Algebra 2 · S.ID.A.4

Normal calculations: areas and percentiles

A normal model describes a real measurement in grams or seconds. How do we get from it the share of values in a given range, or the value that cuts off a given share?

You need

The standard normal distributionAlgebra 2 Density curves: area, mean and medianAlgebra 2 PercentilesAlgebra 1

Unlocks

Critical valueStatistics

Where you are 8 in Distributions

Statistics7

Distributions8

▴ 3 above Sampling distributions7 z-scores and the empirical rule The central limit theorem The standard normal distribution

Normal calculations: areas and percentiles

When you reach for it

A normal model N(\mu, \sigma) has been fitted to a real measurement: bag weights, brewing times, fish lengths. Two kinds of question then come up, and they run in opposite directions.

From a value to a share. What share of bags weigh under 1{,}000 g? What share of brews take between 141 and 168 seconds? The range is given in the measurement’s own units, and the answer is an area under the density curve.
From a share back to a value. What weight do only the heaviest 10\% of bags exceed? Where do the middle 50\% of brewing times lie? Now the share is given, and the answer is a percentile in the measurement’s units.

No normal curve has a simple formula for its areas, and no table could be printed for every choice of \mu and \sigma. The standard normal distribution has one table that serves them all, because every normal model turns into it when its values are written as z-scores and the empirical rule. So both kinds of question are answered by passing through z: into it on the way to a share, and out of it on the way back to a value. This lesson takes two days, one for each direction.

The procedure

The table below holds every standard normal area this lesson uses. Each entry is \Phi(z) = P(Z < z), the area to the left of z under N(0, 1), rounded to four places; where it overlaps the table in The standard normal distribution, it is the same table.

z	\Phi(z)	z	\Phi(z)
0.25	0.5987	1.40	0.9192
0.50	0.6915	1.50	0.9332
0.60	0.7257	1.60	0.9452
0.67	0.7486	1.645	0.9500
0.75	0.7734	1.96	0.9750
0.80	0.7881	2.00	0.9772
0.84	0.7995	2.33	0.9901
1.00	0.8413	3.00	0.9987
1.25	0.8944
1.28	0.8997

For a negative z, symmetry gives \Phi(-z) = 1 - \Phi(z). A calculator or spreadsheet with a normal function gives the same areas, and can work with the model’s own \mu and \sigma directly; the reasoning below is what it does inside.

Part 1: from a value to a share

Let X be the measurement, modeled by N(\mu, \sigma). A question about a range of X is a question about an area under its curve.

Draw the curve and shade the region. Mark \mu in the middle and the cut or cuts on the axis, then shade the region the question asks about: left of a cut for “less than”, right of it for “more than”, between two cuts for “between”. The sketch decides which of the three area rules below is needed, and it gives a size check at the end: a region that is clearly less than half the curve cannot have an area of 0.8.

Turn each cut into a z-score. A cut a becomes

z = \frac{a - \mu}{\sigma}.

This is allowed because \displaystyle Z = \frac{X - \mu}{\sigma} has distribution N(0, 1), as The standard normal distribution showed, and because subtracting \mu and dividing by the positive number \sigma keeps every inequality the same way round. So X < a happens exactly when \displaystyle Z < \frac{a - \mu}{\sigma}: the same event, and so the same share. A good sketch carries two rows of numbers under the axis, the measurement and its z-score, as in the figures here.

Read the area under N(0, 1). With the cuts as z-scores, the three rules from The standard normal distribution give the share:

\begin{aligned} P(X < a) &= \Phi\!\left(\tfrac{a - \mu}{\sigma}\right), \\ P(X > a) &= 1 - \Phi\!\left(\tfrac{a - \mu}{\sigma}\right), \\ P(a < X < b) &= \Phi\!\left(\tfrac{b - \mu}{\sigma}\right) \\ &\quad - \Phi\!\left(\tfrac{a - \mu}{\sigma}\right). \end{aligned}

The second holds because the regions left and right of a cut make up the whole curve, with area 1. The third holds because the region left of b is the region left of a together with the strip between them.

