Normal Approximation: Calculating Percentages Under the Normal Curve

Normal approximation uses areas under a normal curve to estimate the proportion of observations within a specified interval.

If a variable is reasonably modeled by:\[ X\sim N(\mu,\sigma^2), \]

then any interval on the original measurement scale can be converted to an interval under the standard normal distribution.

The general process is:

verify that a normal model is reasonable;
standardize the interval endpoints;
identify the desired area;
use the standard normal cumulative distribution;
subtract cumulative probabilities;
interpret the result in the original context.

The Example

Suppose father heights are approximately normally distributed with:\[ \mu=68.3\text{ inches} \]

and:\[ \sigma=1.8\text{ inches}. \]

We want to estimate:\[ P(67.4\leq X\leq71.9). \]

In words:

What percentage of the height distribution lies between 67.4 and 71.9 inches?

Why Mean and Standard Deviation Are Sufficient Here

Within the family of normal distributions, the mean and standard deviation completely determine the curve.

Thus, if:\[ X\sim N(68.3,1.8^2), \]

then all model-based probabilities can be calculated from these two parameters.

However, the mean and standard deviation do not determine an arbitrary observed histogram. The calculation is valid only to the extent that the normal model reasonably approximates the data.

Mean and standard deviation are sufficient for computing probabilities after normality has been assumed or justified—not for describing every possible distribution.

Step 1: Standardize the Lower Endpoint

The z-score formula is:\[ z=\frac{x-\mu}{\sigma}. \]

For the lower endpoint:\[ x=67.4. \]

Therefore:\[ z_{\text{lower}} = \frac{67.4-68.3}{1.8}. \]

Calculate the numerator:\[ 67.4-68.3=-0.9. \]

Then:\[ z_{\text{lower}} = \frac{-0.9}{1.8} = -0.5. \]

Thus, 67.4 inches is half a standard deviation below the mean:\[ \boxed{ 67.4\text{ inches} \longleftrightarrow z=-0.5 } \]

Step 2: Standardize the Upper Endpoint

For the upper endpoint:\[ x=71.9. \]

Therefore:\[ z_{\text{upper}} = \frac{71.9-68.3}{1.8}. \]

Calculate:\[ 71.9-68.3=3.6. \]

Then:\[ z_{\text{upper}} = \frac{3.6}{1.8} = 2. \]

Thus, 71.9 inches is two standard deviations above the mean:\[ \boxed{ 71.9\text{ inches} \longleftrightarrow z=2 } \]

Transforming the Probability Statement

The original probability is:\[ P(67.4\leq X\leq71.9). \]

After standardization:\[ P(67.4\leq X\leq71.9) = P(-0.5\leq Z\leq2). \]

The original height interval and standardized interval represent the same region of their respective distributions.

			
Original scale
67.4             68.3                         71.9
  |----------------|-----------------------------|
 z=-0.5           mean                           z=2
Standard normal scale
-0.5               0                              2
  |----------------|------------------------------|

		

Step 3: Identify the Desired Area

Under the standard normal curve, we want the area between:\[ z=-0.5 \]

and:\[ z=2. \]

Conceptually:

			
Density
   |
   |                         /\
   |                       /████\
   |                     /████████\
   |                  __/███████████\__
   |_______________ /███████████████████\________
                  -0.5       0          2
                    |<--- desired area -->|

		

Because probability under a continuous distribution corresponds to area:\[ P(-0.5\leq Z\leq2) \]

is the shaded region.

The Standard Normal Cumulative Distribution

The standard normal cumulative distribution function is:\[ \boxed{ \Phi(z)=P(Z\leq z) } \]

It returns the area under the standard normal curve to the left of \(z\).

For example:\[ \Phi(2)=P(Z\leq2). \]

Modern statistical software can often calculate interval probabilities directly, but expressing the calculation with \(\Phi\) makes the underlying logic clear and works with traditional cumulative tables.

Step 4: Express the Interval Using Left-Tail Areas

The area to the left of 2 includes both:

the desired interval from \(-0.5\) to 2;
the unwanted area below \(-0.5\).

Therefore:\[ P(-0.5\leq Z\leq2) = P(Z\leq2)-P(Z\leq-0.5). \]

Using cumulative notation:\[ \boxed{ P(-0.5\leq Z\leq2) = \Phi(2)-\Phi(-0.5) } \]

This is the general interval formula:\[ \boxed{ P(a\leq Z\leq b) = \Phi(b)-\Phi(a) } \]

for \(a<b\).

Step 5: Look Up the Cumulative Probabilities

From the standard normal distribution:\[ \Phi(2)\approx0.97725. \]

Thus, approximately 97.725% of the standard normal distribution lies to the left of \(z=2\).

