Finding Percentiles with the Normal Distribution

Normal approximation can be used in two directions.

In the first direction, a measurement is given and we calculate the percentage below or above it:\[ x \longrightarrow z \longrightarrow \text{probability}. \]

In the reverse direction, a percentage is given and we find the corresponding measurement:\[ \text{probability} \longrightarrow z \longrightarrow x. \]

The second process is used to calculate percentiles and cutoff values.

What Is a Percentile?

The \(p\)th percentile is a value below which approximately \(p\%\) of the distribution lies.

If \(x_p\) is the \(p\)th percentile, then:\[ \boxed{ P(X\leq x_p)=\frac{p}{100} } \]

For example, the 30th percentile satisfies:\[ P(X\leq x_{30})=0.30. \]

This means:

• approximately 30% of observations lie at or below \(x_{30}\);
• approximately 70% lie above it.

For a continuous distribution, the distinction between “below” and “at or below” does not affect the probability because:\[ P(X=x)=0. \]

The Standard Normal Percentile

Let:\[ Z\sim N(0,1). \]

The 30th percentile of the standard normal distribution is the value \(z_{0.30}\) satisfying:\[ P(Z\leq z_{0.30})=0.30. \]

Using the standard normal inverse cumulative distribution:\[ z_{0.30} = \Phi^{-1}(0.30). \]

Numerically:\[ \boxed{ z_{0.30}\approx-0.5244 } \]

Rounded to two decimal places:\[ z_{0.30}\approx-0.52. \]

Why the Z-Score Is Negative

The standard normal distribution is symmetric around zero:\[ P(Z\leq0)=0.50. \]

The 30th percentile lies below the median because:\[ 0.30<0.50. \]

Therefore:\[ z_{0.30}<0. \]

A z-score of approximately \(-0.52\) means that the 30th percentile lies about 0.52 standard deviations below the mean.

Height Example

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

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

We want to find \(x_{30}\), the 30th percentile of this modeled height distribution.

Reversing the Standardization Formula

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

Multiply both sides by \(\sigma\):\[ z\sigma=x-\mu. \]

Add \(\mu\):\[ \boxed{ x=\mu+z\sigma } \]

This is the inverse-standardization formula.

For a percentile:\[ \boxed{ x_p = \mu+ z_p\sigma } \]

where:\[ z_p=\Phi^{-1}(p). \]

Here, \(p\) is written as a decimal, so the 30th percentile uses \(p=0.30\).

Calculating the 30th Percentile

Substitute:\[ \mu=68.3 \]\[ \sigma=1.8 \]

and:\[ z_{0.30}\approx-0.5244. \]

Then:\[ x_{30} = 68.3+(-0.5244)(1.8). \]

Calculate the standardized displacement:\[ (-0.5244)(1.8)\approx-0.9439. \]

Therefore:\[ x_{30} = 68.3-0.9439 \approx67.3561. \]

Rounded to one decimal place:\[ \boxed{ x_{30}\approx67.4\text{ inches} } \]

Under the normal approximation, approximately 30% of the modeled heights lie at or below 67.4 inches.

Intuitive Calculation

The same calculation can be understood without first rearranging the formula.

The z-score:\[ z=-0.5244 \]

means:

The value lies 0.5244 standard deviations below the mean.

One standard deviation is:\[ 1.8\text{ inches}. \]

Therefore, 0.5244 standard deviations correspond to:\[ 0.5244(1.8) \approx0.9439\text{ inches}. \]

Move this distance below the mean:\[ 68.3-0.9439 \approx67.4\text{ inches}. \]

This is mathematically identical to:\[ x=\mu+z\sigma. \]

Verifying the Answer

Standardize the result:\[ z = \frac{67.3561-68.3}{1.8}. \]

Therefore:\[ z \approx-0.5244. \]

Then:\[ P(X\leq67.3561) = P(Z\leq-0.5244). \]

By construction:\[ P(Z\leq-0.5244)\approx0.30. \]

Thus, the inverse calculation is consistent.

General Normal-Percentile Formula

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

the \(p\)th quantile is:\[ \boxed{ x_p = \mu+ \sigma\Phi^{-1}(p) } \]

where:\[ 0<p<1. \]

For a percentile expressed as a number from 0 to 100, first divide by 100.

