Statistics Reference Sheet

01

The normal curveSymbol Glossary

Symbol	Name	What it means
x	—	Any value on your measurement scale (a tree height, an age, a score)
μ	mu	The mean (average) of the data
σ	sigma (lowercase)	The standard deviation, the typical distance from the mean
σ²	—	The variance, the standard deviation squared
Σ	sigma (capital)	"Run this calculation once for each value of the counter, then add the results." An instruction, not a number. The counter (usually n or k) is the only thing that changes between runs. Every other letter stays fixed.
n	—	A counter (0, 1, 2, 3…), or a count of data points
N	—	Size of a whole population
e	—	2.718281828…, the natural growth constant
π	pi	3.14159265…, the circle constant
√	radical	Square root
!	factorial	4! = 4 × 3 × 2 × 1 = 24
z	—	How many standard deviations x sits from the mean
f(x)	—	The height of the curve at x
Φ(z)	Phi (capital)	The area to the left of z, a proportion between 0 and 1

Two traps

Σ and σ are the same Greek letter in different cases doing completely different jobs. Same for φ (height) and Φ (area).

02

The normal curveNormal Distribution Formulas

↑ Contents

The normal distribution has a small family of formulas. Each one answers a different question, and each builds on the one before it:

Formula	What it gives you	Use it when you need
Standard deviation, σ	How spread out the values are	The bell's width, before you can draw anything
z-score, z	Where one value sits, counted in standard deviations	To compare values from different bells, or to use a z-table
The curve, f(x)	The bell itself, and its height at any point	To graph it, or see how crowded values are at one spot (a height is not a probability)
Area to the left, Φ(z)	The probability of landing below a value	"What percent scored under 110?"
Area to the right, 1 − Φ(z)	The probability of landing above a value	"What percent scored over 110?"
Area between, Φ(z_b) − Φ(z_a)	The probability of landing in a range	"What percent scored between 90 and 110?"

Work through them in order below. The formula sits right under the graph, with the next one peeking out beneath it: scroll, swipe, or tap to move to it. The boxes under the formula hold its values, stay put as you move between formulas, and update the graph as you go.

Step 1 of 6 · Measure the spread

$$\sigma = \sqrt{\frac{\sum (x-\mu)^2}{N}}$$

Scroll, swipe, or tap the formula for the next one. It loops, so scrolling up from the first takes you to the last.

Standard deviation (σ)

Mean (μ) Where the center of the bell sits

Slide, type, or tap the graph to move the points. Your values are saved in this browser.

Every bell is set by two numbers: where its center sits (μ) and how spread out it is (σ). Here's what σ is actually doing:

Distance from the mean: x − μ for every value. Some come out negative, some positive.
Square each one: added raw, the positives and negatives cancel to exactly 0 every time, which tells you nothing. Squaring makes every distance count. Squared distances are also what add cleanly when you combine independent values, which is where the √n rule comes from.
Average the squares: dividing by N gives the variance, in squared units.
Square root: brings it back to the original units. That's σ.

Coin check: every flip lands exactly 1 away from its mean of 0, so σ = √(100 ÷ 100) = 1.

Your numbers

x is the raw measurement: an IQ of 110, a coin total of +6. z is the same point, measured in standard deviations from the mean.

z = 1 is one standard deviation above the mean.
z = −2 is two standard deviations below.
z = 0 is the mean itself.

Measured in z, every bell curve is exactly the same shape. That's why one z-table works for every problem. Switch the graph's scale between z and raw value to see both rulers.

Your numbers

This is the line you type into a graphing calculator. Read it in two pieces:

The shape, e^−z²/2: the exponent is just −z²/2, because (x − μ)² ÷ σ² is z². At the mean, z = 0 and e⁰ = 1, the peak. Moving away, z² grows and the height collapses fast: 0.61 at 1 standard deviation, 0.14 at 2, 0.011 at 3. Squaring is what makes both sides match.
The scale, 1 ÷ (σ√(2π)): makes the total area exactly 1, or 100%. The √(2π) is the area under e^−z²/2. The σ is there because a wider bell has to be shorter to hold the same area.

Your numbers

The orange bullseye is point A, with its pin sitting right on the curve. Plug that one x into the curve and you get the height there. Same thing another way: find its z, read the standard bell φ(z), then divide by σ.

