Module 1 · Foundations

Variation and basic statistics

Every tool in this course is a way of describing variation, and every one of them rests on a handful of ideas: a centre, a spread, a shape, and the difference between the parts you measured and the process that made them. This module builds those ideas from printed data with the arithmetic shown, so that when a piece of software prints a standard deviation you know exactly what it did.

Learning objectives

Why this matters

In 1973 Francis Anscombe published four small datasets in The American Statistician. Each has eleven pairs of x and y. All four have the same mean of x, the same mean of y, the same variances, the same correlation, the same fitted straight line and the same R². Plotted, they are four different stories: a clean linear relationship, a curve, a line with one outlier, and ten identical x-values plus one point that decides everything.[7][9] The full data are printed in Section 4 and every summary statistic is recomputed there; you will see the numbers agree to the second decimal while the pictures disagree completely.

Anscombe's point is the point of this module. Summary statistics are compressions. They are indispensable, because nobody can read 200 measurements, but they throw away the shape of the data, and the shape is often where the engineering is. A mean and a standard deviation describe a symmetric, single-peaked distribution well and describe a skewed, bounded or two-mode distribution badly, and a capability index computed from the wrong description predicts a defect rate that is simply wrong. So the module works in both directions: the arithmetic of the summaries, so that you can trust them, and the habit of looking at the data first, so that you know when not to.

The second idea that runs through the module is Deming's distinction between an enumerative study, which asks about the items in front of you, and an analytic study, which aims at action on the process that will make the items of the future.[10] Almost everything you will do as an engineer is analytic. The sample of sixty shafts in Section 2 is not interesting as sixty shafts; it is interesting as evidence about the grinder, and that changes what the statistics are allowed to claim.

1. Types of data

The first question about any measurement is what kind of number it is, because that decides which summaries, which distribution and which control chart apply.

Table 1. The three kinds of data an engineer records, and what each supports.
KindExamplesWhat you can computeWhere it goes later
Continuous (variables)Diameter in mm, weight in g, torque in N·m, temperature in °C, pull strength in NMean, standard deviation, histogram, capability index; the richest kind, because every part carries a valueCapability (Modules 7 and 8), X̄-R and I-MR charts (Module 16)
Count (discrete)Defects per board, scratches per panel, leaks per assemblyRates and averages; the Poisson distributionDPU and DPMO (Module 6), c and u charts (Module 17)
Attribute (pass or fail)Leak test passed, go/no-go gauge accepted, visual inspection rejectedProportions; the binomial distributionAttribute capability (Module 8), p and np charts (Module 17)

Continuous data carry far more information per part than the other two. A bore measured to 0.001 mm tells you how close it came to the limit; a plug gauge tells you only which side of it the bore fell on. When you have a choice of gauge, measure the value. A rule of thumb from the amount of information involved: a study on attribute data needs something like ten times as many parts as one on variables data to reach the same confidence about the process, which is why Modules 8 and 17 treat attribute data as the fallback rather than the norm.

2. Where the data sit: mean and median

The data below is a constructed example generated by a seeded script, not a real production run. Setting: a ground shaft of Ø8.000 ± 0.015 mm, measured with a micrometer reading to 0.001 mm, 60 consecutive parts in production order.

Table 1. Shaft diameter, mm, 60 consecutive parts in production order, read left to right then down (constructed data).
Values +0+1+2+3+4+5+6+7+8+9
1–108.0048.0097.9948.0108.0027.9987.9987.9978.0018.006
11–208.0058.0057.9947.9958.0098.0068.0128.0077.9978.008
21–308.0058.0037.9988.0028.0018.0068.0038.0097.9998.004
31–408.0047.9987.9978.0078.0048.0018.0047.9998.0037.999
41–508.0038.0068.0008.0067.9978.0017.9968.0028.0028.001
51–608.0018.0088.0087.9978.0077.9947.9998.0018.0027.997
x̄ = Σxi / n The arithmetic mean: add every value and divide by the number of values. For the 60 shafts, Σx = 480.131 mm and x̄ = 8.0022 mm.

