normal Distribution, Area Shaded to the Right of z=-1.25

Recall that the area (probability) under the normal distribution is 1 (100%). If we add the left area to the right area, we should get 1.

Left Area + Right Area = 1

In other words, if we subtract the whole area (in this case, it is 1) from the left area, we will be left with only the right area.

Right Area = 1 - Left Area

Kelly’s z-score = -1.25

To find the left area, we go down -1.2 and move over to the right 0.05. The left area is 0.1056.

Right Area = 1 - 0.1056 = 0.8944

\(P(z>-1.25)=0.8944\)=89.44%

Thus, the probability of students who scored more than Kelly on the statistics exam is 0.8944 (89.44%).

Figure \(\PageIndex{9}\): Standard Normal Distribution Table. Describes How to Find Probability of \(z\le-1.25\)

Note: Computing the Area to the Right of a Z-Value Using the TI-84+ Calculator

1. Press the [2nd] button.
2. Press [VARS] to open the [DISTR] menu.
3. Select [2: normalcdf(] by either scrolling down and pressing [ENTER] or simply pressing 2.
4. Enter the lower bound -1.25 (the given z-score).
5. Enter the upper bound, which for a standard normal distribution extends to positive infinity. Since the calculator cannot input infinity, use E99 (which represents a very large negative number). To enter E99, press [2nd], then [EE], and finally type 99.
6. Enter the mean as 0 and the standard deviation as 1 for the standard normal distribution.
7. Press [ENTER] to compute the probability.

The calculator will display 0.8944.

Finding Probability Between Two z-scores

Using the standard normal distribution and z-table, the z-score can also tell us roughly the percentage of individuals in the group that were between two values. This can be done as long as the data is roughly normally distributed.

Example \(\PageIndex{7}\)

For Exam #1 in a statistics class at Citrus College, students' scores follow a normal distribution. The professor wants to analyze how many students performed within a specific range compared to the class average. Using the standard normal distribution, determine the percentage of students whose z-scores fall between z = -1.53 and z = 2.17.

Solution

In this case, we are trying to find \(P\left(-1.53<z<2.17\right)\)

We know we need to shade the middle because it asks to find the area (probability) between two z values (z = -1.53 and z = 2.17).

Figure \(\PageIndex{10}\): Standard normal distribution, area shaded between z=-1.53 and z=2.17

We can find our probability to the left of each z-score by using the standard normal distribution table. If we subtract the lower probability value from the higher probability value, we will find the area (probability) between z = -1.53 and z = 2.17.

\(P\left(-1.53<z<2.17\right)=P\left(z<2.17\right)-P\left(z<-1.53\right)\)

To find \(P\left(z<2.17\right)\) and \(P\left(z<-1.53\right)\) we use the z-table.

z = 2.17

To find the area or probability to the left of z = 2.17, we use the standard normal distribution table provided below. We go down 2.1 and move over to the right 0.07 to get the area of 0.9850. This area (probability) represents the area to the left of z=2.17.

Figure \(\PageIndex{11}\): Standard Normal Distribution Table. Describes How to Find Probability of \(z<2.17\)

\(P\left(z<2.17\right)=0.9850\)

z = -1.53

To find the area or probability to the left of z = -1.53, we use the standard normal distribution table provided below. We go down -1.5 and move over to the right 0.03 to get the area of 0.0630. This area (probability) represents the area to the left of z=-1.53.

Figure \(\PageIndex{12}\): Standard Normal Distribution Table. Describes How to Find Probability of \(z<-1.53\)

\(P\left(z<-1.53\right)=0.0630\)

Now that we found \(P\left(z<2.17\right)=0.9850\) and \(P\left(z<-1.53\right)=0.0630\), we can use the following formula:

\(P\left(-1.53<z<2.17\right)=P\left(z<2.17\right)-P\left(z<-1.53\right)\)

\(=0.9850-0.0630\)

\(=0.9220\)

Thus, the probability of students with z-scores between -1.53 and 2.17 is 0.9220 (92.20%).

Note: Computing the Area Between Two Z-Values Using the T-84+ Calculator

1. Press the [2nd] button.
2. Press [VARS] to open the [DISTR] menu.
3. Select [2: normalcdf(] by either scrolling down and pressing [ENTER] or simply pressing 2.
4. Enter the lower bound -1.53 (the given z-score).
5. Enter the upper bound 2.17 (the given z-score).
6. Enter the mean as 0 and the standard deviation as 1 for the standard normal distribution.
7. Press [ENTER] to compute the probability.

