Reason with range and standard deviation

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 27 minutes
Domains: Problem Solving and Data Analysis
Techniques: Measures-of-spreadRangeStandard-deviationData-transformations

What you’ll learn

Find the range from a list or a display.
Use known bounds to find the most the range can be.
Compare standard deviations and explain what they say about spread around the mean.
Predict how adding the same amount to every value, or multiplying every value by the same amount, changes spread.

Prerequisites

Why this matters on the SAT

Separate center from spread

SAT questions often ask which data set is more spread out, or what happens to the spread when every value follows the same rule. So start by checking which measure the question asks about. Range uses only the two endpoints. Standard deviation looks at how all the values spread out around their mean.

Solution to the example

Both lists balance around $20$ . Now look at how far each value sits from $20$ :

\begin{aligned} P &: 20-6,\ 20-3,\ 20,\ 20+3,\ 20+6,\\[1.4em] Q &: 20-12,\ 20-6,\ 20,\ 20+6,\ 20+12. \end{aligned}

Each distance in Q is twice the matching distance in P. Q is more spread out around the same mean, so Q has the greater standard deviation. Choice B is correct, and you didn’t need a formula.

SAT example

Data sets P and Q are shown.

P: 14,\ 17,\ 20,\ 23,\ 26

Q: 8,\ 14,\ 20,\ 26,\ 32

Which choice correctly compares the standard deviations of the two data sets?

A
The standard deviation of P is greater than the standard deviation of Q.
B
The standard deviation of Q is greater than the standard deviation of P.
C
The standard deviations are equal because the means are equal.
D
The standard deviations cannot be compared without using a formula.

See the spread before you calculate

Before you name a statistic, look at the two distributions.

Same center, but data set B is more spread out.

Both sets have mean $8$ , so their centers match. Their spreads don’t. Set aside the three $8$ s in each set and look at the other values:

In data set A, they’re $4,6,10,12$ , which sit $4,2,2,4$ away from $8$ .
In data set B, they’re $3,4,12,13$ , which sit $5,4,4,5$ away from $8$ .

B’s values reach farther from $8$ , so B has the greater standard deviation.

Standard deviation measures how spread out the values are around their mean. Think of it as a typical distance from the mean, where the far-out values count more heavily. It’s a way to picture it, though: it doesn’t mean every value sits exactly one standard deviation from the mean.

A smaller standard deviation means the values cluster more tightly around their mean.
A larger standard deviation means the values are more widely spread around their mean.
A standard deviation of $0$ means every value equals the mean.

The mean alone doesn’t settle the spread. Two data sets can have the same mean and different standard deviations, like A and B above. They can also have different means and the same standard deviation, when one is a shifted copy of the other.

Check your understanding:

Data set R is $2,5,8,11,14$ . Data set S is $32,35,38,41,44$ . How do their means and standard deviations compare?

Common mistake:

Equal means tell you the centers match, not the spreads. When you catch yourself comparing means to answer a spread question, switch to comparing how far the values sit from each set’s own mean.

Find the range, or the most it can be

The range is the distance from the minimum to the maximum:

\text{range}=\text{maximum}-\text{minimum}.

For

7,\ 11,\ 11,\ 15,\ 18,\ 24,

the minimum is $7$ and the maximum is $24$ , so

\text{range}=24-7=17.

The four values in between don’t matter for the range. They do matter for standard deviation.

Sometimes the SAT doesn’t list the data. It only tells you the lowest and highest values possible. If every value $x$ satisfies

L\le x\le U,

then the range can be no greater than

U-L.

The range equals $U-L$ only if the data actually includes both $L$ and $U$ . For example, if every measurement is at least $12$ and at most $31$ , then

\text{range}\le 31-12=19.

The range could be less than $19$ if one or both of those boundary values never shows up. If the measurements were $14$ , $20$ and $31$ , the range would be $31-14=17$ .

Check your understanding:

Every value in a data set is greater than or equal to $18$ and less than or equal to $45$ . The data set contains both $18$ and $45$ . What is its range?

Common mistake:

The range is a distance between two numbers. It isn’t how many values there are, or how many gaps sit between them in the list. Find the actual largest and smallest values, subtract the smallest from the largest, and check that your answer isn’t negative.

Example: Change every value with one rule

Sometimes every value in a data set goes through the same linear rule, like “multiply by $2$ , then add $5$ .” You don’t need the data itself for these. You can apply the change straight to the statistics.

Worked example

A data set has a mean of $14$ , a range of $8$ , and a standard deviation of $3$ . A new data set is created by replacing every value $x$ with

\mathbf{y=-1.5x+6}.

Which choice gives the mean, range, and standard deviation of the new data set?

A
Mean $-15$ ; range $6.5$ ; standard deviation $1.5$
B
Mean $-15$ ; range $12$ ; standard deviation $10.5$
C
Mean $-15$ ; range $12$ ; standard deviation $4.5$
D
Mean $27$ ; range $12$ ; standard deviation $4.5$

Step 1

The mean follows the whole rule

Every value goes through $y=-1.5x+6$ , so the mean does too:

-1.5(14)+6=-21+6=-15.

The new mean is $-15$ .

Step 2

Spread only feels the multiplier

Multiplying by $-1.5$ stretches every distance by

|-1.5|=1.5.

The negative sign flips the order of the values, so the smallest becomes the largest. But a distance can’t be negative. So

\text{new range}=1.5(8)=12

and

\text{new standard deviation}=1.5(3)=4.5.

Adding $6$ then slides every value by the same amount, so it doesn’t change either spread.

Step 3

Match all three

The new statistics are

\text{mean}=-15,\qquad \text{range}=12,\qquad \text{SD}=4.5.

