BMS 1110 · Common Core (Tronc commun)Section 5.3 · Centre, spread and rates

Section 24 of 30

Standard deviation and the normal rule

Compute a sample standard deviation as the square root of the variance, say what it means in the original unit, and use the empirical rule to judge whether a patient value is unusual.

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Why this matters

Two wards can report the same mean and still be describing very different patients. A mean serum calcium of 2.30 mmol/L means one thing in a ward where every result sits within 0.10 mmol/L of it, and something quite different in a ward where results swing 0.50 mmol/L either way. The standard deviation is the single number that carries that difference, and because it is measured in the same unit as the data it can be compared with the mean and with a patient’s own result without any conversion. The empirical rule then turns it into a range: about 68 per cent of observations fall within one standard deviation of the mean, 95 per cent within two and 99.7 per cent within three. That is how a nurse decides whether a reported value is ordinary for this population or unusual enough to act on.

What you need to know first

You need three things you already have from earlier sections: adding and dividing to find a mean, squaring a number, and the idea of a squared unit such as mmol²/L². You also need to be comfortable reading √ as “the positive number that multiplies by itself to give this”: √9 = 3 because 3 × 3 = 9. If a calculator is allowed, use it for the final square root; the reasoning is what matters here.

Words to know (5)

Standard deviation: The non-negative square root of the variance, measured in the same unit as the original observations and read as the typical distance of an observation from the mean. Pulse rates with mean 80 beats/min and SD 4.24 beats/min differ from that mean by about 4.24 beats/min as a rule.
Square root: The operation that undoes squaring. It takes a squared quantity back to the original unit, which is why it is the last move in a standard deviation. A variance of 18 beats²/min² becomes √18 ≈ 4.24 beats/min once the root is taken.
Empirical rule: For data whose distribution is approximately normal, about 68 per cent of observations lie within 1 SD of the mean, 95 per cent within 2 SD and 99.7 per cent within 3 SD. Mean 2.30 mmol/L with SD 0.10 puts about 68 per cent of results between 2.20 and 2.40 mmol/L.
Outlier: An observation that lies unusually far from the rest of the data. It may be a true clinical finding, a measurement problem or a data-entry error, so it is investigated before any decision to remove it. In 10, 11, 12, 13, 14 and 35, the value 35 is the outlier; it must be checked, not deleted automatically.
Variance: The average of the squared deviations from the mean. Useful in the working, but expressed in squared units such as mmol²/L², which is why it is not usually reported to clinicians. A sample variance of 18 beats²/min² belongs to the same data as an SD of √18 ≈ 4.24 beats/min.

One idea at a timeIdea 1 / 6

In one sentence Standard deviation is the non-negative square root of variance: it returns spread to the original unit and is zero when all values are identical.

The standard deviation is the square root of the variance, and the root restores the unit

The variance gives you the spread of a dataset, but it gives it in squared units: a set of calcium results in mmol/L produces a variance in mmol²/L². Nobody at the bedside can picture a square millimole, so the variance is a working quantity rather than a reporting one. Standard deviation (SD) is defined as the non-negative square root of the variance. Because the variance squared every deviation, taking the root reverses that squaring and brings the answer home in the unit you started with. That single step is what makes the SD usable: it can be written beside the mean, compared with a patient’s own result, and read as a distance. The formulas are s² = Σ(x − x̄)² ÷ (n − 1) for a sample variance and s = √s² for the sample SD, with x̄ the mean, x each observation and n the number of observations.

For example

A sample of four calcium results gives a variance of 0.09 mmol²/L². The SD is √0.09 = 0.30 mmol/L. Because 0.09 is a perfect square, the root is exact, and the answer goes back to the original unit.

Pause and check yourself

A dataset of patient ages in years has a sample variance of 10 years². What is the standard deviation, and what is its unit?

Show the answer

The standard deviation is √10 ≈ 3.16 years. The unit is years, not years², because the square root removed the square that the variance had added.

