5.6: Normal Distribution and Percentiles

Last updated
Save as PDF

Page ID: 206902

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$\newcommand{\longvect}{\overrightarrow}$

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

When you take an exam, what is often as important as your actual score on the exam is the way your score compares to other students’ performance. If you made a $70$ but the average score (whether the mean, median, or mode) was $85$, you did relatively poorly. If you made a $70$ but the average score was only $55$ then you did relatively well. In general, the significance of one observed value in a data set strongly depends on how that value compares to the other observed values in a data set. Therefore, we wish to attach a number to each observed value that measures its relative position.

As we know, quartiles in the previous section. Now we will introduce the percentile.

Percentiles

Anyone who has taken a national standardized test is familiar with the idea of being given both a score on the exam and a “percentile ranking” of that score. You may be told that your score was $625$ and that it is the $85^{th}$ percentile. The first number tells how you actually did on the exam; the second says that $85\%$ of the scores on the exam were less than or equal to your score, $625$.

A percentile is a measure of ranking. It represents a location measurement of a data value to the rest of the values. Many standardized tests give the results as a percentile. Doctors also use percentiles to track a child’s growth.

Definition: percentile of data

The kth percentile is the data value that has $k\%$ of the data at or below that value.

The P^th percentile cuts the data set in two so that approximately $P\%$ of the data lie below or equal to it and $(100−P)\% $ of the data lie above it. In particular, the three percentiles that cut the data into fourths are called the quartiles of a data set. The quartiles are the three numbers $Q_1$, $Q_2$, $Q_3$ that divide the data set approximately into fourths.a

Given an observed value $x$ in a data set, $x$ is the $P^{th}$ percentile of the data if $P\%$ of the data are less than or equal to $x$. The number $P$ is the percentile rank of $x$.

Example $\PageIndex{1}$: Interpreting Percentile

What does a score of the $90$^thpercentile mean? If a sample consists of $800$ test scores, how many of them would be at or below the $90$^thpercentile?
How many scores are above the $90$^th percentile?

Answer

This means that $90\%$ of the scores were at or below this score. (A person did better than $90\%$ of the test takers.)\[0.90\times800=720\nonumber\] So $720$ score would be below or at $90$^th percentile.
$800-720=80$ of the scores were above $90$^th percentile.

Example $\PageIndex{2}$: Percentile Vs Score

If the test was out of $100$ points and you scored at the $80$^th percentile, what was your score on the test?

Answer: You don’t know! All you know is that you scored the same as or better than $80\%$ of the people who took the test. If all the scores were low, you could have still failed the test. On the other hand, if many of the scores were high, you could have gotten.

Your Turn $\PageIndex{2}$: Interpret Percentile

$z$-Scores and Normal Distribution

You have probably seen a "bell-shaped curve" before. It turns out that many data sets, particularly when the sample size is sufficiently large, are distributed symmetrically, with most data values around the center (mean) and few outliers, resulting in a bell-shaped curve. This is called a "normal distribution." For instance, the weights of all the people in a certain country are likely distributed normally, and so are their heights. Normal distributions can describe all sorts of data involving people's cholesterol levels, blood pressure, duration of pregnancy, IQ (intelligence quotient) scores, credit scores, and many more. For instance, the pregnancy (gestation) period for humans is known to have an average of 266 days (38 weeks) with a standard deviation of 16 days. The average IQ score is defined as $100$ with a standard deviation of $15$ . Your SAT and other standardized test scores are defined in similar ways.

But what does it mean, and how does this information help us?

When data values are distributed normally, given their mean and standard deviation, we can then standardize the normal distribution and calculate various percentages related to scores/values. For example, given that the IQ scores have a mean of $100$ and a standard deviation of $15$ , one can conclude that only about$2.3\%$ of the population has an IQ of $130$ or higher.

Here is how this works. First, let us define the "Standard Normal Distribution."

Definition: Standard Normal Distribution

The standard normal distribution is a normal distribution (bell-shaped curve) with a mean of 0 and a standard deviation of 1.