For example:\[ \text{90th percentile} \longrightarrow p=0.90. \]

Four-Step Percentile Process

Step 1: Convert the percentile to a probability

For the 30th percentile:\[ p=\frac{30}{100}=0.30. \]

Step 2: Find the corresponding standard normal quantile

\[ z_p=\Phi^{-1}(p). \]

For \(p=0.30\):\[ z_p\approx-0.5244. \]

Step 3: Convert to the original scale

\[ x_p=\mu+z_p\sigma. \]

Step 4: Interpret in context

Approximately \(p\times100\%\) of the modeled distribution lies at or below \(x_p\).

Common Standard Normal Percentiles

Because the standard normal distribution is symmetric:\[ z_{1-p}=-z_p. \]

For example:\[ z_{0.70}=-z_{0.30}\approx0.5244. \]

Finding the Median

For a normal distribution, the 50th percentile has:\[ z_{0.50}=0. \]

Therefore:\[ x_{0.50} = \mu+(0)\sigma = \mu. \]

Thus:\[ \boxed{ \text{median}=\mu } \]

for a normal distribution.

In the height example:\[ x_{0.50}=68.3\text{ inches}. \]

Finding the 90th Percentile

The standard normal 90th percentile is:\[ z_{0.90}\approx1.2816. \]

For the height distribution:\[ x_{0.90} = 68.3+(1.2816)(1.8). \]

Therefore:\[ x_{0.90} \approx68.3+2.3069 \approx70.6069. \]

Rounded:\[ \boxed{ x_{0.90}\approx70.6\text{ inches} } \]

Approximately 90% of the modeled distribution lies at or below 70.6 inches, while approximately 10% lies above it.

Finding a Lower-Tail Cutoff

Suppose we want a cutoff below which only 5% of observations lie.

The relevant z-score is:\[ z_{0.05}\approx-1.6449. \]

Then:\[ x_{0.05} = 68.3+(-1.6449)(1.8). \]

Therefore:\[ x_{0.05} \approx65.34\text{ inches}. \]

Under the normal model:\[ P(X\leq65.34)\approx0.05. \]

Finding an Upper-Tail Cutoff

Suppose we want a threshold exceeded by only 5% of observations.

If 5% lies above the threshold, then 95% lies below:\[ p=0.95. \]

Use:\[ z_{0.95}\approx1.6449. \]

Then:\[ x_{0.95} = 68.3+(1.6449)(1.8) \approx71.26. \]

Thus:\[ P(X>71.26)\approx0.05. \]

A common mistake is to use \(z_{0.05}\) for an upper 5% cutoff. Upper-tail language must first be converted into a left cumulative probability.

Finding the Central 95% Interval

A central 95% interval leaves:\[ 5\% \]

outside the interval.

Symmetry places:\[ 2.5\% \]

in each tail.

The endpoints are therefore the 2.5th and 97.5th percentiles:\[ z_{0.025}\approx-1.96 \]

and:\[ z_{0.975}\approx1.96. \]

On the original scale:\[ x_{\text{lower}} = \mu-1.96\sigma \]

and:\[ x_{\text{upper}} = \mu+1.96\sigma. \]

For the height example:\[ x_{\text{lower}} = 68.3-1.96(1.8) \approx64.77 \]

and:\[ x_{\text{upper}} = 68.3+1.96(1.8) \approx71.83. \]

Thus, the central 95% of the modeled height distribution lies approximately between:\[ \boxed{ 64.8\text{ and }71.8\text{ inches} } \]

The empirical rule gives the nearby interval:\[ \mu\pm2\sigma = 64.7\text{ to }71.9. \]

Quartiles Under a Normal Model

The first and third quartiles correspond to:\[ p=0.25 \]

and:\[ p=0.75. \]

Their standard normal values are:\[ z_{0.25}\approx-0.6745 \]

and:\[ z_{0.75}\approx0.6745. \]

Therefore:\[ Q_1 = \mu-0.6745\sigma \]

and:\[ Q_3 = \mu+0.6745\sigma. \]

The normal-model IQR is:\[ \operatorname{IQR} = Q_3-Q_1. \]

Thus:\[ \operatorname{IQR} = (0.6745-(-0.6745))\sigma \]\[ \boxed{ \operatorname{IQR}\approx1.349\sigma } \]

For \(\sigma=1.8\):\[ \operatorname{IQR} \approx1.349(1.8) \approx2.43\text{ inches}. \]

Percentile Rank Versus Percentile Value

These concepts are related but opposite.