Also:\[ \Phi(-0.5)\approx0.30854. \]

Thus, approximately 30.854% lies to the left of \(z=-0.5\).

Subtract:\[ P(-0.5\leq Z\leq2) = 0.97725-0.30854. \]

Therefore:\[ P(-0.5\leq Z\leq2) \approx0.66871. \]

As a percentage:\[ \boxed{ P(67.4\leq X\leq71.9) \approx66.87\% } \]

Using rounded table values of 97.7% and 30.9% gives approximately 66.8%.

Visual Interpretation of the Subtraction

The calculation can be visualized as:

			
Area to the left of 2
██████████████████████████████████████████████░
= 0.97725
minus area to the left of -0.5
███████████████░
= 0.30854
equals area between -0.5 and 2
               ███████████████████████████████░
= 0.66871

		

The subtraction removes everything below the lower endpoint, leaving only the interval of interest.

General Formula on the Original Scale

If:\[ X\sim N(\mu,\sigma^2), \]

then:\[ \boxed{ P(a\leq X\leq b) = \Phi \left( \frac{b-\mu}{\sigma} \right) – \Phi \left( \frac{a-\mu}{\sigma} \right) } \]

For the height example:\[ P(67.4\leq X\leq71.9) \]

equals:\[ \Phi \left( \frac{71.9-68.3}{1.8} \right) – \Phi \left( \frac{67.4-68.3}{1.8} \right). \]

Therefore:\[ = \Phi(2)-\Phi(-0.5) \approx0.6687. \]

Left-Tail Probabilities

To calculate:\[ P(X\leq x), \]

standardize \(x\):\[ z=\frac{x-\mu}{\sigma}. \]

Then:\[ \boxed{ P(X\leq x)=\Phi(z) } \]

For example:\[ P(X\leq71.9) = \Phi(2) \approx0.97725. \]

Thus, approximately 97.7% of the modeled height distribution lies at or below 71.9 inches.

Right-Tail Probabilities

To calculate:\[ P(X>x), \]

use the complement rule:\[ P(X>x) = 1-P(X\leq x). \]

Therefore:\[ \boxed{ P(X>x) = 1- \Phi \left( \frac{x-\mu}{\sigma} \right) } \]

For 71.9 inches:\[ P(X>71.9) = 1-\Phi(2). \]

Thus:\[ P(X>71.9) = 1-0.97725 = 0.02275. \]

As a percentage:\[ \boxed{ P(X>71.9)\approx2.28\% } \]

Intervals Centered at the Mean

For a symmetric interval:\[ P(\mu-k\sigma\leq X\leq\mu+k\sigma), \]

the standardized interval is:\[ P(-k\leq Z\leq k). \]

Using symmetry:\[ P(-k\leq Z\leq k) = \Phi(k)-\Phi(-k). \]

Since:\[ \Phi(-k)=1-\Phi(k), \]

we obtain:\[ \boxed{ P(-k\leq Z\leq k) = 2\Phi(k)-1 } \]

For \(k=2\):\[ P(-2\leq Z\leq2) = 2\Phi(2)-1. \]

Therefore:\[ = 2(0.97725)-1 = 0.9545. \]

Thus, the exact normal-curve area within two standard deviations is approximately 95.45%.

The Empirical Rule as a Normal Approximation

The empirical rule states that approximately 95% of a normal distribution lies between:\[ \mu-2\sigma \]

and:\[ \mu+2\sigma. \]

On the standardized scale:\[ P(-2\leq Z\leq2)\approx0.95. \]

The exact normal value is:\[ P(-2\leq Z\leq2)\approx0.9545. \]

The empirical rule rounds this to 95% for convenient mental calculation.

Why the Empirical Rule Gives 97.5% Below Two Standard Deviations

Using the rounded empirical rule:

95% lies between \(-2\) and 2;
5% lies outside;
symmetry places 2.5% in each tail.

Therefore, the approximate area below \(z=2\) is:\[ 95\%+2.5\%=97.5\%. \]

Using the exact normal distribution:\[ \Phi(2)\approx97.725\%. \]

Both values are correct in their intended contexts:

97.5% is an empirical-rule approximation;
97.725% is the more precise standard normal value.

Symmetry of the Normal Curve

The standard normal distribution is symmetric around zero:\[ \Phi(-z)=1-\Phi(z). \]

For \(z=0.5\):\[ \Phi(-0.5) = 1-\Phi(0.5). \]

Since:\[ \Phi(0.5)\approx0.69146, \]

we get:\[ \Phi(-0.5) = 1-0.69146 = 0.30854. \]

This provides another way to obtain the lower cumulative probability.