The median is the middle value when the data are sorted: with 60 values it is the average of the 30th and 31st, here 8.002 mm. The mean and the median agree to a few microns because the data are symmetric. They separate when the data are skewed, and the gap between them is the first and cheapest test of shape: in Section 6 a flatness dataset has a mean of 7.875 µm and a median of 7.2 µm, and that difference is the tail talking.

The mean is the right centre for a symmetric distribution and for anything that will be added up (weights, costs, times). The median is the right centre when a few extreme values would drag the mean somewhere no part actually sits. Report both when they differ, and then look at the histogram.

3. How far the data spread: range, variance, standard deviation

The range is the largest value minus the smallest, 8.0127.994 = 0.018 mm for the shafts. It is simple, it is what a machinist sees, and it is what the R chart in Module 16 uses for small subgroups. Its weakness is that it grows with the sample size and depends on only two values.

The standard deviation uses every value. It is built in three steps: the deviation of each value from the mean, the average of the squared deviations, and the square root to get back to the original units.

s² = Σ(xi − x̄)² / (n − 1)      s = √s² s² is the sample variance, s the sample standard deviation. The numerator, the sum of squared deviations, is written SS. For the shafts SS = 0.0011610 mm², so s² = SS / 59 = 0.00001968 mm² and s = 0.00444 mm.

Why n − 1 and not n

If you knew the true process mean µ, the average squared deviation from µ over n values would be an unbiased estimate of the process variance, and you would divide by n. You do not know µ; you use x̄, and x̄ is the one number that makes the sum of squared deviations of this sample as small as it can be. The deviations from x̄ are therefore, on average, a little smaller than the deviations from µ, and dividing by n underestimates the variance. Dividing by n − 1 corrects the bias exactly. The usual way of saying it is that one degree of freedom was spent estimating the mean.

For the shafts the two versions are s = 0.00444 mm with n − 1 and 0.00440 mm with n, a difference of less than 1 % at n = 60. At n = 5, the subgroup size of most X̄-R charts, the difference is about 12 %, which is why the choice matters on small samples and why spreadsheet functions come in two versions (in Excel, STDEV.S and STDEV.P). This course always means the n − 1 version when it writes s, and it says so whenever a within-subgroup estimate such as R̄/d₂ is used instead (Module 7).

Two more habits. The variance is in squared units and is the quantity that adds: if two independent sources of variation act on a part, their variances add, their standard deviations do not (Module 15 uses this for tolerance stacks, Module 4 for gauge error). And the standard deviation of the mean of n values, called the standard error, is s/√n; for the shafts 0.00057 mm. That √n is the central limit theorem in Section 8, and it is the reason a control chart on subgroup means has such narrow limits.

Calculator: descriptive statistics and histogram

Pre-loaded with the 60 shaft diameters and the drawing limits. Every value in the output table is one of the statistics defined above, computed from the data you see in the box.

Descriptive statistics and histogram

Paste any list of numbers. The specification limits are optional and only draw lines on the histogram and count the values outside them. Nothing leaves your browser.

4. Looking at the data: histograms, quartiles, and Anscombe

A histogram divides the range of the data into bins of equal width and draws a bar whose height is the number of values in each bin. It is the single most useful picture in this course and the first thing to draw with any new dataset. Three things to know about it.

Quartiles give a compact numerical version of the shape. The first quartile Q1 has a quarter of the data below it, the third quartile Q3 three quarters; for the shafts Q1 = 7.998 and Q3 = 8.006 mm, so the middle half of the parts lie within 0.008 mm of each other. The box plot, which draws the quartiles as a box with the median inside and whiskers to the extremes, is Tukey's invention and is the tool of choice for comparing several groups side by side (Module 9).[12] Several conventions exist for interpolating a quartile between two data values; the calculators in this course use the one that Minitab and Excel's QUARTILE.EXC use, so that your numbers agree with theirs. Skewness, a signed number that is zero for a symmetric distribution, is 0.03 for the shafts and 1.40 for the flatness data in Section 6; the calculator prints it, and a value beyond about ±1 is a strong hint that the normal model will fail.

Worked example 1: Anscombe's quartet, four datasets with one set of statistics

Real data, published by Anscombe in 1973.[7] The values are as reproduced in two secondary sources,[8][9] because the original article is available only as a scanned image; the transcription was checked by recomputing every property the paper reports. No units are given in the source.