Say it in the measurement’s terms. The area is the share of all values in the range. Read as chance, it is the probability that one value chosen at random falls there. Multiplied by a count, it is how many of that many values the model expects in the range. Because a single point has no area, “less than” and “at most” give the same share, and so do “more than” and “at least”.

Part 2: from a share back to a value

Now a share is given and a value is wanted. The same bridge is crossed the other way.

Turn the share into an area to the left. The table lists areas to the left, so every question is put in that form first. “The heaviest 10\%” has 0.10 to its right, so 0.90 to its left. “The middle 50\%” leaves 0.25 in each tail, so its lower cut has 0.25 to its left and its upper cut has 0.75.

Find the z-score with that area. Read the table backwards: find the area, read off its z. If the area lies between two entries, take the nearer. For an area below 0.5 the cut is negative, and symmetry finds it: the cut with area p to its left is the negative of the cut with area p to its right.

Turn the z-score into a value. Solving \displaystyle z = \frac{x - \mu}{\sigma} for x gives

x = \mu + z\sigma.

In words: start at the mean and go z standard deviations, up for a positive z and down for a negative one. The value found has the same share below it under N(\mu, \sigma) as z has under N(0, 1), since the two are the same event written in different units.

Working back to the model itself. The same equation x = \mu + z\sigma links four numbers. A target such as “only 1\% of bags under 1{,}000 g” fixes x and the share, so it fixes z; then the equation tells us the \mu needed for a given \sigma, or the \sigma needed for a given \mu. Two percentiles known at once give two such equations, enough to find both \mu and \sigma.

Worked examples

Example 1

Bags under the label weight

A machine fills bags of rice labeled 1{,}000 g. The fill weights are modeled by N(1005, 4), in grams. What share of bags weigh less than 1{,}000 g? What is the chance that a bag chosen at random is underweight?

Solution

Sketch the curve centered at 1{,}005 and shade the region left of 1{,}000, a cut a little below the center. The region is a left tail, clearly well under half the curve.

The cut as a z-score:

z = \frac{1000 - 1005}{4} = \frac{-5}{4} = -1.25.

So 1{,}000 g lies 1.25 standard deviations below the mean, and the share of bags under it is the standard normal area left of -1.25. The table lists only positive z, so use symmetry:

\begin{aligned} P(X < 1000) &= \Phi(-1.25) = 1 - \Phi(1.25) \\ &\approx 1 - 0.8944 = 0.1056. \end{aligned}

About 0.1056 of bags, a little over 10\%, weigh less than 1{,}000 g, and that is also the chance that one bag picked at random is underweight. The share is small, as the sketch said it would be.

Figure 1Bags under 1{,}000 g. The cut 1{,}000 g sits at z = -1.25, so the shaded share is the standard normal area left of -1.25, about 0.1056.

Example 2

Brews between two times

A coffee machine’s brewing times for a full jug are modeled by N(150, 12), in seconds. What share of brews take between 141 and 168 seconds?

Solution

Shade the strip from 141 to 168. It contains the center, 150, and reaches further to the right than to the left, so it should hold well over half the area.

Both cuts become z-scores:

\begin{aligned} \frac{141 - 150}{12} &= \frac{-9}{12} = -0.75, \\ \frac{168 - 150}{12} &= \frac{18}{12} = 1.5. \end{aligned}

The strip’s area is the area left of the upper cut take away the area left of the lower one. The lower one is negative, so symmetry gives \Phi(-0.75) = 1 - 0.7734 = 0.2266, and

\begin{aligned} P(141 < X < 168) &= \Phi(1.5) - \Phi(-0.75) \\ &\approx 0.9332 - 0.2266 = 0.7066. \end{aligned}

About 70.7\% of brews take between 141 and 168 seconds: well over half, as expected.

Figure 2Brews between 141 and 168 seconds. Both cuts are turned into z-scores, and the strip between them is a difference of two left areas.

