The same theorem stated for totals rather than averages. If random samples of size n are drawn from any population with mean mu and standard deviation sigma, then as n increases the sum of the sample tends to be normally distributed, with a mean equal to the population mean multiplied by the sample size and a standard deviation equal to the population standard deviation multiplied by the square root of the sample size. The asymmetry between those two is the section's whole content: the centre of a total grows in proportion to n while its spread grows only in proportion to the square root of n, so a total becomes proportionally more predictable as more values are added. Probabilities and percentiles for a sum are then ordinary normal questions asked on that distribution, and because a sum is simply a mean multiplied by n, either form can be converted into the other as a check.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 7 — The Central Limit Theorem
The Central Limit Theorem for Sums
Objectives
Five outcomes, and the second explains why insurance works.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-373 — the section these objectives are drawn from
Warm-up
Section 7.1 gave the distribution of a sample mean: centre unchanged, spread divided by the square root of n.
Discussion prompt
Eighty values are drawn from a population with mean 90 and standard deviation 15. Their mean has a spread of 15 over root 80, about 1.68. What is the spread of their total?
Hint: The total is eighty times the mean. What does multiplying a random quantity by 80 do to its spread?
Answer:
Multiplying every value of a quantity by 80 multiplies its standard deviation by 80 as well, so the total's spread is 80 times 1.68, which is about 134.
That is the answer, and it is worth noticing what it is NOT. The population's spread was 15, and 80 times 15 would be 1,200 — nearly ten times too large. The total is far more tightly concentrated than simply adding up eighty independent spreads would suggest.
The reason is that 80 times sigma over root 80 simplifies to root 80 times sigma. The square root that shrank the mean's spread reappears here as growth, but slower growth than the centre's — and that mismatch is the whole of this section.
Concept
If you draw random samples of size n from a population with mean mu and standard deviation sigma, then as n increases the sum of the values tends to be normally distributed, with a mean equal to n times mu and a standard deviation equal to the square root of n times sigma.
the central limit theorem for sums — If you repeatedly draw samples of a given size and calculate the sum of each, those sums tend to follow a normal distribution whose mean is the original mean multiplied by the sample size and whose standard deviation is the original standard deviation multiplied by the square root of the sample size.
\[ \sum X \sim N\left(n\mu, \sqrt{n}\,\sigma\right) \]
The theorem is not a separate result but the same one restated, since a sum is a mean multiplied by n. Multiplying the mean's distribution by n multiplies its centre by n, giving n times mu, and multiplies its spread by n, giving n times sigma over root n — which simplifies to root n times sigma. Everything in this section follows from that single simplification.
Figure (svg): A card giving the central limit theorem's distribution for a sum, with the mean multiplied by n and the standard deviation by the square root of n
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-371
Section
Section 1
Concept
The normal distribution for a sum has a mean equal to the original mean multiplied by the sample size, and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
the distribution of a total — Written as the sum of X following N of n mu and root n sigma. Both parameters grow with n, but at different rates, which is why totals become proportionally more predictable.
\[ \mu_{\sum X} = n\mu, \qquad \sigma_{\sum X} = \sqrt{n}\,\sigma \]
It is worth writing both forms side by side once. For a mean, root n sits in the denominator and shrinks the spread; for a sum it sits in the numerator and grows it. The centre behaves differently again: unchanged for a mean, multiplied by n for a sum. Four behaviours across two statistics, and mixing them up is the section's characteristic error.
Figure (svg): A card giving the central limit theorem's distribution for a sum, with the mean multiplied by n and the standard deviation by the square root of n
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-371 — the statement of the theorem for sums
Picture it
How the centre and the spread of a total grow as values are added.
Figure (svg): A straight line rising steeply for the mean of a sum and a slowly curving line beneath it for the standard deviation of that sum
The straight line and the curve diverge steadily, and that divergence is the point. At a sample of 100 the total is ten times its own standard deviation, while at a sample of 4 it is only twice — so larger totals are relatively tighter, even though they are absolutely more variable.
Worked example
Example 7.5's setup, from an unknown population.
\[ \mu = 90, \; \sigma = 15, \; n = 80 \]
Multiply the mean by n
Why: Eighty times ninety.
\[ 7, 200 \]
Take the square root of n
Why: Root eighty.
\[ \text{about } 8.944 \]
Multiply sigma by it
Why: 8.944 times 15.
\[ 134.16 \]
Write the distribution
Why: By the theorem.
\[ N(7200, 134.16) \]
Figure (svg): The solution to Worked example the sum of 80 values shown as a ladder of expressions, one row per legal move
\[ \sum X \sim N\left(7200, \sqrt{80}\cdot 15\right) \approx N(7200, 134.16) \]
Verify: confirm the spread against the naive alternative
Why: Adding eighty independent spreads of 15 would suggest 1,200, which is nearly nine times too large. The reason it is wrong is that the values vary independently, so their departures from the mean partly cancel rather than accumulating — and the root n captures exactly that cancellation. Any spread for a sum that comes out at n times sigma has missed the square root.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 371-372
Faded example
App engagement has mean 8.2 minutes and standard deviation 1 minute; take a sample of 70.