Area Between the Mean and a Positive Z-Score

Sometimes a table reports the area between 0 and \(z\) rather than the area to the left.

For positive \(z\):\[ P(0\leq Z\leq z) = \Phi(z)-0.5. \]

For \(z=2\):\[ P(0\leq Z\leq2) = 0.97725-0.5 = 0.47725. \]

For \(z=0.5\):\[ P(0\leq Z\leq0.5) = 0.69146-0.5 = 0.19146. \]

The area from \(-0.5\) to 2 is then:\[ 0.19146+0.47725 = 0.66871. \]

This agrees with the cumulative-subtraction method.

Different Normal Tables Use Different Conventions

Before using a printed normal table, check what it reports.

A table may provide:

area to the left of \(z\);
area between 0 and \(z\);
area to the right of \(z\);
upper-tail critical values.

For example, the entry corresponding to \(z=2\) could be:

\(0.97725\) for left cumulative area;
\(0.47725\) for area between 0 and 2;
\(0.02275\) for right-tail area.

The numbers differ because the tables answer different questions.

Computing the Result with Software

The calculation can be performed using a normal cumulative distribution function:

			
from math import erf, sqrt
def standard_normal_cdf(z):
    return 0.5 * (1 + erf(z / sqrt(2)))
lower_z = -0.5
upper_z = 2.0
probability = (
    standard_normal_cdf(upper_z)
    - standard_normal_cdf(lower_z)
)
print(probability)

		

The result is approximately:

0.668712

The important part is not the particular function used. It is translating the original question into the correct cumulative-area calculation.

Approximation Versus Exact Sample Percentage

The normal calculation estimates the percentage under a theoretical model:\[ P(67.4\leq X\leq71.9) \approx66.87\%. \]

If all observed measurements are available, the empirical sample percentage is:\[ \frac{ \text{number of observed heights between 67.4 and 71.9} }{ \text{total number of observed heights} }. \]

These two quantities need not be identical.

Differences can arise because:

the data are not exactly normal;
the sample is finite;
the mean and standard deviation are estimated;
measurements are rounded;
the sample may contain outliers or subgroups.

The model-based percentage and observed percentage answer related but different questions.

Checking the Normal Model

Before applying the approximation, examine:

a histogram;
a density plot;
a boxplot;
a normal Q–Q plot;
the difference between mean and median;
skewness;
tail behavior;
potential subgroups.

A normal approximation is questionable when the distribution is:

strongly skewed;
multimodal;
heavy-tailed;
bounded with observations concentrated near a boundary;
dominated by extreme values.

Common Mistakes

Forgetting to standardize both endpoints

An interval requires two z-scores.

Subtracting in the wrong order

The interval probability is:\[ \Phi(z_{\text{upper}}) – \Phi(z_{\text{lower}}). \]

Reversing the order produces a negative number.

Using density height instead of cumulative area

The normal density \(\phi(z)\) is not the percentage below \(z\). Use the cumulative function \(\Phi(z)\).

Forgetting the complement for a right tail

\[ P(Z>z)=1-\Phi(z). \]

Assuming mean and standard deviation are sufficient for arbitrary data

They determine the normal model, not every distribution.

Treating the approximation as an exact observed percentage

A finite dataset will generally differ from its fitted curve.

Confusing 95% with the exact area between \(-2\) and 2

The exact normal value is approximately 95.45%; 95% is the empirical-rule approximation.

Rounding z-scores too early

Premature rounding can noticeably alter small tail probabilities.

A Reliable Normal-Approximation Process

Verify that a normal model is reasonable.
Write the probability statement on the original scale.
Standardize the lower endpoint.
Standardize the upper endpoint.
Sketch or identify the desired region.
Express it using cumulative probabilities.
calculate the upper cumulative area.
subtract the lower cumulative area.
convert the result to a percentage.
State that the result is model-based and approximate.

Key Takeaway

To find the percentage between two values under a normal model, standardize both endpoints and subtract their cumulative standard normal areas:\[ P(a\leq X\leq b) = \Phi \left( \frac{b-\mu}{\sigma} \right) – \Phi \left( \frac{a-\mu}{\sigma} \right). \]

For heights with mean 68.3 inches and standard deviation 1.8 inches:\[ 67.4\longrightarrow z=-0.5 \]

and:\[ 71.9\longrightarrow z=2. \]

Therefore:\[ P(67.4\leq X\leq71.9) = \Phi(2)-\Phi(-0.5) \approx66.87\%. \]

The calculation is meaningful only when the normal curve provides a reasonable approximation to the distribution.