Example 3

A share above a cut, as a count

The heights of a maize variety at harvest are modeled by N(180, 15), in centimeters. What share of plants are taller than 210 cm? In a field of 500 plants, about how many does the model expect to be that tall?

Solution

The cut as a z-score is

z = \frac{210 - 180}{15} = \frac{30}{15} = 2.

The share above it is the whole take away the share below:

P(X > 210) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228.

About 2.28\% of plants are taller than 210 cm. A share applies to any number of plants, so in a field of 500 the model expects about

500 \times 0.0228 = 11.4,

that is, about 11 plants. The count is an expectation, not a promise: a real field might have 9 or 14, but the model’s best figure is about 11.

Example 4

Two models, two shapes of answer

Two apple varieties are graded by diameter. Variety A is modeled by N(72, 5) and variety B by N(70, 2.5), both in millimeters. Apples under 65 mm cannot be sold as dessert fruit, and apples over 80 mm are sold as large. For each variety, find the share too small to sell and the share large.

Solution

Each cut must be standardized with its own variety’s mean and standard deviation, because a cut of 65 mm is a different number of standard deviations out in each model.

Variety A. The cuts are \displaystyle \frac{65 - 72}{5} = -1.4 and \displaystyle \frac{80 - 72}{5} = 1.6, so

\begin{aligned} P(A < 65) &= 1 - \Phi(1.4) \\ &\approx 1 - 0.9192 = 0.0808, \\ P(A > 80) &= 1 - \Phi(1.6) \\ &\approx 1 - 0.9452 = 0.0548. \end{aligned}

Variety B. The cuts are \displaystyle \frac{65 - 70}{2.5} = -2 and \displaystyle \frac{80 - 70}{2.5} = 4, so

\begin{aligned} P(B < 65) &= 1 - \Phi(2) \\ &\approx 1 - 0.9772 = 0.0228, \\ P(B > 80) &= 1 - \Phi(4) \\ &\approx 0.00003. \end{aligned}

The value \Phi(4) is beyond the table; a calculator gives 1 - \Phi(4) \approx 0.00003, three in a hundred thousand.

Variety A has the larger mean, yet it has more apples too small to sell: about 8.1\% against 2.3\%. Its larger standard deviation spreads its curve wider in both directions, so it puts more fruit in both tails, and it is also the only one of the two that gives a real share of large apples, about 5.5\%. A mean alone cannot compare two models’ tails; the standard deviation decides how far out each cut really is.

Example 5

The value only the heaviest 10% exceed

For the rice bags of the first example, modeled by N(1005, 4), find the fill weight that only the heaviest 10\% of bags exceed.

Solution

This is the second direction: the share is known and the value is not. The heaviest 10\% lie to the right of the cut, so 0.90 lies to its left, and the cut is the 90th percentile.

Under N(0, 1), the area 0.9000 falls between \Phi(1.28) \approx 0.8997 and \Phi(1.29) \approx 0.9015, nearer the first, so the cut is about z = 1.28, as found in The standard normal distribution. Go 1.28 standard deviations above the mean:

\begin{aligned} x &= \mu + z\sigma = 1005 + 1.28(4) \\ &= 1005 + 5.12 = 1010.12. \end{aligned}

Only the heaviest 10\% of bags weigh more than about 1{,}010.1 g. Check by going forwards: \displaystyle \frac{1010.12 - 1005}{4} = 1.28, and 1 - \Phi(1.28) \approx 0.10.

Figure 3Working backwards. The heaviest 10\% lie beyond the 90th percentile, z \approx 1.28, which is 1{,}005 + 1.28(4) \approx 1{,}010.1 g.

Example 6

The middle half of the brewing times

For the coffee machine modeled by N(150, 12) seconds, find the range that holds the middle 50\% of brewing times.