Fill in the blanks
\mu_574 = 70 \times 8.2 = 8.37, \qquad \sigma____ = \sqrt___ \times 1 \approx ___
Why: The mean multiplies by 70 to give 574 minutes, while the standard deviation multiplies only by the square root of 70, about 8.37 — Example 7.7's values.
Worked example
Example 7.6(a), with the book's own values.
\[ \mu = 34, \; \sigma = 15, \; n = 50 \]
Multiply the mean by n
Why: Fifty times 34.
\[ 1, 700 \]
Take the square root of n
Why: Root fifty.
\[ \text{about } 7.071 \]
Multiply sigma by it
Why: 7.071 times 15.
\[ 106.07 \]
State the shape
Why: By the theorem for sums.
Figure (svg): The solution to Worked example the sum of 50 ages shown as a ladder of expressions, one row per legal move
\[ \mu_{\sum X} = 1700, \quad \sigma_{\sum X} = \sqrt{50}\cdot 15 \approx 106.07 \]
Verify: confirm the spread is a plausible fraction of the total
Why: The spread of 106 is about 6 percent of the total of 1,700, so totals of fifty ages should typically land within a few hundred of 1,700 — which is a much tighter relative range than individual ages, whose spread of 15 is 44 percent of their mean of 34. That relative tightening is the practical content of the theorem for sums.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, p. 373
Trap
\[ \sigma_{\sum X} = 80 \times 15 = 1\,200 \]
Scale both parameters by n
Why: The mean was multiplied by 80, so the spread should be too.
\[ \text{a spread of } 1\,200 \text{ on a total of } 7\,200 \]
That would make totals vary by a sixth of their own size, when in reality they vary by under two percent.
\[ \sigma_{\sum X} = \sqrt{80} \times 15 \approx 134.16 \]
Multiply the mean by n and the standard deviation by ROOT n
Why: The two parameters scale at different rates.
The reason is cancellation: eighty independent departures from the mean do not all point the same way, so they partly offset. Multiplying by n would be right only if every value moved in lockstep, which independence rules out. The check is that the spread should be a small fraction of the total for any reasonably large n, and 1,200 out of 7,200 is not small.
Two truths and a lie
All three concern the distribution of a sum.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and is the section's defining error. Only the mean is multiplied by n; the standard deviation is multiplied by the square root of n, because independent departures from the mean partly cancel rather than accumulating.
Prediction
Commit before reasoning.
Predict first
As n grows, what happens to the standard deviation of a sum as a FRACTION of that sum's mean?
Correct: It falls, like one over the square root of n.
Why: The ratio is root n sigma over n mu, which simplifies to sigma over root n mu — so it falls with the square root of n. The absolute spread does grow, which is why the second option is tempting, but it grows more slowly than the total does. This is precisely why insurers prefer many small independent policies to a few large ones.
Sorting
Ask whether the question is about an average or a total.
Sort into buckets
Sort each question.
The words total, combined and altogether signal a sum; average and mean signal the other. Section 7.3 makes this the first question to ask of any problem in the chapter, ahead of any arithmetic at all.
Section
Section 2
Concept
The centre of a total grows in proportion to n because every value contributes its full average. The spread grows only in proportion to the square root of n because the values vary independently, so their departures from the mean partly offset one another rather than accumulating.
the square root law — The rule that the spread of a sum of n independent quantities grows like the square root of n rather than like n. It is the same square root that appears, inverted, in the standard error of a mean.
\[ \frac{\sigma_{\sum X}}{\mu_{\sum X}} = \frac{\sqrt{n}\,\sigma}{n\mu} = \frac{\sigma}{\sqrt{n}\,\mu} \]
The consequence is the one that matters commercially. A single insurance policy is close to unpredictable — the claim is either nothing or a great deal. A portfolio of a hundred thousand independent policies has a total that is predictable to within a fraction of a percent, because the total grew by a factor of a hundred thousand while its spread grew only by about 316. Diversification is this formula.
Figure (svg): A straight line rising steeply for the mean of a sum and a slowly curving line beneath it for the standard deviation of that sum
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-371 — the mean and standard deviation of a sum
Picture it
The total's centre against its spread, as n rises.