A height is not a probability by itself. It's probability per unit of x, so a thin strip has a probability of about height × width.

Coin check: the height at a total of 0 is 0.03989. Totals move in steps of 2, so the chance of landing exactly on 0 is about 0.03989 × 2 = 0.0798. The real chance of exactly 50 heads is 7.96%.

This height is sometimes called the density. It isn't a t-value. The t in the next step is only a placeholder, and the t-value you'll meet later in the course is a separate tool.

Your numbers

To turn heights into a probability, add up thin strips (height × width) from the far left up to your point.

½ is the entire left half, everything below the mean.
The integral sweeps a slider t from the center out to your z, adding each thin strip's area as it goes. t is only the slider's position, and it's gone once the sweep finishes.
e^−t²/2 has no simple antiderivative, so the sweep is done with the series under "The area formula, worked by hand" below, a z-table, or a calculator.

Need the area to the right? Switch the shading to the right. The bell is symmetrical, so fold it along the center line and the right tail of z lands exactly on the left tail of −z. That makes the right side a left-side problem again: area right of z = 1 − Φ(z) = Φ(−z). The formula itself only ever finds area to the left.

Your numbers

The formula only finds area to the left, so the area between two points takes two of those areas and subtracts one from the other.

Solid blue: everything left of the higher point.
Striped: everything left of the lower point. It sits inside the blue, so it gets subtracted.
The solid blue left over is the slice between the two points.

Try A = −1 and B = 1 and you get 68.3%. ±2 gives 95.4% and ±3 gives 99.7%. Those hold for every bell curve, because in z every bell is the same shape. That's the empirical rule.

Your numbers

Graph settings

Bottom scale

A z value of 1 is 1 standard deviation from the mean. On this scale every bell looks the same; changing σ only rescales the height.

Zoom: ±4 SD

Type into Desmos or any graphing calculator

Coin flips: where the 10 comes from

Heads counts +1 and tails counts −1. One flip has a mean of 0 and always lands exactly 1 away from it, so one flip's standard deviation is 1.

One sample of 100 flips

Each bar is one flip's deviation: +1 or −1, always exactly 1 from the mean. The line is the running total, what's left over after heads and tails cancel.

Squared deviations: 100 flips × 1 = 100. Standard deviation of one flip = √(100 ÷ 100) = 1.

Many samples: how much the total wobbles

Bars are your samples' totals. The line is the exact count you'd expect from Pascal's triangle, which is the bell with σ = 1 × √100 = 10.

Dice: why a pair makes a bell

What to notice

The area formula, worked by hand

First convert your value to z:

$$z = \frac{x - \mu}{\sigma}$$

Then, general form:

$$\Phi(z) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \sum_{n=0,1,2,\ldots} \frac{(-1)^n \, z^{2n+1}}{2^n \cdot n! \cdot (2n+1)}$$

Written out, with the structure intact in every slot:

$$\Phi(z) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \times \left[ \frac{z^1}{2^0 \cdot 0! \cdot 1} - \frac{z^3}{2^1 \cdot 1! \cdot 3} + \frac{z^5}{2^2 \cdot 2! \cdot 5} - \frac{z^7}{2^3 \cdot 3! \cdot 7} + \frac{z^9}{2^4 \cdot 4! \cdot 9} - \cdots \right]$$

What each piece does

Piece	Job
$(-1)^n$	Produces the alternating + − + − signs
$z^{2n+1}$	Gives the odd exponents 1, 3, 5, 7, 9
$2^n$	Comes from the ½ in the exponent raised to the nth power
$n!$	Comes from the exponential series
$(2n+1)$	Comes from integrating; always matches the exponent above it

Nothing in this formula is a rounded decimal. It gives the proportion of all values falling below x.

03

The normal curveWhere Each Value Comes From

↑ Contents

Value	How you get it
x	Given in the problem. The cutoff you care about.
μ	Add every data point, divide by how many there are. Often given directly.
σ	Computed from the data (see Section 8). Often given directly.
z	Computed: (x − μ) ÷ σ. Never given.
1/√(2π)	An exact constant. Never changes. Decimal value below.
1/2	Exact. The entire left half of the curve.
2·1·3, 4·2·5, 8·6·7, 16·24·9	Exact. Built from 2ⁿ · n! · (2n+1). See Section 5.

Decimal value of 1/√(2π)

Use this only when you reach for a calculator. The formula itself keeps the exact form.