Table 2. Anscombe's quartet (real data, 1973). Four pairs of columns; sets I to III share the same x-values.
RowxIyIxIIyIIxIIIyIIIxIVyIV
110.08.0410.09.1410.07.468.06.58
28.06.958.08.148.06.778.05.76
313.07.5813.08.7413.012.748.07.71
49.08.819.08.779.07.118.08.84
511.08.3311.09.2611.07.818.08.47
614.09.9614.08.1014.08.848.07.04
76.07.246.06.136.06.088.05.25
84.04.264.03.104.05.3919.012.50
912.010.8412.09.1312.08.158.05.56
107.04.827.07.267.06.428.07.91
115.05.685.04.745.05.738.06.89

Section 3's statistics, plus the correlation and fitted line that Module 12 explains, come out the same for all four sets to the precision Anscombe reported. The verification scripts and the regression engine behind Module 12's calculator produce Table 3 from the data above.

Table 3. Summary statistics of the four Anscombe datasets, computed from Table 2. Anscombe reported the shared values as mean x = 9.0, mean y = 7.5, regression y = 3 + 0.5x, correlation 0.82; the fourth set's correlation is 0.8165, which rounds to 0.817 at three decimals.
StatisticSet ISet IISet IIISet IV
n11111111
Mean of x9.009.009.009.00
Mean of y7.507.507.507.50
Slope of fitted line0.5000.5000.5000.500
Intercept3.003.003.003.00
Correlation r0.8160.8160.8160.817
0.670.670.670.67
Residual standard deviation1.241.241.241.24
Scatter plots of the four Anscombe datasets with the common fitted line Four panels. Set I: points scattered evenly about a rising straight line. Set II: points on a smooth arch, curving up then down, with the straight line cutting through it. Set III: ten points on a perfect straight line and one point well above it, which tilts the fitted line. Set IV: ten points stacked vertically at x equals 8 and one point far to the right at x equals 19 that alone determines the line. Set I 5 10 15 20 4 8 12 Set II 5 10 15 20 4 8 12 Set III 5 10 15 20 4 8 12 Set IV 5 10 15 20 4 8 12 x (no units given in the source); vertical axis y; the line is y = 3.00 + 0.500x in every panel
Figure 2. Anscombe's quartet plotted from Table 2, each panel with its own fitted line (they coincide to two decimals). Same statistics, four different processes.

Read the four panels against Table 3. Set I is what the statistics describe. Set II is a curve, and a straight line through it is the wrong model however good its R². Set III is a perfect straight line with one point off it, and that one point has bent the fitted line away from the other ten. Set IV has no relationship at all among ten of its points; the eleventh creates the correlation single-handed. An engineer who receives Table 3 without Table 2 or the plots cannot tell which of the four processes is in front of them. This is why every analysis in this course starts with a picture, why Module 9 exists, and why the calculators draw before they compute.

5. The normal distribution and what it lets you do

Many continuous characteristics, when the process is stable and the variation comes from many small independent causes, have a histogram close to the familiar bell. The normal distribution is the mathematical form of that bell. It is defined by two numbers, its mean µ and its standard deviation σ, and its density is

f(x) = (1 / (σ√(2π))) · exp[ −(x − µ)² / (2σ²) ] The curve is symmetric about µ, and σ sets its width: the points of inflection, where the curve changes from bulging out to curving in, sit exactly one σ either side of the mean.[1]

What makes the normal useful is that the fraction of the distribution beyond any point depends only on the distance from the mean measured in standard deviations, the Z-value of Module 0: Z = (x − µ)/σ. There is no closed formula for that fraction,[1] which is why textbooks carry a Z table, but every calculator in this course computes it directly to better than one part in ten million (the algorithm and its test against scipy are described in the code), so you never need the table. Three values are worth memorising:

Table 4. The empirical rule: fraction of a normal distribution within ±Z standard deviations of the mean. Tail areas from the sigma-level table verified in Module 0.
±ZOutside, both tails (PPM)InsideUsed for
1317,31168.27 %The standard error interval; zone C on a control chart
245,50095.45 %The "95 %" of a confidence interval is 1.96σ, not 2
32,70099.73 %Control limits (Module 16); the 6σ "natural" width in Cp (Module 7)
The normal curve with the one, two and three sigma bands A bell curve on a horizontal axis marked from minus four to plus four standard deviations. Vertical lines at plus and minus one, two and three sigma divide it into bands, labelled with the fraction of the area inside each: 68.27 percent within one sigma, 95.45 within two, 99.73 within three. −4σ−3σ−2σ−1σ µ +1σ+2σ+3σ+4σ 68.27 % 95.45 % within ±2σ 99.73 % within ±3σ 0.135 % beyond −3σ 0.135 % beyond +3σ Distance from the mean in standard deviations
Figure 1. The empirical rule. The area within ±1σ, ±2σ and ±3σ is 68.27 %, 95.45 % and 99.73 %; each 3σ tail holds 0.135 %, or 1,350 parts per million.

Applied to the shafts: the mean is 8.0022 mm and s = 0.00444 mm, so the upper limit of 8.015 sits Z = (8.015 − 8.0022) / 0.00444 = 2.89 standard deviations above the mean, and the lower limit 3.87 below. Both are beyond 2.5σ, so a normal model puts well under 1 % of shafts outside the tolerance in total; the sample shows 0 of 60 outside. Module 7 turns that Z into a capability index. Module 0's Table 1 turns it into a sigma level.

Where the normal model stops

The normal distribution is a model, and three kinds of characteristic break it as a matter of physics, not of bad luck.

The test this course uses to ask "is a normal model reasonable?" is the Anderson-Darling test, whose p-value the calculator prints; Module 11 explains what a p-value is, and for now read a p below 0.05 as "the data depart from the normal shape by more than chance would explain, find out why". The shafts give p = 0.38. The flatness data in the next section give p below 0.001. A histogram and the skewness usually tell you the same thing before the test does, and the physical reason is what you act on.

6. The other distributions an engineer meets

Lognormal, and worked example 2: a bounded characteristic

Constructed data, not a real production run. Setting: the flatness of a milled face on an aluminium housing, measured on a CMM to 0.1 µm, 200 consecutive parts. Flatness cannot be negative and a perfect face would read zero.

Table 5. Flatness, µm, 200 consecutive parts in production order, read left to right then down (constructed data).
Values +0+1+2+3+4+5+6+7+8+9
1–104.95.85.69.07.28.511.711.89.96.5
11–205.75.56.97.85.24.89.23.57.49.7
21–305.14.85.710.77.53.46.612.612.510.6
31–407.23.58.67.48.316.08.79.110.57.1
41–505.95.711.310.06.413.48.74.410.57.5
51–604.410.16.96.710.43.04.56.68.95.9
61–705.15.07.610.84.23.17.06.75.310.5
71–8011.56.26.910.36.64.37.85.39.36.4
81–907.310.17.66.05.117.413.24.15.010.4
91–1006.18.111.58.57.56.57.79.517.512.9
101–1107.312.18.44.17.612.56.112.27.08.1
111–1204.720.39.812.66.54.113.85.816.87.4
121–13016.65.83.611.07.88.012.85.44.54.7
131–1403.83.54.55.26.54.67.511.17.19.5
141–1505.28.99.65.83.97.07.614.46.315.6
151–1602.66.56.06.25.87.215.36.912.57.0
161–1707.72.57.69.64.45.65.018.84.75.3
171–1802.23.68.04.12.16.47.68.28.65.5
181–19025.25.78.84.15.53.78.06.14.73.0
191–2009.310.510.57.09.910.99.213.96.88.1

The summary statistics: mean 7.875 µm, median 7.2 µm, s = 3.566 µm, minimum 2.1, maximum 25.2, Q1 = 5.425, Q3 = 9.675 µm, skewness 1.40. The mean is above the median, the upper quartile is further from the median than the lower, and the skewness is well above 1: three signs of a right-hand tail before you draw anything. The Anderson-Darling p-value is below 0.001. And a normal model with this mean and standard deviation would place its lower 3σ point at −2.8 µm, a negative flatness, which is the model telling you it does not apply.