Read detailed explanations

How variance and standard deviation divide the labour, and why the root is exact: The standard deviation and the variance describe the same dispersion, but they split the work of a statistical report between them, and that division is the reason both exist. The variance is the quantity that behaves well algebraically: it is built from squared deviations, it adds neatly when independent sources of variation are combined, and it is the natural currency of analysis of variance, of regression and of the pooled estimates used in meta-analysis. The standard deviation is the quantity that behaves well descriptively: because it is the non-negative square root of the variance, it is expressed in the original measurement unit, so it can be placed on the same axis as the observations, quoted next to the mean, and compared with a single patient’s result. In practice this means the variance is the intermediate answer you keep on the worksheet and the standard deviation is the final answer you report at the handover. The notation follows the same split: the Greek letter sigma names a population parameter (σ and σ²), while the Latin letter s names a sample statistic (s and s²). The population variance divides the sum of squared deviations by N, the size of the whole population; the sample variance divides by n − 1, with n the sample size. That extra step in the denominator costs a little and is called Bessel’s correction: it acknowledges that the sample mean was itself estimated from the same data and compensates for the fact that a sample tends to sit slightly closer to its own mean than the population does. A useful check on the whole business is that the root really does undo the square: if the standard deviation of a set of heights in centimetres is 12.7 cm, then squaring it returns a variance of 161.29 cm², and no other square root would restore the centimetre. That is why the SD cannot be negative, why the symbol √ is defined as the positive root, and why a report containing a negative standard deviation contains an error somewhere in the arithmetic.
Standardising a single value: the z-score and the thresholds it creates: The empirical rule gives intervals around a mean, but clinical work usually arrives from the other direction: a single result is in hand and the question is how far it lies from the centre. Standardising converts that question into a short division. Subtract the mean from the observation, divide the difference by the standard deviation, and the answer is a z-score, the number of standard deviations between the value and the mean. The formula in words is z = (observation − mean) ÷ standard deviation, and it has no unit, because the unit above the line cancels the unit below it. A z-score of +2 means the observation sits two standard deviations above the mean; a z-score of −0.5 means it sits half a standard deviation below. The standardisation works because a z-score is comparable across variables even when the units are not: a calcium result expressed in a mmol/L scale and a systolic pressure expressed in mmHg can both be placed on the same z-scale. Clinically useful z-score thresholds fall straight out of the empirical rule. About 68 per cent of values lie between z = −1 and z = +1, about 95 per cent between z = −2 and z = +2, and about 99.7 per cent between z = −3 and z = +3; only about 5 per cent of values lie beyond ±2 and only about 0.3 per cent beyond ±3. A z-score beyond ±2 therefore marks a value that is unusual for the distribution it came from, and beyond ±3 marks one that is rare enough to demand an explanation. Two cautions belong with the tool. First, the sign carries direction, so z = −3 is not the same clinical situation as z = +3. Second, the thresholds inherit every assumption of the rule: if the underlying distribution is not approximately normal, the z-score is still a legitimate standardised distance but 2 no longer corresponds to the same tail probability, so the cut-off must be treated as a rough flag rather than a probability statement.
When the empirical rule does not apply: The empirical rule is a description of one particular shape, the symmetric bell, and several common clinical variables do not have that shape. Length of stay, waiting time, time to readmission and the concentration of a drug in the blood after a single dose are typically skewed: many small values crowd together on one side and a long tail of large values stretches away on the other. In such a distribution the mean and the standard deviation still exist and can still be computed, but the percentages no longer hold. A skewed dataset can easily place 80 per cent of its observations within one standard deviation of the mean while leaving a heavy tail beyond three, and a nurse who applied the 68–95–99.7 percentages to it would systematically misjudge which results are unusual. The warnings are practical rather than theoretical. Report the shape, not just the two numbers: say that the distribution is skewed, and quote the median with the interquartile range instead of the mean with the standard deviation. Look at the histogram before trusting the rule. Investigate any outlier rather than deleting it, since an outlier in a skewed distribution is often the most informative observation in the set. The most important habit is to read the statistical sentence, not just the number: “approximately normally distributed, mean 2.30, SD 0.10 mmol/L” licenses the use of the empirical rule, whereas “right-skewed, median 4 days, interquartile range 2 to 9 days” does not, and the second sentence is a complete report even though it contains no standard deviation at all.

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Standard deviation, one step at a time

A normal curve with aligned standard-deviation intervals, a worked sample calculation, and two spreads compared on one scale.

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01 / 4Where the 68–95–99.7 bands come from

If the drawing is wider than this panel, scroll it sideways. Labels stay readable; the numbered explanations below show the same ideas.

Where the 68–95–99.7 bands come from

The bell is the shape the empirical rule describes. A mean of 2.30 mmol/L with an SD of 0.10 mmol/L puts about 68 per cent of calcium results in the one-standard-deviation band from 2.20 to 2.40, about 95 per cent in the two-standard-deviation band from 2.10 to 2.50, and about 99.7 per cent in the three-standard-deviation band from 2.00 to 2.60. Each bound is computed as mean ± k × SD, so the SD must be in the same unit as the mean. The rule applies to approximately normal data, not to every dataset.