Solution

The middle 50\% leaves 50\% for the two tails, and the curve is symmetric, so each tail holds 25\%. The upper cut has 0.75 to its left; the table gives \Phi(0.67) \approx 0.7486, the nearest entry, so its z-score is about 0.67. The lower cut has 0.25 to its left, which by symmetry is the mirror image: about z = -0.67.

Turn both into seconds:

\begin{aligned} 150 - 0.67(12) &= 150 - 8.04 = 141.96, \\ 150 + 0.67(12) &= 150 + 8.04 = 158.04. \end{aligned}

The middle half of brews take between about 142 and 158 seconds. These two cuts are the model’s first and third quartiles, so the model’s interquartile range is about 16 seconds.

Example 7

Hitting a target: move the mean or tighten the spread

The rice company of the first example wants only 1\% of bags to weigh under 1{,}000 g.

(a) Keeping the standard deviation at 4 g, what mean fill weight should the machine be set to?

(b) Keeping the mean at 1{,}005 g, what standard deviation would the machine need?

Solution

Either way, 1{,}000 g must become the 1st percentile of the fill weights. The cut with 0.01 to its left is the mirror image of the cut with 0.01 to its right, which is the 99th percentile, z \approx 2.33 (the table has \Phi(2.33) \approx 0.9901). So 1{,}000 g must sit at z \approx -2.33, and the equation x = \mu + z\sigma becomes

1000 = \mu - 2.33\sigma.

(a) With \sigma = 4:

\mu = 1000 + 2.33(4) = 1000 + 9.32 = 1009.32.

The machine should be set to fill about 1{,}009.3 g on average, about 4.3 g more than now.

(b) With \mu = 1005:

\begin{aligned} 2.33\sigma &= 1005 - 1000 = 5, \\ \sigma &= \frac{5}{2.33} \approx 2.15. \end{aligned}

The standard deviation would have to come down from 4 g to about 2.15 g.

Both give the target, by different means: (a) gives away more rice in every bag, while (b) needs a more precise machine. Check (b) forwards: \displaystyle \frac{1000 - 1005}{2.15} \approx -2.33, whose area to the left is about 1 - 0.9901 = 0.0099, about 1\%.

Boundaries

The model must fit. Every share here is a share of the model, and it is a share of the real measurements only as far as the model fits them. A skewed variable, or one with a hard floor near its mean, gives wrong tail areas when treated as normal. Whether a normal model is fair is a separate question, answered by checking the data against it.

Tails far out are where models fail first. A normal curve never reaches the axis, so it gives a small but positive share to any value at all, including impossible ones such as a negative weight. Shares many standard deviations out, like the 0.00003 in the fourth example, describe the model more than the world.

\sigma must be positive. Standardizing divides by \sigma, and keeping the inequality the same way round needs \sigma > 0. A target such as 1000 = \mu - 2.33\sigma with \mu below 1{,}000 would ask for a negative \sigma, which means the target cannot be met by tightening the spread alone.

No share of exactly 0 or 1. Every normal model puts some area beyond every cut, so a target of “no bags under 1{,}000 g” cannot be met by any normal model; only a small share can be.

Table answers are rounded. Cuts read from a table are the nearest listed z, and areas are rounded to four places, so the last digit of an answer can differ from a calculator’s.

Where this goes next

Normal calculations are the working tool for anything modeled as normal. The The central limit theorem says that averages of many independent values are close to normal, so these same calculations answer questions about a sampling distribution, such as how often a sample mean lands far from the population mean. Working backwards from a share to a cut is exactly how a Critical value is found: 1.96 for the middle 95\%, 1.645 for a 90\% middle, and these multiply a standard error to make a margin of error. For raw data rather than a model, a Cumulative frequency graphs and percentiles answers the same two questions: read up from a value to the share below it, or across from a share back to a value.

Exercises

Use the table in the lesson; give shares to four decimal places.