Figure (svg): A straight line rising steeply for the mean of a sum and a slowly curving line beneath it for the standard deviation of that sum
The gap between the two lines is what makes aggregation useful. Neither quantity is falling — a bigger total genuinely is more variable in absolute terms — but the variability is being outpaced, and that is what predictability means in practice.
Worked example
The same population, aggregated at four scales.
\[ \mu = 15, \; \sigma = 15 \]
At n = 1
Why: Spread 15, total 15.
\[ 100 \% \]
At n = 4
Why: Spread 30, total 60.
\[ 50 \% \]
At n = 25
Why: Spread 75, total 375.
\[ 20 \% \]
At n = 100
Why: Spread 150, total 1,500.
\[ 10 \% \]
Figure (svg): The solution to Worked example how the ratio changes shown as a ladder of expressions, one row per legal move
\[ \frac{\sqrt{n}\,\sigma}{n\mu} = \frac{\sigma}{\sqrt{n}\,\mu} \]
Verify: confirm the pattern against the formula
Why: Each hundredfold rise in n should divide the relative spread by ten, and going from n equal to 1 to n equal to 100 does exactly that. The absolute spread meanwhile rose from 15 to 150 — tenfold growth — so both statements are true at once: the total is more variable and more predictable, in different senses. Keeping those two senses apart is the point of the exercise.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-371
Estimation
A population has mean 15 and standard deviation 15, so single values are wildly variable.
Predict first
For a total of 10,000 values, roughly what is the spread as a fraction of the total?
Correct: About 1 percent.
Why: The relative spread is sigma over root n times mu, which is 15 over 100 times 15, or 1 percent. A quantity that is completely unpredictable individually becomes predictable to within about a percent once ten thousand of them are added — which is the whole basis of insurance and of any actuarial calculation.
Worked example
The same sample, described both ways.
\[ \mu = 90, \; \sigma = 15, \; n = 80 \]
The mean's centre
Why: Unchanged.
\[ 90 \]
The mean's spread
Why: Fifteen over root 80.
\[ 1.677 \]
The sum's centre
Why: Eighty times 90.
\[ 7, 200 \]
The sum's spread
Why: Eighty times 1.677.
\[ 134.16 \]
Figure (svg): The solution to Worked example comparing a mean and a sum shown as a ladder of expressions, one row per legal move
\[ 80 \times 90 = 7200, \qquad 80 \times 1.677 \approx 134.16 \]
Verify: confirm this reproduces the root n formula
Why: Eighty times sigma over root 80 is 80 over root 80 times sigma, and 80 over root 80 is root 80 — so the multiplication gives root 80 times sigma, exactly the theorem's formula. Deriving it this way makes the square root inevitable rather than arbitrary, and it also gives a second route to any sum's parameters if the formula is misremembered.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-372
Error analysis
The correct value is about 134.16.
Annotate
On: \( \begin{aligned} &(1)\; 80 \times 15 = 1\,200 \\ &(2)\; \frac{15}{\sqrt{80}} \approx 1.68 \\ &(3)\; \sqrt{80 \times 15} \approx 34.6 \\ &(4)\; \sqrt{80} \times 15 \approx 134.16 \end{aligned} \)
Error (2) is the one to watch, because it is a correct formula applied to the wrong question, so nothing about it looks like a mistake. The check is scale: a sum of eighty values around 90 is in the thousands, so a spread of 1.68 is far too small to describe it.
Two truths and a lie
All three concern the two growth rates.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and confuses the sum with the mean. It is the MEAN whose spread falls absolutely; a sum's spread grows, just more slowly than the sum itself. Both statements about the sum are true simultaneously, in the two different senses, and holding them together is the idea here.
Faded example
A population has mean 20 and standard deviation 8; take a sample of 64.
Fill in the blanks
\frac64 \times 8}0.05 = \frac___}___ = ___
Why: The spread of the sum is 8 times 8, which is 64, against a total of 1,280 — five percent. A single value's spread of 8 against a mean of 20 is 40 percent, so aggregating 64 values has cut the relative variability eightfold, the square root of 64.
Explain it
A classmate insists that adding 80 values each varying by 15 must give a total varying by 80 times 15.
Discussion prompt
In two sentences or fewer, explain why not.
Hint: Ask whether all eighty values would be high at once.
Answer:
Ask how often all eighty values would happen to be high together: almost never, because they vary independently, so the high ones are usually offset by low ones.
A spread of 80 times 15 would be the answer if the values always moved in lockstep, and the square root is exactly the discount for their not doing so.