Teaching values, not clinical limits. The 68–95–99.7 rule requires an approximately normal distribution. The curve and its ticks share one scale; both blood-pressure rows share another. SD describes spread, not the average absolute distance.

Key points to remember

The standard deviation is the non-negative square root of the variance: s = √s². The root is the step that turns squared units back into the unit of the data.
To compute a sample SD: mean, deviations, squares, total, divide by n − 1, then take the square root. The deviations must add to zero before you square them.
The SD is read as the typical distance of an observation from the mean, which is why it is quoted beside the mean as mean ± 1 SD.
For approximately normal data, about 68 per cent of observations lie within 1 SD of the mean, 95 per cent within 2 SD and 99.7 per cent within 3 SD — computed as mean ± k × SD.
Two groups with the same mean and different SDs are different populations. An outlier sits far from the rest and is investigated before it is removed, because it may be a real finding, a measurement problem or a data-entry error.

Go deeper

Worked example

The sample SD of five children’s heights, and what the number means. A school nurse records the heights of five children in centimetres: 118, 131, 137, 142 and 152. The nurse wants to report the average height of this small group together with its spread, and to work in the original unit rather than in squared centimetres.

Find the mean. The heights add to 118 + 131 + 137 + 142 + 152 = 680 cm, and there are five children, so x̄ = 680 ÷ 5 = 136 cm.
Subtract the mean from each observation to get the five deviations: −18, −5, +1, +6 and +16 cm. They add to zero, which confirms the mean is right.
Square each deviation so the signs disappear: 324, 25, 1, 36 and 256 cm². These add to 324 + 25 + 1 + 36 + 256 = 642 cm².
Divide by n − 1, which is 5 − 1 = 4: the sample variance is 642 ÷ 4 = 160.5 cm². The unit is centimetres squared, which is why this figure would not be reported to a reader.
Take the non-negative square root: s = √160.5 ≈ 12.7 cm. The unit is back to centimetres, so the spread can be written beside the mean as 136 ± 12.7 cm.

The mean height is 136 cm and the sample standard deviation is about 12.7 cm. Read with the rule, the SD says that a child in this group typically differs from the average by about 12.7 cm, so most of the children’s heights lie roughly between 123 cm and 149 cm. The child at 118 cm sits about 1.4 standard deviations below the mean, which is within the ordinary spread, while a value of 100 cm would be nearly three standard deviations out and would call for a look at the record. The unit matters throughout: the variance of 160.5 cm² could not be compared with a height, but 12.7 cm can.

Tutorial 1 — a complete deviation table for five pulse rates

Five patients on the same ward have resting pulse rates of 74, 78, 80, 84 and 84 beats/min. A learner is asked for the sample mean and the sample standard deviation of these five measurements, working through a complete deviation table rather than jumping to a formula.

Rule used: Subtract the mean from each observation to obtain its deviation, square every deviation, add the squares, divide by n − 1 to get the variance in squared units, and take the non-negative square root to return to the original unit.

Add the five pulse rates and divide by the number of patients to find the mean.
Reveal this step
74 + 78 + 80 + 84 + 84 = 400, and 400 ÷ 5 = 80. The mean resting pulse is 80 beats/min.
What is the deviation of each patient from the mean, and what should their total be?
Reveal the answer
With the mean at 80, the deviations are 74 − 80 = −6, 78 − 80 = −2, 80 − 80 = 0, 84 − 80 = +4 and 84 − 80 = +4. They add to −6 − 2 + 0 + 4 + 4 = 0, and they must: the mean is the balance point, so the positive and negative gaps cancel exactly.
Square each deviation and add the five squares.
Reveal this step
(−6)² = 36, (−2)² = 4, 0² = 0, 4² = 16 and 4² = 16. The total is 36 + 4 + 0 + 16 + 16 = 72.
Why is the total of 72 in squared units, and what divisor do you use to turn it into the sample variance?
Reveal the answer
Each squared deviation is in beats²/min² because a deviation in beats/min was multiplied by itself. The divisor for a sample is n − 1, here 5 − 1 = 4, because the mean was estimated from the same five values. The variance is 72 ÷ 4 = 18 beats²/min².
Take the square root of the variance and write the answer with its unit.
Reveal this step
s = √18 ≈ 4.24 beats/min. The root removes the square, so the answer is back in beats/min, the unit of the original measurements and of the mean.
State the mean and the SD as a single summary and interpret the SD in words. Is a resting pulse of 78 unusual for this group?
Reveal the answer
The summary is 80 ± 4.24 beats/min. The SD means a patient in this group typically differs from the mean by about 4.2 beats/min. A pulse of 78 is only 2 beats/min below the mean, well inside one standard deviation, so it is entirely ordinary. By the empirical rule, about 68 per cent of similar patients would fall between about 75.8 and 84.2 beats/min.