Part 1: shares below and above

The lengths of a cucumber variety are modeled by N(30, 2.5), in centimeters. What share of cucumbers are shorter than 27.5 cm?
For the cucumbers in Exercise 1, what share are longer than 35 cm?
The times a phone takes to charge fully are modeled by N(96, 8), in minutes. What share of charges take more than 90 minutes?
The lengths of trout in a stocked pond are modeled by N(58, 4), in centimeters. What share of trout are under 60 cm?
For the rice bags modeled by N(1005, 4) grams, at about which percentile does a bag of 1{,}008 g sit?

Part 1: shares between, and counts

For the cucumbers modeled by N(30, 2.5), what share are between 28 cm and 33.75 cm long?
For the phone charges modeled by N(96, 8), what share take between 86 and 112 minutes?
For the trout modeled by N(58, 4), what share are between 59 cm and 62 cm long?
A farm picks 2{,}000 of the cucumbers modeled by N(30, 2.5). About how many does the model expect to be longer than 33.75 cm?

Part 1: comparing two models

Tomato variety P has masses modeled by N(120, 10) and variety Q by N(110, 5), in grams. For each, find the share over 125 g and the share under 100 g.
Two discus throwers’ distances are modeled by N(52, 1.5) for Hedda and N(50.5, 3) for Wiebke, in meters. Which thrower is more likely to throw beyond 55 m on a given throw, and by how much?

Part 2: from a share back to a value

For the cucumbers modeled by N(30, 2.5), find the length that only the longest 25\% exceed.
For the phone charges modeled by N(96, 8), find the time below which the quickest 5\% of charges fall.
Find the 20th percentile of the masses of tomato variety Q, modeled by N(110, 5) grams.
Find the range that holds the middle 90\% of the trout lengths modeled by N(58, 4).

Part 2: working back to the model

A dairy fills cartons labeled 500 ml. The fill volumes are normal with standard deviation 3 ml. What mean should the filler be set to so that only 2.5\% of cartons hold less than 500 ml?
A mill cuts steel bars with mean length 250 mm. A customer needs 95\% of bars to lie between 245 mm and 255 mm. If the lengths are normal, what is the largest standard deviation that meets this?
A cyclist’s times on a training loop are modeled by a normal model whose 10th percentile is 41 minutes and whose 90th percentile is 59 minutes. Find the model’s mean and standard deviation.

Reasoning

Explain why the value at a given percentile of N(\mu, \sigma) is \mu + z\sigma, where z is the same percentile of N(0, 1), and why that value lies below \mu for every percentile under the 50th.
In Exercise 11 the thrower with the lower mean is more likely to pass 55 m. Explain how a model with a smaller mean can give a larger share above a high cut.
A rice company says it will set its machine so that no bag at all weighs under 1{,}000 g. Explain why no normal model of the fill weights can do this, and what such a model can promise instead.
Two normal models have the same standard deviation and means d apart. Explain why every percentile of the second is exactly d more than the same percentile of the first.