Section
Section 3
Concept
Once the sum's distribution is written down, probabilities are computed exactly as in section 6.2. The only thing that catches people out is the scale: a total of many values lives in a range far removed from the individual values, so estimates have to be recalibrated.
the z-score for a sum — The sum minus n mu, divided by root n sigma. It has the same structure as every other z-score, with the sum's own two parameters in place of the population's.
\[ z = \frac{\sum x - n\mu}{\sqrt{n}\,\sigma} \]
The recalibration is worth taking seriously. In Example 7.7 a total of 600 minutes sounds ordinary — it is ten hours across seventy sessions, and each session averages 8.2 minutes. But on the sum's scale it sits over three standard deviations above the mean of 574, so its probability is 0.0009. Intuitions formed on individual values are simply not transferable to totals.
Figure (svg): A normal curve for the sum of fifty ages with the region between fifteen hundred and eighteen hundred shaded
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 371-373 — Examples 7.5(a), 7.6(b) and 7.7(c)
Picture it
Example 7.6(b): the total of fifty ages, between 1,500 and 1,800.
Figure (svg): A normal curve for the sum of fifty ages with the region between fifteen hundred and eighteen hundred shaded
The interval runs from about 1.9 standard deviations below the mean to about 0.9 above, which is why it captures roughly four fifths of the distribution. Reading the endpoints in standard deviations rather than in years is what makes an answer like 0.7974 recognisable as reasonable.
Worked example
Example 7.6(b).
\[ \sum X \sim N(1700, 106.07); \text{ find } P(1500 < \sum x < 1800) \]
Left area at 1,800
Why: Just under one above.
\[ 0.8281 \]
Left area at 1,500
Why: Nearly two below.
\[ 0.0307 \]
Subtract
Why: The interval.
\[ 0.7974 \]
Interpret
Why: Totals of fifty ages.
\[ \text{about } 80 \% \]
Figure (svg): The solution to Worked example a total between 1,500 and 1,800 shown as a ladder of expressions, one row per legal move
\[ P(1500 < \sum x < 1800) = 0.7974 \]
Verify: confirm the interval's position in standard deviations
Why: The endpoints sit about 1.89 below and 0.94 above the mean, so the region should hold rather more than half but well short of everything — and 0.80 fits. It is also worth checking the endpoints are plausible as totals: fifty ages averaging 30 to 36 gives 1,500 to 1,800, which brackets the population mean of 34 sensibly.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, p. 373
Faded example
For mu = 90, sigma = 15 and n = 80, the sum follows N(7200, 134.16).
Fill in the blanks
z = \frac134.162.24} \approx ___
Why: Three hundred above the mean, divided by a spread of 134.16, gives about 2.24 standard deviations — so a total above 7,500 has probability about 0.0127, a little over one percent.
Worked example
Example 7.7(c), where the units have to be converted first.
\[ \sum X \sim N(574, 8.37); \text{ find } P(\sum x \ge 600) \]
Convert the units
Why: Ten hours into minutes.
\[ 600\text{ minutes} \]
Locate it
Why: 600 minus 574, over 8.37.
\[ 3.11\text{ sd above} \]
Take the right tail
Why: One minus the left area.
\[ 0.0009 \]
Interpret
Why: Totals of 70 sessions.
\[ \text{about } 1\text{ in } 1, 100 \]
Figure (svg): The solution to Worked example at least ten hours of engagement shown as a ladder of expressions, one row per legal move
\[ P(\sum x \ge 600) \approx 0.0009 \]
Verify: confirm the unit conversion came first, and why it must
Why: Had 10 been used directly, the answer would have been essentially 1, since 10 minutes is far below a total of 574 — a completely different and absurd conclusion. Converting hours to minutes before anything else is the step that makes the comparison meaningful, and problems mixing units in this way are common precisely because totals often get quoted in larger units than the individual values.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, p. 373
Trap
\[ 600 \text{ minutes across } 70 \text{ sessions sounds ordinary} \]
Judge the total by what a single session looks like
Why: Each session averages 8.2 minutes, and 600 over 70 is only 8.57.
\[ \text{but } P(\sum x \ge 600) = 0.0009 \]
A mean of 8.57 is close to 8.2 on the individual scale, but on the scale of a 70-session total it is over three standard deviations out.
\[ z = \frac{600 - 574}{8.37} \approx 3.11 \]
Place the total on the SUM's distribution before judging it
Why: The sum has its own centre and spread.
This is the same lesson section 7.1 taught about means, in a sharper form: averaging and totalling both concentrate a distribution, so small-looking departures become large ones. The habit that protects against it is to convert any boundary into a z-score before forming an opinion about whether it is unusual.