A characteristic bounded at zero with a long right tail is often described well by the lognormal distribution: X is lognormal if ln X is normal.[4] The recipe is to take the natural logarithm of every value and look again. For these 200 values, ln x has mean 1.970, standard deviation 0.435, skewness -0.05 and an Anderson-Darling p-value of 0.98: on the log scale the data are as normal as anyone could ask. The mean of the logs converted back, e1.970 = 7.17 µm, is the geometric mean, and it sits near the median (7.2 µm) rather than the arithmetic mean (7.875 µm); for a skewed distribution the geometric mean or the median is the honest "typical part".

Module 8 shows what to do with capability for such a characteristic: fit the lognormal or transform, or better, report the fraction beyond the limit from the percentiles of the data themselves. The one thing not to do is to compute mean ± 3s and read a defect rate off it.

Example 2 in the calculator (inputs collapsed; expand to edit)

Weibull

The Weibull distribution has a shape parameter that lets one family cover distributions from steeply right-skewed to nearly symmetric, and a scale parameter that sets the size; both the Weibull and the lognormal are, in NIST's words, used extensively in reliability applications to model failure times.[5][6] You will meet it in fatigue and life testing, in the cycles-to-failure of a spring or a switch, and occasionally as a fit for a bounded process characteristic (Module 8). Its arithmetic is not needed in this course; what is needed is to recognise it as the standard answer to "what distribution do lifetimes follow?".

Binomial and Poisson, and worked example 3: what a clean sample proves

Arithmetic on two standard distributions, computed by the verification scripts; not data. The settings are illustrative.

Attribute data follow the binomial distribution: if each of n parts independently has probability p of being defective, the number of defectives in the sample has mean np and standard deviation √(np(1 − p)), and the probability of exactly x defectives is C(n, x) px (1 − p)n − x, where C(n, x) counts the ways of choosing which x parts.[2] Counts of defects follow the Poisson distribution: if defects occur at an average rate λ per unit, the probability of exactly x on a unit is e−λ λx / x!, and the mean and the variance are both λ.[3] The p and np charts of Module 17 are built on the first, the c and u charts on the second, and Module 6's yield formula e−DPU is the Poisson probability of zero defects.

Suppose a supplier's lot is running at 2 % defective and you pull a sample of 50. The expected number of defectives is np = 1.0, with standard deviation 0.99. The probabilities:

Table 6. Binomial probabilities for n = 50, p = 0.02, and Poisson probabilities for λ = 1.5 defects per board.
xBinomial P(x defectives)Binomial P(≤ x)Poisson P(x defects)Poisson P(≤ x)
00.3640.3640.2230.223
10.3720.7360.3350.558
20.1860.9220.2510.809
30.0610.9820.1260.934
40.0150.9970.0470.981
50.0031.0000.0140.996

Read the first row. A sample of 50 from a lot that is 2 % defective contains no defectives 36.4 % of the time. "We checked 50 and found nothing" is therefore weak evidence that the lot is clean: it is what you would see more than a third of the time from a lot that is not. The probability of three or more, 1 − 0.922 = 0.078, is about 8 %, so three defectives in fifty is unusual but not remarkable at 2 %. Sample sizes for attribute decisions are large for exactly this reason. The Poisson columns tell the same story for defects: a board with an average of 1.5 defects comes out clean 22.3 % of the time, and shows three or more 19.1 % of the time.

7. Sample, population, and process

Statistics textbooks speak of a sample drawn from a population, and estimate the population's mean and standard deviation from the sample's. That picture fits a warehouse: 5,000 shafts in stock, 60 pulled at random, and the question "how many of the 5,000 are out of tolerance?" Deming called this an enumerative study: the population exists, is finite, and could in principle be measured completely.[10]

An engineer's question is almost never that one. It is "what will the grinder make tomorrow, and what should we change?" The 60 shafts are not a sample from a population; they are a sample from a process in time, and the population they are supposed to represent, next week's output, does not exist yet. Deming's term is an analytic study: one that aims at action on the cause system to improve the product of the future.[10] Two consequences follow, and the rest of the course keeps returning to them.