Answers

About 0.1587. The cut is \displaystyle \frac{27.5 - 30}{2.5} = -1, and \Phi(-1) = 1 - 0.8413 = 0.1587.
About 0.0228. The cut is \displaystyle \frac{35 - 30}{2.5} = 2, and the share above it is 1 - 0.9772 = 0.0228.
About 0.7734. The cut is \displaystyle \frac{90 - 96}{8} = -0.75, and by symmetry P(Z > -0.75) = \Phi(0.75) \approx 0.7734.
About 0.6915. The cut is \displaystyle \frac{60 - 58}{4} = 0.5, and \Phi(0.5) \approx 0.6915.
About the 77th percentile. The z-score is \displaystyle \frac{1008 - 1005}{4} = 0.75, and \Phi(0.75) \approx 0.7734, so about 77\% of bags weigh less.
About 0.7213. The cuts are \displaystyle \frac{28 - 30}{2.5} = -0.8 and \displaystyle \frac{33.75 - 30}{2.5} = 1.5, so \Phi(-0.8) = 1 - 0.7881 = 0.2119, so the share is 0.9332 - 0.2119 = 0.7213.
About 0.8716. The cuts are \displaystyle \frac{86 - 96}{8} = -1.25 and \displaystyle \frac{112 - 96}{8} = 2, so the share is 0.9772 - 0.1056 = 0.8716.
About 0.2426. The cuts are \displaystyle \frac{59 - 58}{4} = 0.25 and \displaystyle \frac{62 - 58}{4} = 1, so the share is 0.8413 - 0.5987 = 0.2426.
About 134 cucumbers. The cut is z = 1.5, the share above it is 1 - 0.9332 = 0.0668, and 2000 \times 0.0668 = 133.6.
Variety P: about 0.3085 over 125 g and 0.0228 under 100 g. Variety Q: about 0.0013 over 125 g and 0.0228 under 100 g. For P the cuts are z = 0.5 and z = -2; for Q they are z = 3 and z = -2, so 1 - 0.9987 = 0.0013. The two varieties share the same small tail below 100 g, though P has far more heavy fruit.
Wiebke, about 0.0668 against 0.0228, a difference of about 0.0440. For Hedda the cut is \displaystyle \frac{55 - 52}{1.5} = 2, giving 1 - 0.9772 = 0.0228; for Wiebke it is \displaystyle \frac{55 - 50.5}{3} = 1.5, giving 1 - 0.9332 = 0.0668.
About 31.7 cm. The cut has 0.75 to its left, so z \approx 0.67, and 30 + 0.67(2.5) = 31.675.
About 82.8 minutes. The cut has 0.05 to its left, so z \approx -1.645, and 96 - 1.645(8) is 96 - 13.16 = 82.84.
About 105.8 g. The 80th percentile of Z is about 0.84, so the 20th is about -0.84, and 110 - 0.84(5) = 110 - 4.2 = 105.8.
From about 51.4 cm to about 64.6 cm. Each tail holds 0.05, so the cuts are z \approx \pm 1.645, and 58 \pm 1.645(4) = 58 \pm 6.58.
About 505.9 ml. The cut 500 ml must have 0.025 to its left, so z \approx -1.96; then 500 = \mu - 1.96(3) gives \mu = 500 + 5.88 = 505.88.
About 2.55 mm. The middle 95\% lies within 1.96 standard deviations of the mean, so 1.96\sigma = 5 and \displaystyle \sigma = \frac{5}{1.96} \approx 2.55. A smaller \sigma puts even more than 95\% inside.
Mean 50 minutes, standard deviation about 7.0 minutes. The two percentiles are mirror images, z \approx \pm 1.28, so the mean is halfway: \displaystyle \frac{41 + 59}{2} = 50. Then 59 = 50 + 1.28\sigma gives \displaystyle \sigma = \frac{9}{1.28} \approx 7.0.
A value x and its z-score \displaystyle z = \frac{x - \mu}{\sigma} describe the same event, since X < x happens exactly when Z < z; so they have the same share below them, and the p percentile of X corresponds to the p percentile of Z. Solving for x gives x = \mu + z\sigma. Below the 50th percentile, \Phi(z) < 0.5 = \Phi(0), and \Phi only increases, so z < 0; with \sigma > 0, z\sigma < 0 and x < \mu.
The share above a cut depends on how many standard deviations above the mean the cut is, not on the distance alone. A larger standard deviation makes each standard deviation longer, so the same cut can be fewer standard deviations out even from a lower mean. For Hedda 55 m is 2 standard deviations up; for Wiebke it is only 1.5, so Wiebke’s tail beyond it is larger.
A normal curve is above the axis everywhere, so for every cut there is some positive area to its left, whatever \mu and \sigma are. The share under 1{,}000 g can be made as small as wanted, by raising the mean or reducing the spread, but never 0. The model can promise a small named share, such as 1\% or 0.1\%, instead.
The p percentile of each model is \mu + z\sigma with the same z (the p percentile of N(0, 1)) and the same \sigma. The only difference is \mu, which differs by d, so the two percentiles differ by exactly d. The second curve is the first slid d along the axis.