Two truths and a lie
All three concern probabilities for sums.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false as stated. Near has to be measured in standard deviations of the sum, and those shrink relative to the total as n grows — so for a large n a total that looks proportionally close to n mu can still be many standard deviations out. Example 7.7's 600 against 574 is exactly that case.
Estimation
Seventy sessions average 8.2 minutes each, so the expected total is 574 minutes.
Predict first
A total of 590 minutes corresponds to an average session of about 8.4 minutes. Is that unremarkable?
Correct: No: nearly two standard deviations above.
Why: The total's spread is 8.37 minutes, so 590 is about 1.9 standard deviations above 574 — around the 97th percentile. On the individual scale 8.4 against 8.2 looks like nothing, which is exactly the trap: aggregation shrinks the spread relative to the centre, so proportionally tiny departures become statistically large.
Discrimination
A population has mu = 34 and sigma = 15.
Sort into buckets
Sort each question by which distribution answers it.
Section
Section 4
Concept
A percentile for a sum is found by reversing the cumulative function on the sum's distribution, exactly as for a mean or an individual value. The interpretation must say that it describes totals of n values.
a percentile of the sum — The value k such that a stated proportion of all totals of n values falls below it. The book asks for these to be interpreted in a complete sentence, as it does throughout chapter 6 and 7.
\[ P\left(\sum X < k\right) = p \;\Longrightarrow\; k = \text{invNorm}\left(p, n\mu, \sqrt{n}\,\sigma\right) \]
Example 7.7(b) supplies the model interpretation: ninety-five percent of the sums of app engagement times are at most 587.76 minutes. Notice that it says sums rather than sessions — a single session's 95th percentile would be about 9.8 minutes, a completely different number describing a completely different thing.
Figure (svg): A normal curve for the total app engagement of seventy users with the far right tail beyond six hundred minutes shaded
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 372-373 — Examples 7.6(c) and 7.7(b)
Picture it
Example 7.7(c): the chance seventy sessions total at least ten hours.
Figure (svg): A normal curve for the total app engagement of seventy users with the far right tail beyond six hundred minutes shaded
The shaded sliver is barely visible, which is the honest picture of a probability of 0.0009. Drawing the region before computing gives a rough expectation of the answer's size, and a sliver this thin should never produce an answer in the tenths.
Worked example
Example 7.7(b), with the book's own interpretation.
\[ \sum X \sim N(574, 8.37); \text{ find the } 95\text{th percentile} \]
Write the distribution
Why: Mean 574, spread 8.37.
\[ N(574, 8.37) \]
Set the area
Why: Ninety-five percent below.
\[ 0.95 \]
Reverse the cumulative
Why: On that distribution.
\[ 587.76 \]
Interpret
Why: In the book's words.
Figure (svg): The solution to Worked example the 95th percentile of totals shown as a ladder of expressions, one row per legal move
\[ k = \text{invNorm}(0.95,\; 574,\; 8.37) \approx 587.76 \]
Verify: confirm the distance from the mean in standard deviations
Why: The answer is 13.76 above 574, which divided by 8.37 gives about 1.645 — exactly the z-score for a 95th percentile, and a number that recurs throughout chapters 8 and 9. Checking a percentile against its expected z-score works for any distribution and catches a wrong spread instantly.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, p. 373
Faded example
For the sum of fifty ages, N(1700, 106.07), find the 80th percentile.
Fill in the blanks
k = 1700 + (0.842)(106.07) \approx 1789.3
Why: The 80th percentile of any normal distribution sits about 0.842 standard deviations above its mean, and here that is 0.842 times 106.07 added to 1,700.
Worked example
Example 7.6(c).
\[ \sum X \sim N(1700, 106.07); \text{ find the } 80\text{th percentile} \]
Set the area
Why: Eighty percent below.
\[ 0.80 \]
Recall the z-score
Why: For an 80th percentile.
\[ \text{about } 0.842 \]
Scale and shift
Why: 1700 plus 0.842 times 106.07.
\[ 1789.3 \]
Interpret
Why: Totals of fifty ages.
Figure (svg): The solution to Worked example the 80th percentile of a total of ages shown as a ladder of expressions, one row per legal move
\[ k = \text{invNorm}(0.80,\; 1700,\; 106.07) \approx 1789.3 \]
Verify: confirm the answer as an average per person
Why: Dividing 1,789.3 by fifty gives about 35.8 years per user, only 1.8 above the population mean of 34 — which is a sensible-looking average and confirms the total is not absurd. Converting a percentile of a sum back into a per-value figure is a good habit, because totals in the thousands are hard to judge directly while averages are not.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, p. 373
Trap
\[ k = 587.76 \;\Rightarrow\; \text{95 percent of sessions are under } 587.76 \text{ minutes} \]
Read the percentile as describing one session
Why: It is a number of minutes, so it sounds like a session length.
\[ \text{but a single session averages } 8.2 \text{ minutes} \]
Almost no session lasts ten hours; the number describes the total of seventy of them.
\[ \text{95 percent of the SUMS of } 70 \text{ sessions are at most } 587.76 \text{ minutes} \]
Name the sum in the interpreting sentence
Why: The percentile belongs to the total's distribution.