Graph of the standard normal distribution with mean (μ) at 0 and standard deviation (σ) at 1, highlighted area under the curve. — Figure $\PageIndex{1}$: Standard Normal Distribution with Mean $0$ and Standard Deviation $1$.

It turns out that there is a simple way to "convert" any normal distribution to the standard one. It involves what is known as the "$z$-score."

Another way to locate a particular observation $x$ in a data set is to compute its distance from the mean in units of standard deviation. The $z$-score indicates how many standard deviations an individual observation $x$ is from the center of the data set, its mean. It is used on distributions that have been standardized, which allows us to better understand its properties. If $z$ is negative, then $x$ is below average. If $z$ is $0$ then $x$ is equal to the average. If $z$ is positive, then $x$ is above the average.

What is z-score?

A z-score tells you how many standard deviations a value is away from the mean (average)

Positive Z-score: The value is above average.
Negative Z-score: The value is below average.
Z-score of $0$: The value is exactly the average.

The following formula is used to calculate the z-score (z) of a data value (x) when the mean, standard deviation, and data value (x) are provided

FORMULA
\[z-\text{score of data} = \dfrac{\text{data value($x$)} - \text{mean (M)}}{\text{standard deviation (SD)}} \nonumber \]

Sometime $z$ score of data value $x$ is denoted by $z_x$.

Another notation:

Mean: $\mu$ or M

Standard Deviation: $\sigma$ or SD

Example $\PageIndex{3}$: Finding z-Score

On a nationwide math test, the mean was $65$ and the standard deviation was $10$. If Robert scored $81$, what was his $z$−score?

Answer

\[\begin{array}{l}
z&=\dfrac{x-\text{~M}}{\text{~SD}} \\
&=\dfrac{81-65}{10} \\
&=\dfrac{16}{10} \\
&=1.6\\
\end{array} \nonumber \]

This means Robert’s score was $1.6$ standard deviations above the mean.

Your Turn $\PageIndex{3}$: Find z-score

The following formula is used to calculate the data value (x) when the mean, standard deviation, and z-score are provided.

FORMULA

\[\text{ data } = \text{mean}+z-\text{data value}\times\text{standard deviation} \nonumber \]

Example $\PageIndex{4}$: Find the Score or Data Value

A math test has mean $70$ and standard deviation $ 8$. A student has a z-score of $1.5$ on this test. What is the student’s actual score?

Answer

Here, we need to find the student's score (data value), denoted as $x$.

\[\begin{array}{l}
\text{data($x$)}&=\text{M}+z-\text{score}\times\text{SD}\\
&=70+1.5\times8\\
&=82\\
\end{array} \nonumber \]

This means Robert’s score was $82$ corresponding to z-score $1.5$.

Your Turn $\PageIndex{4}$: Find the Score or Data Value

Example $\PageIndex{5}$: Compare Using z-Score

A server at a restaurant tracks their weekly tips earned and hours worked. Over several weeks, the average weekly tips earned is $\$500$ with a standard deviation of $\$100$. This week, the server earned $\$650$ in tips. Meanwhile, the average number of hours worked per week is $30$ hours with a standard deviation of $5$ hours, and this week, the server worked $400$ hours.

Using z-scores, determine whether the server's weekly tips or weekly hours worked deviate more from the average, indicating where their performance was stronger relative to past trends.

Answer: Using the z-score formula:
z-score for weekly tips: \[z = \dfrac{650 - 500}{100} = \dfrac{150}{100} = 1.5\nonumber\]
z-score for weekly hours worked:
\[z = \dfrac{40 - 30}{5} = \dfrac{10}{5} = 2.0 \nonumber\]The z-score for hours worked $92$ is higher than the z-score for tips earned $1.5$. This means that, relative to their average, the server worked significantly more hours than usual, compared to the amount of tips they earned. Therefore, their performance in terms of hours worked deviated more from the average than their tips.