π = 3.1415927
2π = 6.2831853
√6.2831853 = 2.5066283
1 ÷ 2.5066283 = 0.3989423

What it does

Scales the curve so its total area equals exactly 1, which is what makes the output a valid percentage. Rounding it to 0.4 throws the 4th decimal off.

Deriving z, worked

Data: μ = 251, σ = 50, cutoff x = 187

z = (187 − 251) ÷ 50 = −64 ÷ 50 = −1.28

Negative means x is below the mean. Positive means above.

04

The normal curveExplanation of Each Part

↑ Contents

The curve formula, read from the inside out

Part	What it does
$(x - \mu)$	How far your value sits from the mean
$(x - \mu)^2$	Squared: removes the sign so left and right are symmetric, and makes distance punishing rather than linear
$\div\, 2\sigma^2$	Converts raw distance into distance measured in standard deviations
minus sign	Makes the exponent most negative when you are farthest out
$e^{(\ldots)}$	Exponential decay: maximum at the mean (e⁰ = 1), dropping fast as you move away
$\frac{1}{\sigma\sqrt{2\pi}}$	Scaling constant so total area equals 1

Why π appears

To find the total area you square it, which turns it into a 3D volume whose height depends on x² + y², the distance from the origin squared. That makes the surface circular, and circles bring π.

The area formula, piece by piece

Part	What it does
$z$	First term: the rough area, before corrections
$-\,z^3/(2\cdot1\cdot3)$	First correction, pulls the estimate down
$+\,z^5/(4\cdot2\cdot5)$	Second correction, pushes it back up
$-\,z^7/(8\cdot6\cdot7)$	Third correction, smaller
$+\,z^9/(16\cdot24\cdot9)$	Fourth correction, smaller still
$\times\, 1/\sqrt{2\pi}$	Scales so the full curve totals 1
$+\,1/2$	Adds the entire left half of the curve

The terms alternate sign and shrink fast, so the running total closes in from both sides.

05

The normal curveWhere the Denominators Come From

↑ Contents

Term rule (n = which term you're on, starting at 0):

$$\frac{z^{2n+1}}{2^n \cdot n! \cdot (2n+1)}$$

n	2ⁿ	n!	(2n+1)	Denominator
0	1	1	1	1
1	2	1	3	6
2	4	2	5	40
3	8	6	7	336
4	16	24	9	3456
5	32	120	11	42240
6	64	720	13	599040

Why three factors, using n = 3 as the example

The exponential series gives term u³/3!, where u = −t²/2
Cubing the ½ produces 2³ = 8
The factorial 3! = 6
Integrating t⁶ gives t⁷/7, producing the 7
8 × 6 × 7 = 336

Why exponents jump by 2

The t in the exponent is squared, so powers of u give t⁰, t², t⁴, t⁶… Integration then adds 1 to each, making them odd: z¹, z³, z⁵, z⁷…

Shortcut: chain from the previous term

No powers or factorials needed:

$$\text{next} = \text{previous} \times \left(\frac{-z^2}{2n}\right) \times \frac{2n-1}{2n+1}$$

Accuracy: 5 terms is enough up to about |z| = 1. Use 7 terms out to |z| = 1.5.

06

The normal curveWorked Example: Full Arithmetic

↑ Contents

Problem: μ = 251, σ = 50. What proportion falls between 187 and 229?

Value 187

Convert: z = (187 − 251) ÷ 50 = −1.28

Substitute 1.28 into every z slot:

$$\Phi = \tfrac{1}{2} + \tfrac{1}{\sqrt{2\pi}} \times \left[ 1.28 - \frac{1.28^3}{2\cdot1\cdot3} + \frac{1.28^5}{4\cdot2\cdot5} - \frac{1.28^7}{8\cdot6\cdot7} + \frac{1.28^9}{16\cdot24\cdot9} \right]$$

Powers (each is the prior × 1.6384, which is 1.28²):

Power	Value
1.28¹	1.2800000
1.28³	2.0971520
1.28⁵	3.4359738
1.28⁷	5.6294995
1.28⁹	9.2233720
1.28¹¹	15.1115727

Divide each:

Division	Term
1.2800000 ÷ 1	+1.2800000
2.0971520 ÷ 6	−0.3495253
3.4359738 ÷ 40	+0.0858993
5.6294995 ÷ 336	−0.0167545
9.2233720 ÷ 3456	+0.0026688
15.1115727 ÷ 42240	−0.0003578
Add	1.0019305

× 1/√(2π) = 1.0019305 × 0.3989423 = 0.3997125
+ 1/2 = 0.8997
Flip (z was negative): 1 − 0.8997 = Φ(−1.28) = 0.1003

Value 229

Convert: z = (229 − 251) ÷ 50 = −0.44

Powers (each is the prior × 0.1936):

Power	Value
0.44¹	0.4400000
0.44³	0.0851840
0.44⁵	0.0164916
0.44⁷	0.0031928

Divide each:

Division	Term
0.4400000 ÷ 1	+0.4400000
0.0851840 ÷ 6	−0.0141973
0.0164916 ÷ 40	+0.0004123
0.0031928 ÷ 336	−0.0000095
Add	0.4262055

× 1/√(2π) = 0.4262055 × 0.3989423 = 0.1700315
+ 1/2 = 0.6700
Flip: 1 − 0.6700 = Φ(−0.44) = 0.3300

Subtract

0.3300 − 0.1003 = 0.2297

Answer

About 23%. With 500 trees, that's 0.2297 × 500 ≈ 115 trees.

07

The normal curveThe Three Question Types

↑ Contents

Asked	Do this
Below a value	Φ(z)
Above a value ("at least")	1 − Φ(z)
Between two values	Φ(z_larger) − Φ(z_smaller)
Count instead of percent	Multiply the result by your total

Negative z: Φ(−z) = 1 − Φ(z)

Why "between" takes two steps

Everything below the upper bound includes everything below the lower bound, which you don't want. Subtracting peels it off. Same logic as pages read = end page − start page.

Shortcut when both values are on the same side of the mean

Subtract the brackets directly and skip the 1/2 and the flipping entirely.

bracket(1.28) = 1.0019718 (run to 7 terms)
bracket(0.44) = 0.4262055
difference × 1/√(2π) = 0.2297

08

Spread and shape · UpdatedStandard Deviation Formulas

↑ Contents

Population (you have every value):

$$\sigma = \sqrt{\frac{\Sigma(x - \mu)^2}{N}}$$

Sample (you have a subset, the usual case):

$$s = \sqrt{\frac{\Sigma(x - \bar{x})^2}{n - 1}}$$

In plain English: the square root of the average of the squared deviations from the mean.

Steps

Find the mean
Subtract the mean from each value
Square each difference
Add them all up
Divide by N (population) or n − 1 (sample). This is the variance.
Square root. This is the standard deviation.

Why n − 1 for a sample

The sample mean is computed from your own data, so it sits closer to your points than the true mean does. Σ(x − x̄)² is mathematically the smallest possible sum, so dividing by n understates the spread every time. Dividing by n − 1 corrects that bias.

From one person to a group

Start with the regular standard deviation formula, then simplify from what you already know to get the deviation between groups. All you need is σ and how many people are in each group.

Start with the regular formula.
$$\sigma = \sqrt{\frac{\Sigma(x - \mu)^2}{N}}$$
Simplify the Σ. For a population of N, you're adding the squared deviation N times. Adding something N times is the same as multiplying that same thing by N, and that's where the simplification comes from. Each squared deviation averages out to σ², so the Σ becomes N × σ².
$$\sigma = \sqrt{\frac{N\sigma^2}{N}}$$
The ÷ N stays. That's separate. It lands on the × N, and the two cancel.
$$\sigma = \sqrt{\sigma^2} = \sigma$$
So σ is how far one person typically sits from the mean.
Now a group of n people. Add their values together. Same simplification: adding the variance n times is the same as multiplying it by n.
$$\text{SD of the total} = \sqrt{n\sigma^2}$$
Turn the total into an average. Divide the total by n. That divides the SD by n, so it divides the variance by n², because variance is squared.
$$SE = \sqrt{\frac{n\sigma^2}{n^2}} = \sqrt{\frac{\sigma^2}{n}} = \frac{\sigma}{\sqrt{n}}$$
This is the standard error, the deviation between groups: how much one group's average wanders from the next group's.

Check it: σ = 10, n = 100

√(100 × 10² ÷ 100²) = √(10,000 ÷ 10,000) = √1 = 1. One person wanders by 10, but the average of 100 people wanders by only 1.