Here the error is obvious because the scales differ so wildly, which makes it a good place to build the habit. It is much less obvious for a mean, where the percentile and the population value are in the same range — so practising the careful sentence on sums pays off when the same question is asked about averages.
Two truths and a lie
All three concern percentiles for sums.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The 95th percentile of individual sessions is about 9.8 minutes and n times that is 686, against the sum's actual 95th percentile of 587.76. Percentiles do not scale that way, because the sum's spread grows only like root n while n times the population's would grow like n.
Prediction
Commit before reasoning.
Predict first
For any normal distribution, how many standard deviations above the mean is the 95th percentile?
Correct: About 1.645.
Why: That value recurs constantly from chapter 8 onward, where it supplies the boundary of a 90 percent two-sided interval or a 95 percent one-sided one. The 1.96 is the 97.5th percentile, 2.33 the 99th, and 1.28 the 90th — all four are worth recognising on sight, since they let any percentile answer be checked without a calculator.
Explain it
A classmate writes: ninety-five percent of app sessions last at most 587.76 minutes.
Discussion prompt
In two sentences or fewer, correct it.
Hint: Ask how long 587.76 minutes is.
Answer:
Point out that 587.76 minutes is nearly ten hours, and no app session lasts that long — the number is a TOTAL of seventy sessions.
The sentence should read that ninety-five percent of the sums of seventy sessions are at most 587.76 minutes, which works out at about 8.4 minutes per session.
Section
Section 5
Concept
A sum of n values is exactly n times their mean, so any question about one can be restated as a question about the other by multiplying or dividing the boundary by n. Both routes must give the same probability, because they describe the same event.
equivalent boundaries — A total above n times k and a mean above k are the same event. So a question about a sum can always be answered on the mean's distribution, and the reverse, which makes each a check on the other.
\[ \sum x > c \;\Longleftrightarrow\; \bar{x} > \frac{c}{n} \]
The book keeps the two forms separate and gives each its own formulas, which is reasonable for learning them. But recognising that they are one theorem is worth doing at least once, because it turns a memorised pair of rules into a single idea — and it supplies a free check on any answer, since a probability computed both ways that disagrees reveals an arithmetic error immediately.
Figure (svg): A procedure for converting between a question about a sum and a question about a mean
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-372 — the two forms of the theorem
Picture it
Five steps for moving between the two descriptions.
Figure (svg): A procedure for converting between a question about a sum and a question about a mean
The third line is the one to internalise. A total above 7,500 for eighty values IS a mean above 93.75, and no approximation is involved in saying so — it is arithmetic, not statistics. The probability question that follows can then be answered on whichever distribution is more convenient.
Worked example
Example 7.5(a), computed on both distributions.
\[ \mu = 90, \; \sigma = 15, \; n = 80; \text{ find } P\left(\sum x > 7500\right) \]
On the sum's scale
Why: N(7200, 134.16).
\[ z = 2.236 \]
Convert the boundary
Why: 7,500 over 80.
\[ 93.75 \]
On the mean's scale
Why: N(90, 1.677).
\[ z = 2.236 \]
Compare the z-scores
Why: Identical.
Figure (svg): The solution to Worked example one question, two routes shown as a ladder of expressions, one row per legal move
\[ P\left(\sum x > 7500\right) = P(\bar{x} > 93.75) \approx 0.0127 \]
Verify: confirm why the two z-scores must be identical
Why: Dividing the numerator and the denominator of the sum's z-score by n leaves the mean's z-score unchanged: 300 over 134.16 equals 3.75 over 1.677. That is not a coincidence but the algebra of describing one event two ways, so a disagreement between the routes is always an arithmetic slip rather than a conceptual subtlety.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 371-372
Faded example
A sample of 50 values is drawn. A total above 1,800 is the same event as what mean?
Fill in the blanks
\sum x > 1800 \;\Longleftrightarrow\; \bar50 > \frac36___} = ___
Why: Dividing the total by 50 gives 36, so the two statements describe the same event and must have the same probability. Either distribution may be used to compute it.