Your Turn $\PageIndex{5}$: Who is Taller?

Z-Score Percentile Table

A percentile table (or z-table) is a reference chart that converts a z-score into a percentile rank. A percentile tells you the percentage of people or data points that scored below you. If you want to find the percentile for a given z score. You can use the following table.

Let's say you have a data value, and if you calculate a Z-score for your data value, it is $0.20$. You then look it up in the table and find the corresponding percentile value,$57.93$.

Meaning: This indicates that $57.93\%$ of all observations are less than or equal to your value. In other words, your score is in the $58$th percentile. In other words, only $100%-57.93%=42.07\%$ observations are more than your score.

$Table displaying statistical scores and personality traits with various numerical values across multiple columns and rows.$

Example $\PageIndex{6}$: Find the Percentage

Find the percentage of data items in a normal distribution that lie either above, below, or between.

above $1.20$ and below $1.20$.
below $-1.80$ and above $-1.80$.
between $0.20$ and $1.40$.

Answer

According to the above table

$88.49\%$ below and $(100\%−88.49\%)=11.51\%$ above.
$3.59\%$ below and $(100\%−3.59\%)=96.41\%$ above.
For $z=0.20$; $57.93\%$ are below and for $z=1.4$; $91.92\%$ are below. So $91.92\%−57.93\%=33.99\%$ in between.

Example $\PageIndex{7}$: Find the Percentile

Suppose that the cholesterol levels are normally distributed with a mean of 160 (mg/dl) with a standard deviation of 40 (mg/dl).

What is the $z$-score corresponding to the cholesterol level of $200$ (mg/dl)? How about $260$ (mg/dl)?
What percent of people have a cholesterol level between $160$ and $200$? How about $260$ or higher?

Answer

a. Note that data value is $200$ mg/dl. So its z-score is

\[\text{$z$-score}=\dfrac{200-160}{40}=1\nonumber\]

Similarly, z- score of $260$ is

\[\text{$z$-score}=\dfrac{260-160}{40}=2.5\nonumber\]

b. z- score of $160$ is

\[\text{$z$-score}=\dfrac{160-160}{40}=0\nonumber\]

The percent (percentile ) between $160$ and $200$ would correspond to the z-score between $0$ and $1$.

For $z=0.0$; $50\%$ are below and for $z=1$; $84.13\%$ are below. So $84.13\%−50\%=34.13\%$ in between.

In other words, about $34\%$ of all people have a cholesterol level between $160$ and $200$.

However, $160$ mg/dl corresponds to $z=2.5$. The table shows $99.38\%$ below $z=2.5$. Since $100\%-99.38\%=0.62 \%$. Only $0.62\%$, i.e., less than $1\%$ of the population, has a cholesterol level $160$ or higher. This person with a cholesterol level of $260$ should go see the doctor immediately.

Your Turn $\PageIndex{7}$: Find the Percentage

You probably have a good intuitive grasp of what the average of a data set says about that data set. In the following section, we begin to learn what the standard deviation reveals about the nature of the data set.

The Empirical Rule

We start by examining a specific set of data. Table $\PageIndex{1}$ shows the heights in inches of $100$ randomly selected adult men. A relative frequency histogram for the data is shown in Figure $\PageIndex{1}$. The mean and standard deviation of the data, rounded to two decimal places, are $M = 69.92 $ and $SD = 1.70$.

Table of numerical data arranged in rows and columns, likely representing statistical values or measurements. — Table $\PageIndex{1}$: Height of Men

If we go through the data and count the number of observations that are within one standard deviation of the mean, that is, that are between $69.92-1.70=68.22$ and $69.92+1.70=71.62$ inches, there are $69$ of them.

If we count the number of observations that are within two standard deviations of the mean, that is, that are between $69.92-2(1.70)=66.52$ and $69.92+2(1.70)=73.32$ inches, there are $95$ of them.

All of the measurements are within three standard deviations of the mean, that is, between $69.92-3(1.70)=64.822$ and $69.92+3(1.70)=75.02$ inches.