The calculator will display 0.9220.

Summary: Finding Probability When the z-score is Known

1. To find the area (probability) left of a z-score (fence), use the z-table.

Figure \(\PageIndex{13}\): Standard Normal Distribution Curve Where the Area to the Left of z-score is Shaded.

2. To find the area (probability) on the right of the z-score (fence), we perform the following expression 1 - (value obtained from the z table)

Figure \(\PageIndex{14}\): Standard Normal Distribution Curve Where the Area to the Right of z-score is Shaded.

3. To find the area (probability) between two z-scores (fences), we subtract the corresponding areas (take the higher area and subtract the lower area).

   

   Figure \(\PageIndex{15}\): Standard Normal Distribution Curve Where the Area is Between Two z-values

Finding z-score When the Probability is Known

Finding z-score of Probability to the Left

We can also find the z-score (fence) using the z-table if we know the area (probability) to the left of the z-score.

Example \(\PageIndex{8}\) Finding z-score of Probability to the Left

Find the z-score based on the information provided below.

Figure \(\PageIndex{16}\): Standard Normal Distribution Curve where the Area to the Left of the z-score is 0.70

Solution

Based on the diagram above, we have our area (probability) on the left of our z-score (fence). In this case, the area (probability) is 0.70. We can use the z-table to find our z-score.

To find our z-score, we start from the area (probability) region and identify the row and column of the closest number to 0.70. In this case, it will be 0.6985. The row for this value will be 0.5, and the column will be 0.02. If we add the row and column, we get our z-score value of 0.52.

Figure \(\PageIndex{17}\): Explains How to Find the Z-score from a Z-Table where the Probability (Area) to the Left is Given

z = 0.52

For this example, our z-score (fence) will be 0.52.

Finding z-score of Probability to the Right

We can only find the z-score using the z-table if we know the area (probability) on the left of the z-score. If the right area is given, we first need to find the Left Area before we can use the z-table.

Example \(\PageIndex{9}\) Finding z-score of Probability to the Right

Find the z-score based on the information provided below.

Figure \(\PageIndex{18}\): Standard Normal Distribution Curve Where the Area to the Right of z-score is 0.30

Solution

Recall that the area under the standard normal distribution is 1 (100%). If we add the Left Area with the Right Area that should give us the whole area (in this case it will be 1).

Figure \(\PageIndex{19}\): Standard Normal Distribution Curve Where the Left and Right Area are Separated by the z-score.

Left Area + Right Area = 1

In other words, if we subtract the Whole Area (in this case, it is 1) from the Right Area, what will remain will be the Left Area.

Left Area = 1 - Right Area

Left Area = 1 - 0.30 = 0.70

To find our z-score, we start from the area (probability) region and identify the row and column of the closest number to 0.70. In this case, it will be 0.6985. The row for this value will be 0.5, and the column will be 0.02. If we add the row and column we get our z-score value of 0.52.

Figure \(\PageIndex{20}\): Explains How to Find the z-score From a z-Table Where the Probability (Area) to the Left is Given

z = 0.52

For this example, our z-score (fence) will be 0.52.

Finding the z-scores of the Area (Probability) Between Them

Sometimes we will have a question where that asks us to find two z-scores when the area (probability) of the middle is given. The same rule applies to our previous two examples. We can use the z-table to find our z-scores if we know the area (probability) to the left of each z-score.

Example \(\PageIndex{9}\) Finding the z-scores of the Area (Probability) Between Them

Find the z-scores based on the information provided below.

Figure \(\PageIndex{21}\): Standard Normal Distribution Curve Where the Middle Area is 0.90.

Solution

We first need to find the area of all the regions before we can find \(z_1\) and \(z_2\).

Figure \(\PageIndex{22}\): Standard Normal Distribution Curve Where the Left, Middle, and Right Region are Separated by Our z-score Fences.

Recall area under the standard normal distribution is 1. If we subtract the Whole Area (in this case, 1) from the Middle Region, then we are left with the Outside Region (which we will eventually call ).

Outside Region = Whole Area - Middle Region

Outside Region = 1 - 0.90 = 0.10

Note: Outside Region = Left Region + Right Region = 0.10

Figure \(\PageIndex{23}\): Normal Distribution Curve that Shows the Middle Region is 0.90 and the Outside Region (Left and Right Region Together) is 0.10.

Authors

"6.1: Introduction to the Normal Distribution" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0

Attributions

"6.2: Graphs of the Normal Distribution" by Kathryn Kozak is licensed under CC BY-SA 4.0