Worked example
Checking Example 7.6(c) on the mean's scale.
\[ \text{the } 80\text{th percentile of } \sum X \text{ for } n = 50 \]
On the sum's scale
Why: N(1700, 106.07).
\[ 1789.3 \]
Divide by n
Why: 1789.3 over 50.
\[ 35.79 \]
On the mean's scale
Why: N(34, 2.1213).
\[ 35.79 \]
Compare
Why: They agree.
Figure (svg): The solution to Worked example converting a percentile shown as a ladder of expressions, one row per legal move
\[ \frac{1789.3}{50} \approx 35.79 = \text{invNorm}(0.80,\; 34,\; 2.1213) \]
Verify: confirm the two spreads are related by the same factor
Why: The mean's spread is 15 over root 50, about 2.1213, and the sum's is 106.07 — a ratio of exactly 50, as it must be since the sum is 50 times the mean. Checking that the two spreads differ by a factor of n is the fastest way to confirm both were computed correctly, and it takes one division.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-373
Trap
\[ P\left(\sum x > 7500\right) \text{ computed as } P(\bar{x} > 93.75) \text{ on } N(90, 15) \]
Convert the boundary to the mean's scale but keep the population's spread
Why: The boundary is now on the right scale.
\[ z = 0.25 \text{ instead of } 2.236 \]
Half the conversion was done: the boundary moved to the mean's scale but the spread stayed on the population's.
\[ P(\bar{x} > 93.75) \text{ on } N\left(90, \tfrac{15}{\sqrt{80}}\right) = N(90, 1.677) \]
Convert the boundary AND use the matching distribution
Why: Both must describe the same statistic.
This is a good illustration of why writing the distribution on its own line matters. The boundary and the spread have to refer to the same quantity, and the only reliable way to ensure that is to name the distribution explicitly before substituting anything into it.
Two truths and a lie
All three concern converting.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The distribution must be converted too — boundary and spread have to describe the same statistic. Converting only the boundary gives a z-score that is wrong by a factor of the square root of n, which for large n is a very large error.
Prediction
Commit before reasoning.
Predict first
A question asks for the probability that 80 values total more than 7,500. Which route is less error-prone?
Correct: Either, using one as a check on the other.
Why: Since both give the same z-score, the sensible practice is to answer on the scale the question uses — which keeps the interpretation straightforward — and then convert as a verification. That gets a free check for the cost of one division, and it catches exactly the errors that a single route would hide.
Matching
For a sample of size n from a population with mean mu and spread sigma.
Match the pairs
Why: The four entries are the whole of chapter 7's arithmetic. Notice that rows two and four are n times rows one and three respectively — which is the conversion rule, and a way to reconstruct any of the four if one is forgotten.
Comparison
Fill the blanks. Two statistics from one sample, related by a factor of n.
Comparison matrix
| Mean of n | Sum of n | |
|---|---|---|
| Centre | mu | n times mu |
| Spread | sigma over root n | root n times sigma |
| As n grows, the spread | shrinks | grows, but slower than the centre |
| Where root n sits | in the denominator | in the numerator |
The last row is the fastest way to keep the two apart. A mean divides by root n and a sum multiplies by it — and if an answer's scale looks wrong by roughly a factor of n, that is almost always which one went astray.
Pattern
Six steps, and the first two are the ones worth slowing down for.
A quick scale check: the sum's spread should be a small fraction of its centre for any reasonably large n. A spread anywhere near n times sigma has missed the square root.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-372
Check
The distribution of a sum.
Check your understanding
A population has mu = 90 and sigma = 15. For samples of 80, what is the standard deviation of the sum?
Answer: A
Why: The square root of 80 is about 8.944, and multiplying by 15 gives about 134.16.
Check
The centre of a sum.
Check your understanding
For the same population and sample size, what is the mean of the sum?
Answer: A
Why: The mean of a sum is n times mu, which is 80 times 90, or 7,200.
Check
Converting between the two forms.
Check your understanding
For n = 80, the event that the total exceeds 7,500 is the same as which event about the mean?
Answer: A
Why: Dividing the total by the sample size gives 7,500 over 80, which is 93.75. The two statements describe the same event.
Real world
An insurer writes 10,000 independent policies. Each policy's annual claim has a mean of 200 pounds and a standard deviation of 1,500 pounds — most policies claim nothing and a few claim a great deal. The finance director asks how much needs to be held in reserve so that total claims exceed the reserve no more than once in a hundred years.
Discussion prompt
Compute the reserve, and explain why a single policy is uninsurable while ten thousand are routine business.
Hint: The standard deviation of one policy is more than seven times its mean.