These tallies are not coincidences but rather align with the following result, which has been found to be widely applicable.

Bar graph showing frequency distribution, with peaks around the values 70 to 74 and lower counts at extremes. — *Figure $\PageIndex{2}$: Normal (Bell Shaped) Distribution*

Empirical Rule

The Empirical Rule (also called the $68–95–99.7$ rule describes how data are distributed in a normal (bell-shaped) distribution

approximately $68\%$ of the data lie between one standard deviation mean, i.e., within ($\text{M} -1\times\text{SD},\text{M}+1\times\text{SD})$.
approximately $95\%$ of the data lie between one standard deviation mean, i.e., within ($\text{M} -2\times\text{SD},\text{M}+2\times\text{SD})$.
approximately $99.7\%$ of the data lie between one standard deviation mean, i.e., within ($\text{M} -3\times\text{SD},\text{M}+3\times\text{SD})$.

Where M is the mean, and SD is the standard deviation.

Two key points in regard to the Empirical Rule are that the data distribution must be approximately bell-shaped and that the percentages are only approximately true. The Empirical Rule does not apply to data sets with severely asymmetric distributions, and the actual percentage of observations in any of the intervals specified by the rule could be either greater or less than those given in the rule.

A bell curve diagram showing a normal distribution with percentages related to standard deviations (±1, ±2, ±3). — Figure $\PageIndex{3}$: M as a Mean and SD as a Standard Deviation

We see this with the example of the heights of the men: the Empirical Rule suggested 68 observations between $68.22$ and $71.62$ inches, but we counted $69$.

Example $\PageIndex{8}$: Using Empirical Rule/span>

Scores on IQ tests have a bell-shaped distribution with a mean of $100$ and a standard deviation of $10$. Discuss what the Empirical Rule is after sketching a normal distribution curve. What is the percentile of IQ scores of $110$ and $90$?

Answer

A sketch of the IQ distribution is given in Figure $\PageIndex{3}$. The Empirical Rule states that

Normal distribution curve showing percentages for 68%, 95%, and 99.7% confidence intervals marked above the curve. — Figure $\PageIndex{4}$: Distribution of IQ Scores

approximately $68\%$ of the IQ scores in the population lie between $90$ and $110$
approximately $95\%$ of the IQ scores in the population lie between $80$ and $120$
approximately $99.7\%$ of the IQ scores in the population lie between $70$ and $130$.

Since $84\%$ data are less than or equal to $110$, we conclude that the IQ score $110$ is the $84^{th}$ percentile. Similarly, Since $16\%$ data are less than or equal to $90$, we conclude that the IQ score $90$ is the $16^{th}$ percentile.

Your Turn $\PageIndex{8}$: Use Empirical Rule

Example $\PageIndex{9}$: Using Empirical Rule

About what percentage of all such men are between $68.2$ and $71$ inches tall?
What interval centered on the mean should contain about $95\%$ of all such men?
What percentage of all such men are less than $66.8$ inches in tall?

Answer

A sketch of the distribution of heights is given in Figure $\PageIndex{3}$. Since the interval from $68.2$ to $71.0$ has endpoints $\text{~Mean} -1\times\text{~SD}$ and $\text{~Mean} +1\times\text{~SD}$, by the Empirical Rule about $68\%$ of all $18$-year-old males should have heights in this range
By the Empirical Rule the shortest such interval has endpoints \[\text{~Mean} -2\times\text{~SD}=69.6-2(1.4)=66.8 \nonumber \]
\[\text{~Mean}+2\times\text{~SD}=69.6+2(1.4)=72.4 \nonumber \] The interval in question is the interval from $66.8$ inches to $72.4$ inches.
$0.15\%+2.35\%=2.5\%$
1. $Graph depicting a normal distribution curve with percentages indicating standard deviations at various points along the x-axis.$
  Figure $\PageIndex{5}$: Distribution of Heights