8. The central limit theorem

Take any distribution with a mean µ and a standard deviation σ, whatever its shape. Draw subgroups of n values and average each subgroup. The averages have their own distribution, and the central limit theorem says three things about it: its mean is µ; its standard deviation is σ/√n; and its shape approaches the normal as n grows, whatever the parent looked like.[1] The third statement is the remarkable one, and it is the reason the X̄ chart works on data whose individual values are not normal at all.

Worked example 4: the theorem on the flatness data

The same constructed flatness data as worked example 2, taken in consecutive subgroups of 2, 5 and 10.

The 200 individual values are strongly right-skewed (skewness 1.40) with s = 3.566 µm. Averaging consecutive values gives 100 means of 2, 40 means of 5 and 20 means of 10. Table 7 compares each set of means with what the theorem predicts.

Table 7. Means of consecutive subgroups of the 200 flatness values. Every subgroup mean can be reproduced from Table 5; the statistics are computed from those means.
Subgroup size nNumber of means kMean of the means (µm)Standard deviation of the means (µm)Predicted s/√n (µm)RatioSkewness of the meansAnderson-Darling p of the means
1 (individuals)2007.8753.5663.5661.001.40< 0.001
21007.8752.4972.5210.990.520.21
5407.8751.7521.5951.10-0.150.28
10207.8751.1841.1281.050.080.88

All three predictions hold on printed data. The mean of the means is the mean of the individuals, exactly, because averaging averages is averaging. The spread of the means shrinks as √n: the ratio column stays near 1, and it is not exactly 1 because 20 or 40 means give a rough estimate of a standard deviation. And the shape moves toward the normal: skewness falls from 1.40 to about zero by n = 5, and the Anderson-Darling test, which rejects the individuals decisively, finds nothing wrong with the means of 5 or 10. This is the theorem doing exactly what it promises, on a characteristic that could not be less normal.

Two cautions that Module 16 will repeat. The theorem is about the means, not the parts. A control chart of subgroup means will behave well on this flatness process; a histogram of individual parts against a specification will not, and a capability index that assumes normal individuals will be wrong. And the theorem needs the subgroups to be drawn from one distribution; if the process shifts between subgroups, the means carry the shift, which is precisely what the X̄ chart is designed to detect.

Simulator: draw subgroups from a skewed parent and watch the means

The simulator below draws subgroups from a parent distribution of your choice, plots the distribution of the subgroup means beside the parent, and reports their standard deviation against σ/√n. It also lets you add a shift and see how many subgroups a chart of means takes to notice it, which is Module 16's subject. The random numbers are seeded, so the same seed reproduces the same picture.

Central limit theorem and shift-detection simulator

Start with the exponential parent and n = 5, then try n = 2 and n = 10 and compare with Table 7. The parent distributions are scaled to σ = 1 so that the predicted standard deviation of the means is simply 1/√n.

Common mistakes

  1. Reporting a mean and standard deviation without a histogram. Consequence: Anscombe's four processes look identical, and a skewed or two-mode process is described as a bell. Fix: draw first; the calculator does it for you.
  2. Dividing by n on a small sample. Consequence: a standard deviation about 12 % too small at n = 5, and a capability index inflated by the same amount. Fix: n − 1 for a sample; know which function your spreadsheet is using.
  3. Adding standard deviations. Consequence: a tolerance stack or a gauge-plus-process error that is far too pessimistic. Fix: variances add; take the square root afterwards (Modules 4 and 15).
  4. Applying the empirical rule to a bounded characteristic. Consequence: a predicted lower limit below zero and a defect rate that means nothing. Fix: recognise flatness, runout, roughness and leak rate as skewed by physics; use the log scale or the data's own percentiles (Module 8).
  5. Treating a clean small sample as proof. Consequence: "50 checked, none found" is read as zero defectives when a 2 % lot would give the same result 36 % of the time. Fix: compute the binomial probability, and size attribute samples accordingly.
  6. Using bins narrower than the gauge. Consequence: a comb-like histogram whose gaps are artefacts of the resolution. Fix: bins at least as wide as the resolution, about √n of them to start.
  7. Reading a histogram as if time did not exist. Consequence: a drifting process looks like a wide, stable one. Fix: plot in time order first (Module 9), chart before you summarise (Module 16).
  8. Answering an analytic question with an enumerative answer. Consequence: a confidence interval on last week's mean is presented as a forecast of next week's. Fix: show stability before extrapolating, and design the sample to cover the sources of change (Module 5).