09

Spread and shapeThe Empirical Rule

↑ Contents

Applies only when data is bell-shaped / approximately normal. Memorize these.

Within	Percentage
1 standard deviation	68%
2 standard deviations	95%
3 standard deviations	99.7%

Band breakdown (one side of the mean)

Region	%
Mean to 1 SD	34%
1 SD to 2 SD	13.5%
2 SD to 3 SD	2.35%
Beyond 3 SD	0.15%
Half the curve	50%

Those four add to 50%, half the curve. Everything else is built from them.

Common derived values: above 1 SD = 16%. Above 2 SD = 2.5%. Below −1 SD = 16%.

If the data is not bell-shaped

Use Chebyshev's inequality, which works for any distribution: at least 75% within 2 SD, at least 89% within 3 SD.

10

Spread and shape · UpdatedQuartiles, IQR & Boxplots

↑ Contents

Q1 = the median of the lower half (a single number, not a range)
Q2 = the median
Q3 = the median of the upper half
IQR = Q3 − Q1, the width of the box, which holds the middle 50% of the data
Range = max − min (nothing added)

Outlier fences

Lower fence = Q1 − 1.5 × IQR
Upper fence = Q3 + 1.5 × IQR

With n = 10: split 5 and 5. Q1 is the 3rd value, Q3 is the 8th. No data point is shared between halves.

Added Oct 20

A boxplot is built on medians, never the mean

It uses the five-number summary: min, Q1, median, Q3, max. Every one of them is a position in the sorted list. The mean and SD never appear. That's the point: one extreme value can drag a mean around, but it can't move a median.

Splitting into halves

n	Halves	Median	Q1 / Q3
Even (n = 26)	13 and 13	Average of the 13th and 14th values	7th value of each half
Odd (n = 11)	5 and 5, median left out of both	The 6th value	3rd value of each half

Fences vs. min and max

Min and max are the actual smallest and largest data values. Nothing is calculated.
1.5 × IQR sets the fences. The fences are cutoffs, not part of the plot.
A data point outside a fence is an outlier, drawn as a dot or ×.
The whisker stops at the most extreme real data value still inside the fence, never at the fence number itself.
No points outside the fences → the whiskers run all the way to the min and max.

Why 1.5

A convention from John Tukey, who invented the boxplot, not a law of math. On bell-shaped data the fences land at about ±2.7 SD, so only about 0.7% of normal data gets flagged. "Outlier" means rarer than roughly 1 in 140.

Worked example: noon temperatures of 26 cities (°F)

37, 38, 44, 50, 56, 57, 59, 61, 62, 62, 65, 65, 66|67, 69, 69, 70, 70, 71, 71, 75, 79, 85, 85, 87, 93

Step	Work	Result
Median	(13th + 14th) ÷ 2 = (66 + 67) ÷ 2	66.5
Q1	7th value of the lower 13	59
Q3	7th value of the upper 13	71
IQR	71 − 59	12
1.5 × IQR	1.5 × 12	18
Lower fence	59 − 18	41
Upper fence	71 + 18	89
Outliers	below 41 or above 89	37, 38, 93
Whiskers	last real values inside the fences	44 and 87

Trap: off-by-one

The middle of 13 values is position 7 (6 below it, 6 above). Counting to the 8th gives Q1 = 61 instead of 59, and every fence and outlier after it shifts.

11

Counting · UpdatedBinomial: Counting Successes

↑ Contents

Use when you have a fixed number of yes/no trials, each with the same chance of success, and you want the probability of getting a certain number of successes.

The letters

Symbol	Meaning
n	Number of trials. Fixed, never changes.
k	Number of successes you're asking about. This is what Σ varies.
p	Chance of success on one trial
1 − p	Chance of failure on one trial
n − k	Number of failures

The formula

Exactly k successes:

$$P(k) = \frac{n!}{k!\,(n-k)!} \times p^{k} \times (1-p)^{n-k}$$

k or more successes: run it for each k and add.

$$P(X \ge k) = \sum_{k} \frac{n!}{k!\,(n-k)!} \times p^{k} \times (1-p)^{n-k}$$

Three parts, all multiplied (never added inside a term):

p^k: the successes
(1−p)^n−k: the failures. Every trial must be accounted for.
n!/(k!(n−k)!): how many different patterns give that count

My counting method (faster than factorials)

Count the smaller group, successes or failures, whichever is fewer. Multiply down from n, as many factors as items you're placing, then divide by that many factorial.