Percentile rank

Given a measurement \(x\), find the percentage below it:\[ x \rightarrow z \rightarrow \Phi(z). \]

Percentile value

Given a percentage \(p\), find the corresponding measurement:\[ p \rightarrow z_p \rightarrow x_p. \]

Computing a Percentile with Software

The 30th percentile can be calculated using an inverse normal cumulative function:

from statistics import NormalDist
mean = 68.3
standard_deviation = 1.8
probability = 0.30
z = NormalDist().inv_cdf(probability)
height = mean + z * standard_deviation
print(z)
print(height)

This produces approximately:

-0.5244
67.3561

Rounded to one decimal place:

67.4 inches

Theoretical Percentile Versus Empirical Percentile

The normal-model percentile:\[ x_p=\mu+\sigma\Phi^{-1}(p) \]

is based on a theoretical distribution.

An empirical percentile is calculated directly from ordered sample observations.

These may differ because:

• observed data are finite;
• the distribution is not exactly normal;
• parameters are estimated;
• several percentile conventions exist;
• measurements may be rounded;
• the data may contain outliers or subgroups.

If the full dataset is available and the objective is to describe that dataset, an empirical percentile can be calculated directly. The normal approximation is useful when:

• only summary statistics are available;
• a smooth model is desired;
• theoretical calculations are needed;
• the normal distribution is a reasonable approximation.

Normality Must Be Checked

The formula:\[ x_p=\mu+\sigma\Phi^{-1}(p) \]

is valid for a normal distribution. It may produce misleading results for a distribution that is:

• skewed;
• multimodal;
• heavy-tailed;
• bounded;
• affected by extreme observations;
• a mixture of subgroups.

For a right-skewed income distribution, the normal-model 99th percentile may be especially inaccurate because tail behavior differs substantially from the normal curve.

Percentile estimates near the center are often less sensitive to model errors than extreme-tail estimates, but model fit still matters.

Extreme Percentiles Require Extra Caution

Small differences in model assumptions can produce large differences in extreme quantiles.

Examples include:

• 99th percentile;
• 99.9th percentile;
• one-in-a-million threshold;
• rare-event risk limit.

A histogram may appear approximately normal near its center while having heavier tails than a normal curve.

Therefore, central percentile calculations may look reasonable while extreme percentile calculations are seriously wrong.

Tail-sensitive applications should examine:

• Q–Q plots;
• observed tail frequencies;
• alternative distributions;
• uncertainty in estimated parameters;
• sample size in the tails.

Common Mistakes

Using the percentile number as a z-score

The 30th percentile does not mean:\[ z=0.30. \]

It means:\[ \Phi(z)=0.30, \]

which gives:\[ z\approx-0.5244. \]

Forgetting to convert percentages to decimals

Use:\[ 30\%=0.30. \]

Using the z-score formula in the wrong direction

To return to the original scale:\[ x=\mu+z\sigma. \]

Subtracting a negative z-score incorrectly

When \(z<0\):\[ \mu+z\sigma \]

automatically produces a value below the mean.

Confusing an upper-tail percentage with a cumulative percentage

A cutoff exceeded by 10% is the 90th percentile, not the 10th percentile.

Claiming an exact observed percentage

The result is model-based when a normal approximation is used.

Applying the normal percentile formula without checking shape

Mean and standard deviation determine the answer only after the normal model is accepted as reasonable.

A Reliable Percentile Process

1. Express the percentile as a decimal probability \(p\).
2. Determine whether the requested area is below or above the cutoff.
3. Convert upper-tail percentages to left cumulative probabilities.
4. Find:\[ z_p=\Phi^{-1}(p). \]
5. Convert back:\[ x_p=\mu+z_p\sigma. \]
6. Preserve enough digits until the final step.
7. attach the original measurement units.
8. Interpret the percentage below and above the cutoff.
9. State that the result depends on the normal model.

Key Takeaway

To find a percentile under a normal model, first locate the corresponding standard normal z-score and then reverse the standardization:\[ z_p=\Phi^{-1}(p) \]

and:\[ x_p=\mu+z_p\sigma. \]

For the 30th percentile, \(z_{0.30}\approx-0.5244\). With mean 68.3 inches and standard deviation 1.8 inches:\[ x_{0.30} = 68.3+(-0.5244)(1.8) \approx67.4\text{ inches}. \]

Therefore, approximately 30% of the modeled height distribution lies at or below 67.4 inches.