Exercises

Exercise 1: descriptive statistics by hand

Constructed data, not a real production run. Setting: the tightening torque of a bolted joint, specified as 12.0 ± 1.0 N·m, measured with a torque analyser reading to 0.01 N·m, 30 consecutive joints.

Exercise 1 data. Tightening torque, N·m, 30 consecutive joints in production order, read left to right then down (constructed data).
Values +0+1+2+3+4+5+6+7+8+9
1–1012.2012.0012.0311.4712.6512.4712.0612.3712.2311.99
11–2012.4212.0612.0611.9312.2812.1212.3011.9812.1911.90
21–3012.3912.2012.2412.2611.8712.3712.7311.6911.6711.73

Tasks. (a) Compute the sum, the mean and the median. (b) Compute the sum of squared deviations, the sample variance and the sample standard deviation, and the population version of the standard deviation; comment on the difference. (c) Find the minimum, maximum, range and quartiles, and describe the shape from these numbers alone. (d) Using a normal model with the sample mean and standard deviation, how many standard deviations away is each specification limit, and roughly what fraction of joints would be out of tolerance? How many of the 30 are?

Show the worked solution

(a) Σx = 363.86 N·m; x̄ = 363.86 / 30 = 12.1287 N·m. Sorted, the 15th and 16th values average to a median of 12.155 N·m. Mean and median differ by about 0.03 N·m, a tenth of a standard deviation: no sign of skew.

(b) SS = Σ(x − x̄)² = 2.3617 (N·m)². s² = SS / 29 = 0.08144, s = 0.2854 N·m. With n instead of n − 1: √(SS / 30) = 0.2806 N·m, about 1.7 % smaller. The sample version is the right one: the 30 joints are a sample from the process, not the whole population of joints.

(c) Minimum 11.47, maximum 12.73, range 1.26 N·m. Q1 = 11.9675, Q3 = 12.3175 N·m; the median sits close to the middle of the box (Q1 to Q3), the range is about 4.4 standard deviations, which is ordinary for 30 normal values, and the skewness is -0.15. Symmetric and single-peaked as far as the numbers can tell; the calculator's histogram and Anderson-Darling p = 0.91 agree.

(d) Upper limit: Z = (13.0 − 12.1287) / 0.2854 = 3.05; lower limit: Z = (12.1287 − 11.0) / 0.2854 = 3.96. The normal tails beyond those are about 0.11 % above and 0.004 % below, a little over one joint in a thousand in total; 0 of the 30 are out. The process is running about 0.13 N·m above nominal, so the upper limit is the nearer one; a small adjustment of the tool would centre it. Whether this sample of 30 from one run says anything about next week is Section 7's question, and the answer needs a time plot.

Exercise 1 in the calculator

Exercise 2: the central limit theorem on the shaft data

The constructed shaft diameters of Table 1, taken in twelve consecutive subgroups of five (parts 1 to 5, 6 to 10, and so on).

Tasks. (a) Compute the mean of the first subgroup by hand. (b) Compute all twelve subgroup means and their mean and standard deviation. (c) Compare the standard deviation of the means with s/√5 from Section 3. (d) Which of the two, the individual values or the subgroup means, would you expect to be closer to normal, and why does the answer matter for a control chart?

Show the worked solution

(a) Parts 1 to 5 read 8.004, 8.009, 7.994, 8.010, 8.002 mm; their sum is 40.019 and the mean 8.0038 mm.

(b) The twelve means, in order:

Exercise 2. Means of the twelve consecutive subgroups of five shafts, mm (computed from Table 1).
Subgroup123456789101112
Mean8.00388.00008.00168.00608.00188.00428.00208.00128.00248.00048.00427.9986

Their mean is 8.0022 mm, identical to the mean of the 60 individuals, and their standard deviation is 0.00208 mm.

(c) s/√5 = 0.00444 / 2.236 = 0.00198 mm. The observed spread of the means is 1.05 times the prediction. Twelve means give a rough estimate of a standard deviation, so agreement within about 20 % is as good as the sample can show; the point is that the means are roughly √5 = 2.24 times tighter than the parts. This is why an X̄ chart's limits are so much narrower than the tolerance, and why specification limits must never be drawn on it (Module 7 said this once already; Module 16 will say it again).