Items to place	Count
1	n
2	n(n−1) ÷ 2!
3	n(n−1)(n−2) ÷ 3!
4	n(n−1)(n−2)(n−3) ÷ 4!
5	n(n−1)(n−2)(n−3)(n−4) ÷ 5!

Why divide by the factorial

Placing items one at a time finds each finished pattern once per ordering of those items. r items have r! orderings, so every pattern is counted exactly r! times. Uniform overcount, so one clean division fixes it.

Example, 3 failures in 5 trials: 5 × 4 × 3 = 60 pickings ÷ 3! = 6 → 10 patterns

Shortcut check: 4 failures out of 6 = 2 successes out of 6. Count the 2: 6 × 5 ÷ 2 = 15. Much faster than 6 × 5 × 4 × 3 ÷ 24.

Worked example

A machine makes good parts 70% of the time. In 6 parts, what is P(4 or more good)? n = 6, p = 0.7, and k runs 4, 5, 6. n stays 6 in every line.

k	count	p^k	(1−p)^n−k	term
4	6×5÷2! = 15	0.7⁴ = 0.240100	0.3² = 0.09	0.324135
5	6÷1! = 6	0.7⁵ = 0.168070	0.3¹ = 0.30	0.302526
6	6÷0! = 1	0.7⁶ = 0.117649	0.3⁰ = 1.00	0.117649
Total				0.7443

About 74.4%.

Common mistakes

Forgetting the failures. 3 trials, exactly 2 successes, without the failure term: 3 × 0.64² = 1.2288. A probability over 1 is impossible. The failure term pulls it back to 0.4424.
Adding inside a term. p^k and (1−p)^n−k multiply. Addition only happens between the k cases.
Counting outcomes instead of weighting them. With 11 trials, "10 or more" has 12 favorable outcomes out of 2¹¹ = 2048. But 12/2048 = 0.0059 is only correct if p = 0.5. At p = 0.64 the real answer is 0.0530, because success-heavy sequences are far more likely than failure-heavy ones.

Sanity check

Add P(0) through P(n). It must equal exactly 1.00, because one of those outcomes has to happen. It only works if every term carries both successes and failures.

Added Oct 20

Mean and standard deviation of a binomial

$$\mu = n \cdot p \qquad\qquad \sigma = \sqrt{n \cdot p \cdot (1-p)}$$

Where the SD comes from

One trial has variance p(1 − p). The trials are independent, so their variances add: n trials give n · p(1 − p). Square root that for the SD, same last step as Section 8.

Example: the pistons

A machine makes a defective piston 1% of the time. Sample 80.

Mean = 80 × 0.01 = 0.8 defective
SD = √(80 × 0.01 × 0.99) = √0.792 = 0.890

The SD describes how much the count bounces between batches. Run the sample over and over and the counts scatter (0, 1, 0, 2, 1…). Their average is 0.8 and their typical distance from 0.8 is 0.890.

Defective in one batch	Chance
0	44.75%
1	36.16%
2	14.43%
3	3.79%

Mean is not the most likely count

The mean is 0.8, but the single most likely count is 0 (the mode). When a problem says "the mean of the relevant random variable" and "do not round," it wants 0.8, not a whole-number guess.

Absolute vs. relative uncertainty

SD ÷ mean = 0.890 ÷ 0.8 ≈ 111% is the coefficient of variation, the same uncertainty stated relative to the estimate. When a question names "the standard deviation," it wants the absolute version: 0.890.

Try it: win probability

A team wins 64% of its games. Over 12 games, what is the chance it wins exactly 10? Ten or more?

n = 12, p = 0.64, 1 − p = 0.36, k runs 10, 11, 12

Show the work

k	count (smaller group)	p^k	(1−p)^n−k	term
10	2 losses: 12×11÷2! = 66	0.011529	0.36² = 0.1296	0.098616
11	1 loss: 12	0.007379	0.36¹ = 0.36	0.031876
12	0 losses: 1	0.004722	0.36⁰ = 1	0.004722
10 or more				0.135215

Answer

Exactly 10: 0.0986 (about 9.9%). Ten or more: 0.1352 (about 13.5%).