Answer: Mean 80 beats/min, sample variance 18 beats²/min², sample SD ≈ 4.24 beats/min, reported as 80 ± 4.24 beats/min.

Keep this habit: Build the deviation table in order and never divide by n: for a sample the divisor is n − 1. The deviations must total zero before you square them, and the final square root is the step that turns square millimoles back into millimoles and square beats into beats. Always write the unit on the answer line.

Tutorial 2 — using the empirical rule to judge one patient

A clinical chemistry laboratory reports that serum calcium in its adult population is approximately normally distributed with a mean of 2.30 mmol/L and a standard deviation of 0.10 mmol/L. A patient’s result comes back at 2.58 mmol/L. The nurse must give the two ranges that follow from the rule and then decide whether this particular value is unusual.

Rule used: For an approximately normal distribution, about 68 per cent of observations lie within 1 SD of the mean, about 95 per cent within 2 SD and about 99.7 per cent within 3 SD. The bounds are found by computing mean − k × SD and mean + k × SD for k = 1, 2 and 3.

Write down the mean and the standard deviation with their units before doing any arithmetic.
Reveal this step
Mean = 2.30 mmol/L and SD = 0.10 mmol/L. Both are in millimoles per litre, which is what allows them to be added and subtracted directly.
Which interval should you build first, and what are its two bounds?
Reveal the answer
The one-standard-deviation interval, because the rule starts there and because every other bound is a multiple of the same SD. Its bounds are 2.30 − 0.10 = 2.20 mmol/L and 2.30 + 0.10 = 2.40 mmol/L, and about 68 per cent of results lie between them.
Build the two-standard-deviation interval by subtracting and adding twice the SD.
Reveal this step
Twice the SD is 2 × 0.10 = 0.20 mmol/L. The bounds are 2.30 − 0.20 = 2.10 mmol/L and 2.30 + 0.20 = 2.50 mmol/L, and about 95 per cent of results lie between them.
How many standard deviations is 2.58 mmol/L from the mean, and which interval does it fall outside?
Reveal the answer
The distance from the mean is 2.58 − 2.30 = 0.28 mmol/L, and 0.28 ÷ 0.10 = 2.8 standard deviations. That places the value above the two-SD bound of 2.50 mmol/L but still inside the three-SD bound of 2.60 mmol/L, so it is unusual — beyond the 95 per cent band — without being extraordinary.
State the conclusion as a nurse would, including what the rule does and does not claim.
Reveal this step
The result is unusual for this population: fewer than about 5 per cent of results lie beyond two standard deviations in total. The rule gives the probability, not the cause. It says nothing about why this patient’s calcium is high, so the correct action is to check the patient, the sample and the record before drawing a conclusion.

Answer: 68 per cent of results fall between 2.20 and 2.40 mmol/L; 95 per cent between 2.10 and 2.50 mmol/L. The value 2.58 mmol/L is 2.8 SD above the mean, so it is unusual and warrants clinical review.

Keep this habit: Use the rule only when the distribution is stated to be approximately normal, and say so when you report. The range is always mean ± k × SD, so the SD must be in the same unit as the mean. Finish by converting the interval into a clinical judgement about the patient in front of you, not into a verdict on the laboratory.

Now you try

Attempt each one on paper first. Write the formula, substitute the numbers, then check the unit of your answer.