Answer:
A single policy is essentially unpredictable. Its standard deviation of 1,500 is 7.5 times its mean of 200, so no reserve set near 200 would be remotely safe for one policy — the whole difficulty of insurance is contained in that ratio.
\[ \mu_{\sum X} = 10\,000 \times 200 = 2\,000\,000, \qquad \sigma_{\sum X} = \sqrt{10\,000} \times 1500 = 150\,000 \]
For the portfolio the ratio collapses. The expected total is 2 million pounds with a standard deviation of only 150,000 — 7.5 percent of the total rather than 750 percent of it. The square root has divided the relative variability by 100, which is the square root of 10,000.
A once-in-a-hundred-years reserve is the 99th percentile, which sits about 2.326 standard deviations above the mean: 2,000,000 plus 2.326 times 150,000, or about 2,349,000 pounds. So a reserve of roughly 2.35 million covers all but one year in a hundred, which is a premium loading of about 17.5 percent over expected claims.
Two qualifications belong in an honest answer, and both matter commercially. The calculation assumes the policies are INDEPENDENT, which fails badly for correlated risks — a flood or a storm makes thousands of claims arrive together, and for such risks the spread grows closer to n than to root n, which is why catastrophe cover is priced and reinsured quite differently. And the central limit theorem describes the middle of the distribution better than the extreme tail, so a 99th percentile computed this way is indicative rather than exact when the individual claims are as skewed as these.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is the standard deviation of a sum root n times sigma rather than n times sigma?
Correct: Because independent departures partly cancel.
\[ n \cdot \frac{\sigma}{\sqrt{n}} = \sqrt{n}\,\sigma \]
Why: Multiplying by n would be correct only if every value moved in lockstep, so that all n departures pointed the same way at once. Independence makes that vanishingly unlikely — highs offset lows — and the square root is exactly the discount for that offsetting. It is also why the result fails for correlated quantities, where the spread grows much faster.
Explain it
They computed the standard deviation of a sum of 80 values as 80 times 15, giving 1,200.
Discussion prompt
In two sentences or fewer, locate the error.
Hint: Ask when a total would actually be 1,200 away from its centre.
Answer:
A spread of 1,200 would require all eighty values to be unusually high at the same time, and since they vary independently that essentially never happens — the highs and lows offset.
The correct factor is the square root of 80, about 8.94, giving a spread of about 134, which is under two percent of the total of 7,200.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, the centre multiplies by n and the spread by its square root. For the second, independent departures partly cancel. For the third, write the distribution on its own line and convert units before substituting. For the fourth, divide the boundary by n AND use the mean's distribution, not just one of the two. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Twenty minutes.
Draw it
At the top, write the two formulas for a sum — centre equal to n times mu, spread equal to root n times sigma — and beside them the two for a mean, with the root n in the denominator. Circle where the square root sits in each. Below that, plot two curves against n from 0 to 100 for a population with mean 15 and standard deviation 15: a straight line for the total and a curve beneath it for the total's spread, and write the ratio at n equal to 1, 4, 25 and 100. In the middle, work Example 7.5 completely: write N(7200, 134.16) on its own line, compute the probability that the total exceeds 7,500, then convert the boundary to 93.75 and redo it on N(90, 1.677), confirming both give a z-score of 2.236. Below that, work Example 7.7: compute the total's mean and spread, convert ten hours into minutes, find the probability of at least 600, and find the 95th percentile — writing a full interpreting sentence that says sums of seventy sessions rather than sessions. At the bottom, write out the four-row table of mean against sum for centre, spread, direction of change with n, and where root n sits.
Check the middle section by confirming both z-scores are identical to three decimals — if they are not, the likely cause is converting the boundary while leaving the population's spread in place. Check the plot by confirming your ratio at n equal to 100 is exactly a tenth of the ratio at n equal to 1.
Recap
Five things, and the second is the one worth remembering longest.
| If you see | Then |
|---|---|
| Total, combined, altogether, sum | Use the sum's distribution |
| A centre wanted for a total | Multiply mu by n |
| A spread wanted for a total | Multiply sigma by ROOT n |
| A spread near n times sigma | The square root has been missed |
| A spread near sigma over root n | That is the mean's formula, not the sum's |
| Mixed units, such as hours and minutes | Convert before comparing with the total |
| An answer to check | Divide the boundary by n and redo it on the mean |
Section 7.3 puts the two forms together and adds the question that comes before either: whether the problem concerns an individual value, a mean or a sum. It also states the law of large numbers, and works through what happens when the underlying population is strongly non-normal — the case where the theorem is doing the most work.
OpenStax Introductory Statistics 2e, §7.2 The Central Limit Theorem for Sums §7.2, pp. 370-373 — everything on these slides traces back here
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