Cross-check with the mean and SD: expected wins = 12 × 0.64 = 7.68, SD = √(12 × 0.64 × 0.36) = 1.66. Ten wins sits about 1.4 SD above expected, which is why it happens only about 1 time in 7.

12

Tables · NewExpected Frequencies

↑ Contents

A contingency table sorts every person (or sample) by two variables at once. The expected frequency of a cell is the count you'd see if the two variables had nothing to do with each other.

$$E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$$

Why it works

"Independent" means knowing one variable tells you nothing about the other. So each row's share of the grand total should hold inside every column. If 60% of all samples are acidic, then 60% of the sandy ones should be acidic too: 0.60 × the sandy column total. That's row total ÷ grand total × column total, the same formula.

Example: 200 soil samples

Observed counts, by pH and texture:

	Sandy	Loam	Clay	Total
Acidic	38	50	32	120
Neutral / alkaline	12	40	28	80
Total	50	90	60	200

Expected counts, if pH and texture were independent:

	Sandy	Loam	Clay	Total
Acidic	120×50÷200 = 30	120×90÷200 = 54	120×60÷200 = 36	120
Neutral / alkaline	80×50÷200 = 20	80×90÷200 = 36	80×60÷200 = 24	80
Total	50	90	60	200

Check: the expected table keeps the exact same row and column totals as the observed one.
Don't round to whole numbers. Expected counts are averages, so 12.6 is a perfectly good expected count. Keep at least 2 decimals.
The observed values inside the table are never used. Only the totals are.

What it's for: the chi-square test

Compare observed (O) to expected (E) in every cell: $\chi^2 = \Sigma \frac{(O-E)^2}{E}$. The bigger it is, the less the data looks independent. Here: 2.13 + 0.30 + 0.44 + 3.20 + 0.44 + 0.67 ≈ 7.19, with degrees of freedom = (rows − 1) × (columns − 1) = 1 × 2 = 2.

13

The big pictureWhy the Bell Curve Exists

↑ Contents

The Central Limit Theorem. Take values from any distribution (skewed, flat, lumpy) and add many of them together independently. The distribution of that sum converges to a normal curve. The original shape is erased; only its mean and variance survive.

A single coin flip is uniform (0 or 1, equally likely), the least bell-like thing possible. But 100 flips summed is a bell. The shape wasn't in the coin; it emerged from adding.

Why

For a sum to land at an extreme, every term has to cooperate, which is rare. For it to land in the middle, vastly more combinations get there. Counting those combinations produces the hump.

Where it doesn't apply

Income isn't a sum of small independent effects, so it's skewed, not normal. Always check that the problem says "bell-shaped" or "approximately normal" before using the empirical rule.

Why there's no simple area formula: e^−t²/2 has no elementary antiderivative. Liouville proved this in the 1830s. That's why z-tables exist and why the area formula is a series instead of a clean expression.

Piece	Job
\((-1)^n\)	Produces the alternating + − + − signs
\(z^{2n+1}\)	Gives the odd exponents 1, 3, 5, 7, 9
\(2^n\)	Comes from the ½ in the exponent raised to the nth power
\(n!\)	Comes from the exponential series
\((2n+1)\)	Comes from integrating; always matches the exponent above it

Part	What it does
\((x - \mu)\)	How far your value sits from the mean
\((x - \mu)^2\)	Squared: removes the sign so left and right are symmetric, and makes distance punishing rather than linear
\(\div\, 2\sigma^2\)	Converts raw distance into distance measured in standard deviations
minus sign	Makes the exponent most negative when you are farthest out
\(e^{(\ldots)}\)	Exponential decay: maximum at the mean (e⁰ = 1), dropping fast as you move away
\(\frac{1}{\sigma\sqrt{2\pi}}\)	Scaling constant so total area equals 1

Part	What it does
\(z\)	First term: the rough area, before corrections
\(-\,z^3/(2\cdot1\cdot3)\)	First correction, pulls the estimate down
\(+\,z^5/(4\cdot2\cdot5)\)	Second correction, pushes it back up
\(-\,z^7/(8\cdot6\cdot7)\)	Third correction, smaller
\(+\,z^9/(16\cdot24\cdot9)\)	Fourth correction, smaller still
\(\times\, 1/\sqrt{2\pi}\)	Scales so the full curve totals 1
\(+\,1/2\)	Adds the entire left half of the curve