5.6: Application of the Normal Distribution

Now that we’ve explored the normal distribution and how z-scores standardize values, we’re ready to apply these tools to real-world problems.

We’ll work through five example scenarios that involve different types of probability questions using the normal distribution. You may use a z-table, calculator, statistical software, or graphing technology to answer these.

If you are using Excel, you will use these two functions:

Excel function	Output	Example and interpretation
=NORM.DIST( x, , , TRUE)	This gives us the area under the curve to the left of x	Ex: =NORM.DIST(3, 2, 1, TRUE) calculates to be 0.84134475.      This means that the probability that a random value distributed normally with mean 2 and standard deviation 1 is less than 3 is 0.84 or about 84%.
=NORM.INV( p, , )	(Inverse) This will find the x- value with p as the area under the curve to its left	Ex: =NORM.INV(0.25, 2, 1) calculates to be 1.3255.     This means that there is a 25% probability of getting a value less than 1.3255 in the normal distribution with mean 2 and standard deviation 1

Reminder: For all these problems, assume the data is approximately normally distributed unless stated otherwise.

Example 1: Area to the Left (Lower Tail)

The heights of adult women in the U.S. are approximately normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Question: What proportion of adult women are shorter than 60 inches?

Using Technology: Calculate P(x<60) where \(\mu = 64\) and \(\sigma = 3\).  This is the area to the left of 60 in this normal distribution

With Excel: =NORM.DIST(60, 64, 3, TRUE) = 0.091211

Using a z-table: Convert to a z-score: \( z = \frac{60 - 64}{3} \approx -1.33 \) then look up P(Z<-1.33) in the table: 0.1020

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Example 2: Area to the Right (Upper Tail)

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.

Question: What percentage of people have an IQ above 120?

Find the area to the left (from the table), then subtract from 1

Using Technology: Calculate \(P(x>120)\) where \(\mu = 100\) and \(\sigma = 15\).  This is the area to the right of 120 in this normal distribution

With Excel: =1-NORM.DIST(120, 100, 15, TRUE)

Using a z-table: Convert to a z-score: \( z = \frac{120 - 100}{15} \approx 1.33 \) then look up P(Z<1.33) in the table and subtract from 1. 

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Example 3: Area Between Two Values

A college class takes a final exam. The scores are normally distributed with a mean of 72 and standard deviation of 8.

Question: What proportion of students scored a C?

Find the area to the left for both endpoints, then subtract the smaller area from the bigger area.

Using Technology: Calculate \(P(70<x<80)\) where \(\mu = 72\) and \(\sigma = 8\).  This is the area between 70 and 80 in this normal distribution

With Excel: =NORM.DIST(80, 72, 8, TRUE)-NORM.DIST(70, 72, 8, TRUE)

Using a z-table: Convert to z-scores: \( z_1 = \frac{70 - 72}{8} = -\frac{1}{4}, z_2 = \frac{80 - 72}{8} = 1 \) and then look up P(Z<-0.25) and P(Z<1) in the table and subtract \( P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1) \).

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Example 4: Percentile to Value (Reverse Z)

The SAT Math test scores are normally distributed with a mean of 520 and standard deviation of 100.

Question: What score corresponds to the 90th percentile? Recall that a student in the 90th percentile would score higher than 90% of other students. In other words, we have a proportion of 0.90 to the left.

Using Technology: Find \(a\) so that \(P(x<a) = 0.90\) where \(\mu = 520\) and \(\sigma = 100\).  This is the score with an area to the left of 0.90 in this normal distribution

With Excel: =NORM.INV(0.90, 520, 100)

Using a z-table: Look up the area closest to 0.9000 in the z-table. Note: Since the area is greater than 0.5 it will be a positive z-score. The closest z-score is 1.28. Take that z-score and plug into: \( x = \mu + z\sigma = 520 + 1.28(100) = 648\).

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Example 5: Top percentage to Value (Reverse Z and Inverse percentile)

Jordan earns a score that puts them in the top 2% of all test takers. The mean score is 75 and the standard deviation is 6.

Question: What score did Jordan get on the test?

Using Technology: Find \(b\) so that \(P(x>b) = 0.02\) where \(\mu = 75\) and \(\sigma = 6\).  This is the score with an area to the right of 0.02 in this normal distribution

With Excel: =NORM.INV(1-0.02, 75, 6)

Using a z-table: Subtract the area from 1 to get \(1-0.02=0.98\). Look up the area closest to 0.9800 in the z-table. The closest z-score is 2.05. Take that z-score and plug into: \( x = \mu + z\sigma = 75 + 2.05(6)\).

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Normal Distribution Check-In

Use the standard normal table or technology for these questions. Answers rounded to 4 decimal places.

1. Z-score

A student scores 84 on a science test. The mean is 75 and the standard deviation is 5. What is the z-score?

Check Answer

2. Area to the Left

What proportion of data falls below a z-score of 0.85?

Check Answer

3. Area to the Right

What proportion of values are greater than a z-score of 1.2?

Check Answer

4. Area Between

What proportion of values fall between z = -1.0 and z = 1.0?

Check Answer

5. Value from Percentile

SAT Math scores are normally distributed with mean = 500 and standard deviation = 100. What score corresponds to the 84th percentile?

Check Answer

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Tip for Practice

Use a z-table or software to determine areas under the standard normal curve. The goal here is to understand the logic and patterns — not just get numbers. These tools will be used again when we talk about confidence intervals and hypothesis testing.

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