A sample of potassium results has a variance of 6.25 mmol²/L². State the standard deviation with its unit, and explain why the ward report shows the standard deviation rather than the variance.
Hint
The variance is the SD squared, so find the positive number whose square is 6.25. Ask yourself which of the two units a colleague could compare with a patient’s potassium.
Show the worked answer
The standard deviation is the non-negative square root of the variance: √6.25 = 2.5, so s = 2.5 mmol/L. The ward reports 2.5 mmol/L rather than 6.25 mmol²/L² because the SD is in the original unit, so it can be compared with the mean and with a patient’s own result, and it can be read as the typical distance of a potassium result from the mean.
Five patients have body temperatures of 36.6, 36.8, 37.0, 37.2 and 37.4 °C. Compute the sample variance and the sample standard deviation, then interpret the standard deviation in one sentence.
Hint
The mean is exactly 37.0 °C, so the five deviations are −0.4, −0.2, 0, +0.2 and +0.4. Square them, add, and divide by n − 1 = 4 before taking the root.
Show the worked answer
The mean is 37.0 °C and the deviations are −0.4, −0.2, 0, +0.2 and +0.4. The squared deviations are 0.16, 0.04, 0, 0.04 and 0.16, which total 0.40. The sample variance is 0.40 ÷ 4 = 0.10 °C² and the sample SD is √0.10 ≈ 0.32 °C. In words: a patient in this sample typically differs from the mean temperature by about 0.32 °C.
Systolic blood pressure in an adult population is approximately normal with mean 118 mmHg and standard deviation 9 mmHg. Give the ranges containing about 68 per cent and about 95 per cent of measurements, then judge whether a reading of 142 mmHg is unusual.
Hint
One standard deviation on either side of the mean gives the 68 per cent range; two standard deviations give the 95 per cent range. Then count how many standard deviations separate 142 from 118.
Show the worked answer
About 68 per cent of measurements lie between 118 − 9 = 109 and 118 + 9 = 127 mmHg. About 95 per cent lie between 118 − 18 = 100 and 118 + 18 = 136 mmHg. The value 142 is 142 − 118 = 24 mmHg above the mean, and 24 ÷ 9 ≈ 2.7 standard deviations, so it falls outside the 95 per cent range but still inside the three-standard-deviation range of about 91 to 145. It is unusual rather than impossible — fewer than about 5 per cent of readings lie beyond two standard deviations in total — so the rule should prompt a check of the patient and the measurement, not an automatic assumption that the value is wrong.

Common misconception

Beginners often read a large standard deviation as a sign that the measurements are wrong or the care is poor. It is neither. A standard deviation measures how much the observations differ from one another, so a ward with a heterogeneous patient mix will naturally show a larger SD than a ward with similar patients, even when every measurement was taken correctly. Equally, a small SD is not proof of accuracy: a faulty device can return the same wrong number every time and produce a beautifully small SD. The number describes spread, not quality.

Check your reasoning

A colleague reports “mean systolic pressure 118 mmHg” and nothing else. Explain what is missing, what the reader cannot judge without it, and what a standard deviation of 9 mmHg versus 25 mmHg would change about the interpretation.

Reveal an answer

The mean states the centre of the observations but says nothing about their spread, so a reader cannot tell whether 118 describes patients who all sit close to 118 or a wide mixture averaging out to 118. With an SD of 9 mmHg, most readings lie roughly between 100 and 136 mmHg and a value of 142 is about 2.7 standard deviations out, genuinely unusual. With an SD of 25 mmHg, 142 is less than one standard deviation from the mean and entirely ordinary, while readings from 68 to 168 mmHg would all be within the 95 per cent range. The mean alone would have made the two populations look identical. The missing half of the description is the standard deviation, and without it neither an unusual value nor a comparison between groups can be judged.

Where this comes from

NIST: Normal data and standard deviation · Optional English reading; internet required. Checked 6 October 2026 for the diagram and beginner explanation revision. Explains normal-curve spread.
BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · Unit Five “Standard Deviation”, lines 880–914: SD as the non-negative square root of variance and its use of the original unit; population and sample formulas; the age worked example (mean 24 years, variance 10 years², SD ≈ 3.16 years); the blood-pressure example (mean 120 mmHg, variance 62.5 mmHg², SD ≈ 7.91 mmHg); the haemoglobin groups with equal means and different SDs; the variance-versus-SD comparison table; the empirical rule for approximately normal data (68 per cent within 1 SD, 95 per cent within 2, 99.7 per cent within 3, with the mean 100 and SD 10 illustration); and the outlier example 10, 11, 12, 13, 14, 35, which must be investigated rather than automatically removed.
BMS 1110 lecture notes, variance and dispersion context · Unit Five lines 842–879: measures of dispersion and the sample variance s² = Σ(x − x̄)² ÷ (n − 1) that this section takes the square root of. The reasons for dividing by n − 1 (Bessel’s correction) belong to section 5.2; standard error and the coefficient of variation, which the notes introduce at lines 913–919, belong to section 5.4 and are used here only to keep the two quantities apart.
Beyond the supplied notes: standardised values, relative spread and skewed distributions · The z-score as (observation − mean) ÷ standard deviation with the ±1, ±2 and ±3 thresholds it creates; the coefficient of variation CV = (s ÷ x̄) × 100 per cent and its limits when the mean is near zero or values can be negative; the effect of skew on the empirical rule and the median/interquartile-range alternative are standard biostatistics that the supplied notes mention only in passing.

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About this course and its sources

Bilingual lessons organised from the supplied BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (Mr. Ebangan Prosper Achuanoh). Diagrams are original educational schematics; sections that go beyond the supplied notes say so in their sources. Mathematics, clinical and French-language review are pending.