(d) The means. The individuals here are already close to normal (Anderson-Darling p = 0.38), and the means of five are at least as close (p = 0.88, skewness 0.16). For the flatness data the difference was dramatic. A chart of subgroup means can therefore use normal-based limits with confidence on almost any process; a chart of individual values, and any capability calculation on individual values, cannot, and must check the shape first.

Quiz

Ten questions. Score 70 % or more to mark the module complete on this device.

1. The sample standard deviation divides the sum of squared deviations by n − 1 because
2. Five values have a sum of squared deviations of 10. What is the sample variance?
3. A dataset has mean 7.9, median 7.2 and skewness 1.4. The most likely explanation is
4. Anscombe's four datasets share the same
5. What percentage of a normal distribution lies within ±3 standard deviations of the mean? (Two decimals.)
6. A sample of 50 from a lot that is 2 % defective contains no defectives. This
7. Individual values have standard deviation 3.6. Means of subgroups of 9 will have a standard deviation of about
8. The central limit theorem justifies normal-based limits on
9. "What will this grinder produce next week?" is, in Deming's terms,
10. For the flatness data, the mean of the logarithms converted back gave 7.17 µm, close to the median of 7.2 and below the mean of 7.875. The 7.17 is the
Answer key
  1. b. Deviations from the sample mean are too small on average.
  2. 2.5.
  3. c. A right tail.
  4. d. The summary statistics.
  5. 99.73.
  6. a. About 36 % of the time.
  7. c. 1.2.
  8. b. Subgroup means.
  9. d. Analytic; needs stability.
  10. a. Geometric mean.

Key takeaways

References

All web sources accessed 2026-09-09. Sources marked "secondary" were not read in the original by the course author; the claim is taken from the source shown. Books cited by catalogue record are cited only for their existence, edition and general subject.

  1. NIST/SEMATECH. "1.3.6.6.1. Normal Distribution." e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
  2. NIST/SEMATECH. "1.3.6.6.18. Binomial Distribution." e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm
  3. NIST/SEMATECH. "1.3.6.6.19. Poisson Distribution." e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm
  4. NIST/SEMATECH. "1.3.6.6.9. Lognormal Distribution." e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3669.htm
  5. NIST/SEMATECH. "1.3.6.6.8. Weibull Distribution." e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3668.htm
  6. NIST/SEMATECH. "8.1.6. What are the basic lifetime distribution models used for non-repairable populations?" e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/apr/section1/apr16.htm
  7. Anscombe, F. J. "Graphs in Statistical Analysis." The American Statistician 27(1):17 to 21, 1973. https://www.sjsu.edu/faculty/gerstman/StatPrimer/anscombe1973.pdf (scanned copy; the front page was read, the data were transcribed from [8] and checked against the paper's stated properties)
  8. Wikipedia. "Anscombe's quartet." https://en.wikipedia.org/wiki/Anscombe%27s_quartet (secondary: source of the transcribed values and the stated shared statistics)
  9. R Core Team. "anscombe: Anscombe's Quartet of 'Identical' Simple Linear Regressions." R datasets package documentation. https://stat.ethz.ch/R-manual/R-devel/library/datasets/html/anscombe.html
  10. Deming, W. E. "On Probability as a Basis for Action." The American Statistician 29(4):146 to 152, 1975. https://deming.org/wp-content/uploads/2020/06/On-Probability-As-a-Basis-For-Action-1975.pdf
  11. Montgomery, D. C. Introduction to Statistical Quality Control, 8th ed. Wiley, 2019. https://www.wiley.com/en-us/Introduction+to+Statistical+Quality+Control,+8th+Edition-p-9781119399308 (further reading; no single claim on this page rests on it) (catalogue record; the standard textbook treatment of the material in this module, including the derivation of the n − 1 correction)
  12. Tukey, J. W. Exploratory Data Analysis. Addison-Wesley, 1977. https://www.scirp.org/reference/referencespapers?referenceid=1482121 (catalogue record; origin of the box plot)

Further reading