Choice C matches all three. Each wrong choice makes one classic slip. A adds the $-1.5$ instead of multiplying by it ( $8-1.5=6.5$ ). B adds the $6$ to the standard deviation ( $4.5+6=10.5$ ). D drops the negative sign on the mean ( $1.5(14)+6=27$ ).

Check your understanding:

Why does the $+6$ change the mean but not the range or standard deviation?

Why the shift and scale rules work

Say every original value $x$ becomes

y=ax+b.

That’s two moves: multiply by $a$ , then add $b$ . Take them one at a time.

Adding slides the data

If $a=1$ , each value becomes $x+b$ . The minimum and maximum both go up by $b$ :

(\text{maximum}+b)-(\text{minimum}+b) =\text{maximum}-\text{minimum}.

So the range doesn’t change.

The mean goes up by $b$ too. For any value,

(x+b)-(\text{mean}+b)=x-\text{mean}.

Its distance from the mean doesn’t change, so the standard deviation doesn’t either.

Multiplying stretches the data

Multiplying every value by $a$ stretches or shrinks every distance by $|a|$ . If $a$ is negative, the data also flip, so their order reverses. Flipping doesn’t make a distance negative.

\boxed{\text{new range}=|a|(\text{old range})}

and

\boxed{\text{new SD}=|a|(\text{old SD})}.

Adding $b$ afterward changes neither one, as you just saw.

How one rule applied to every value changes spread

Rule applied to every value	Range	Standard deviation
$x+b$	No change	No change
$ax$	Multiply by $	a
$ax+b$	Multiply by $	a

Try it yourself:

A data set has range $18$ and standard deviation $5$ . Every value is replaced by $-2x-9$ . Predict the new range and standard deviation before you open the check.

Check your understanding:

What are the range and standard deviation after the rule $y=-2x-9$ ?

These rules only work when every value gets the same rule. Adding or removing a value, fixing one entry, replacing a few values or dealing with an outlier is a different kind of question, and you’ll handle those in Analyze changed data and outliers.

Reason first, or check with a calculator

Most spread questions take a few seconds of reasoning. Use the quickest method that still fits what the statistic measures.

Pick your first move

What the question gives	Best first move	Why
A short list or marked endpoints	Subtract the smallest value from the largest, by hand.	The range only uses those two values.
Dot plots or symmetric lists where the difference in spread is clear	Compare how far the values sit from each mean, by eye.	You can see which is more spread out without calculating, as with P and Q.
A rule $y=ax+b$ applied to every value	Use the shift and scale rules.	Rebuilding the data is extra work you don’t need.
Two long, awkward lists with similar spread	Enter both lists, check every entry, and compare `stdevp(...)`.	A number can settle a close call.

On the SAT, you’ll usually compare or interpret standard deviation, not calculate it with the formula. So if you can see the answer, don’t turn it into a long calculation.

When a Desmos check is worth it:

Enter each full data set once as a named list, like P=[14,17,20,23,26].
Compare the entries with the question, so no value is missing or doubled.
Enter stdevp(P) and stdevp(Q), using your own list names.
Check that the results match the comparison you predicted.

The calculator here has P and Q from the opening example. It gives about $4.24$ for P and $8.49$ for Q, so Q has the greater standard deviation. Q’s is exactly twice P’s, which matches the doubled distances you saw at the start.

Desmos also has stdev(...), which uses the sample version of standard deviation. For two lists with the same number of values, both versions put the standard deviations in the same order. The shift and scale rules work with either version too. SAT questions like these don’t ask you to choose between them, so don’t let the difference distract you from the comparison.

Check your understanding:

Why is the calculator useful for two long, similar lists but unnecessary for the rule $y=-2x-9$ ?

Calculator loads as you approach

Check both lists and make your prediction before you look at stdevp.

Practice problems

Each problem is a little harder than the last.

Calculate a range

Practice problem

The numbers of minutes needed to complete six repairs were

18,\ 27,\ 14,\ 31,\ 22,\ 37.

What is the range, in minutes, of the data?

Your answer

Calculator loads as you approach

Find the two endpoints by hand. The calculator is here if you want to check.

Recognize equal spread

Practice problem

Data sets A and B are shown.

A: 3,\ 6,\ 9,\ 12,\ 15

B: 43,\ 46,\ 49,\ 52,\ 55

Which statement is true?

Calculator loads as you approach

Look for one rule linking the two lists before you calculate.

Transform a spread bound

Practice problem

Every value in data set X is at least $10$ and at most $28$ . The standard deviation of X is $4$ . Data set Y is created by replacing every value $x$ in X with

y=-2.5x+7.

Which statement must be true?

Calculator loads as you approach

Find the bound and the multiplier first. Use the calculator only for arithmetic.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Range is maximum minus minimum. Only the two endpoints matter.
If every value is from $L$ through $U$ , the range is at most $U-L$ . It equals $U-L$ only when both $L$ and $U$ are in the data.
Standard deviation measures spread around the mean. A smaller one means the values cluster tightly, and a larger one means they’re spread wide.
Equal means don’t guarantee equal standard deviations. A shifted copy has a different mean but the same standard deviation.
Adding the same $b$ to every value slides the data, so the range and standard deviation don’t change.
Multiplying every value by $a$ stretches the data, so the range and standard deviation are multiplied by $|a|$ .
Reason first from the display or the rule. Save stdevp(...) for checking long, awkward lists.

Next lesson

Analyze changed data and outliers

Track what happens when selected observations are added, removed, corrected, or replaced.

Start next lesson

Practice

Practice this lesson

177 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice