Distributions

Stage 16 of 23 Strand 2 of 5 16 lessons

16 illustrated lessons, each teaching the why before the how.

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Random Variables

A number attached to every outcome.

A random variable attaches a number to every outcome of an experiment

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123456

Roll a die and X is the score. Each value has its own probability.

012

Discrete means you can list the values: 0, 1, or 2 heads, with nothing in between.

xy

Continuous means any value in a range, like a height, so its distribution is a curve, not bars.

Now you

012

Is the time a bus takes to arrive discrete or continuous?

012

Is the number of heads in 10 tosses discrete or continuous?

Probability Distributions

Every value with its chance, totaling one.

A probability distribution lists every value with its chance, and they must total one

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ABC

A distribution lists each outcome and how likely that outcome is.

2/10 + 3/10 + 5/10 = 1

The probabilities must total 1, so a missing one is 1 minus the rest.

ABCD

D is blank, so subtract the rest: 1 − 2/10 − 3/10 − 4/10 = 1/10.

Now you

ABC

Outcomes A, B and C have probabilities 4/10, 1/10 and 2/10. What is D?

ABC

Outcomes A, B and C have probabilities 4/10, 2/10 and 1/10. What is D?

Expected Value

The long-run average, weighted by chance.

The expected value is the long-run average, found by weighting each value by its chance

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01050

The prize is 0, 10 or 50, with probabilities 1/2, 2/5 and 1/10.

each value weighted by its own chance
0 + 4 + 5E(X) = 9

Weight each value by its probability and add: 0 + 4 + 5 = 9. That is E(X).

0102030405001050E(X) = 9

E(X) = 9 lands between the prizes, just short of 10. Nobody ever wins 9.

stake 9, so the gain is X − 9E(X − 9) = 9 − 9 = 0neither side profits in the long runa fair game

Charge 9 to play and the expected gain is 0. A game with E = 0 is called fair.

Now you

014

X is 0 with probability 1/2 and 14 with probability 1/2. What is E(X)?

06

X is 0 with probability 1/2 and 6 with probability 1/2. What is E(X)?

Variance

Average squared distance from the mean.

Variance measures spread by averaging the squared distance from the mean

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456

Two variables can have the same mean and still be spread very differently.

159

This one also has mean 5, but its values spread right across the range.

012345678910(−4)² = 164² = 16

Square each distance from the mean so they cannot cancel, then average.

X = 4 or 6, each with chance
each value one step from the mean
mean 5, distances −1 and +1square each distance, then average
Var

X is 4 or 6 with probability 1/2 each: mean 5, squared distances 1, so Var(X) = 1.

Now you

715

X is 7 or 15, each with probability 1/2. What is Var(X)?

511

X is 5 or 11, each with probability 1/2. What is Var(X)?

Transforming a Random Variable

Adding moves the mean and leaves the spread.

Adding a constant shifts the mean but leaves the spread alone

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123

Add 2 to every value and the whole distribution slides 2 to the right.

xy

The mean slides with it: E(X + 2) = E(X) + 2.

double every valuethe mean doubles tooevery distance from the mean doublesvariance averages the squares
every squared distance

Doubling doubles every distance from the mean, and squaring turns that 2 into 4.

xy

A shift leaves the spread alone; doubling quadruples it: Var(2X) = 4Var(X).

Now you

E(3X + 5)what does this come to?

E(X) = 3. What is E(3X + 5)?

Var(2X + 9)what does this come to?

Var(X) = 6. What is Var(2X + 9)?

Bernoulli Trials

Two outcomes, fixed chance, no memory.

A Bernoulli trial has two outcomes and a fixed probability, and each trial is independent of the last

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psuccess1 − pfailure

One trial has two outcomes. Call them success and failure.

pSpS1 − pF1 − pFpS1 − pF

Repeat the trial and p never changes. Each trial is independent of the last.

fixed n, fixed p, independent trialsall three hold, or it is not binomialthe binomial model

Those three assumptions are what every binomial answer rests on.

Now you

pS1 − pF

Are these Bernoulli trials: rolling a die 10 times and counting sixes?

pS1 − pF

Are these Bernoulli trials: tossing the same coin 20 times?

The Binomial Distribution

Counting successes in n independent trials.

The binomial counts successes in a fixed number of independent trials

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012345

A binomial counts the successes in n independent trials, each with probability p.

X ~ B(n, p)the tilde reads “is distributed as”

X ~ B(n, p) names that count: n trials, each with probability p.

P(X = r)r successes
and the rest failures
times the ways to arrange them

Choose which r trials succeed and multiply p for each success, 1 − p for each failure.

01234

The nCr factor is why the bars follow a row of Pascal’s triangle.

Now you

4C22 successes
and 2 failures

X ~ B(4, 1/2). What is P(X = 2)?

5C2what does this come to?

In how many ways can exactly 2 of 5 trials succeed?

Binomial Mean and Variance

np, and np times one minus p.

A binomial has mean np and variance np(1 − p)

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012345

Take 5 trials with probability 1/2. The mean is np = 2.5, right under the peak.

Var(X) = np(1 − p)p(1 − p) is largest at p = 1/2

The variance is np(1 − p): for these five trials, 5 × 1/2 × 1/2 = 1.25.

012345

Take p down to 1/5 and the bars crowd to the left: np(1 − p) falls to 0.8.

n

At p = 1 every trial succeeds: one bar holds everything, and np(1 − p) is 0.

Now you

what does this come to?

X ~ B(30, 1/5). What is the mean?

what does this come to?

X ~ B(70, 1/10). What is the mean?

Modeling with a Binomial

Only as good as its quiet assumptions.

A binomial model is only as good as the assumptions behind it

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0.02faulty0.98fine

Two percent of parts are faulty. In a box of 50, how many fail?

01234

Model the count as a binomial: the expected number faulty is np = 1 per box.

if faults clusterone fault makes the next more likelynot independent

If a bad batch makes faults come in clusters, the trials are not independent and the model no longer holds.

Now you

pS1 − pF

Can a binomial model be used if the sample is drawn without replacement from a small group?

pS1 − pF

Can a binomial model be used if p changes partway through?

Continuous Variables

Probability becomes area under a curve.

For a continuous variable probability is area, so a single exact value has none

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xy

A continuous variable has a curve instead of bars. The curve is called a density.

xy

The whole area underneath is 1, exactly as the bars summed to 1.

xy

A probability is the area over a range of values, so one exact value has none.

Now you

xy

For a continuous variable, what is P(X = 4) exactly?

xy

For a continuous variable, what is P(X = 2) exactly?

The Normal Distribution

The shape a binomial settles into.

The normal curve is symmetric about its mean and its spread is set by sigma

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Increase the number of trials in a binomial and the bars settle into one shape.

xy

That shape is the normal curve: symmetric, with one peak at the mean.

xy

About 68 percent lies within one standard deviation, sigma, of the mean, either side.

xy

Widen the band to two sigmas either side and it holds about 95 percent.

xy

Three sigmas hold about 99.7 percent, so almost nothing is left outside.

Now you

xy

Roughly what percent lies within 3 standard deviations of the mean?

xy

Roughly what percent lies within 1 standard deviation of the mean?

Standardizing

How many sigmas from the mean a value sits.

Standardizing rewrites any value as the number of standard deviations from the mean

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xy

Subtract the mean from x to center it, then divide by sigma: z = (x − μ)/σ.

xy

Now every normal curve becomes the same standard normal curve, with mean 0 and sigma 1.

xy

A z of 1.5 means one and a half standard deviations above the mean.

Now you

xy

The mean is 35 and sigma is 10. What is the z-score of 55?

xy

The mean is 54 and sigma is 10. What is the z-score of 34?

Reading the Z-Table

Turn an area under the curve into a z-value.

A z-table turns an area under the normal curve into the z-value that cuts it off

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xy

A z-table gives the area to the left of a z-value — everything below it.

area to the leftz = 1.280.8997z = 1.6450.9500z = 1.960.9750z = 2.330.9901z = 2.580.9951

Each row pairs a z with the area beneath the curve to its left.

xy

Want 5 percent in the upper tail? Then 95 percent lies to the left. Look up 0.9500.

look upz5% tail0.95001.6452.5% tail0.97501.961% tail0.99002.33

That is where 1.645 and 1.96 come from: they are read from the table, not memorized.

xy

The table lists positive z only; by symmetry −1.645 cuts the same 5 percent below.

Now you

area to the leftz = 1.280.8997z = 1.6450.9500z = 1.960.9750z = 2.330.9901z = 2.580.9951

The area to the left of z is 0.9750. What is z?

xy

You want 1 percent in the upper tail. Which area do you look up?

Reading Normal Probabilities

The area to the left of a z-score.

Reading a normal probability means finding the area to the left of a z-score

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xy

Tables give the area to the left of a z-score. At z = 0 that is a half.

xy

For the right-hand tail, subtract that area from 1.

xy

For a band between two values, subtract the smaller area from the larger.

Now you

area to the leftz = 1.280.8997z = 1.6450.9500z = 1.960.9750z = 2.330.9901z = 2.580.9951

What is P(2.33 < Z < 2.58)?

area to the leftz = 1.280.8997z = 1.6450.9500z = 1.960.9750z = 2.330.9901z = 2.580.9951

What is P(1.645 < Z < 1.96)?

Inverse Normal

Given the area, find the cut-off value.

An inverse normal question gives you the area and asks for the value that cuts it off

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xy

The top 10 percent starts where 90 percent of the area sits to the left.

xy

Read the table backward: 0.90 is not listed, so take the nearest area, 0.8997, at z = 1.28.

zrescalez × sigmashift backmean + z × sigma

Then undo the standardizing to turn that z back into a value on the original scale.

Now you

xy

The mean is 41 and sigma is 5. Which value has z = 2?

xy

The mean is 25 and sigma is 2. Which value has z = -2?

Combining Normals

Means always add; variances only if independent.

Adding independent normal variables adds their means and adds their variances

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xy

Means add: E(X + Y) = E(X) + E(Y). With means 1 and 3, the sum has mean 4.

xy

Var(X + Y) = Var(X) + Var(Y) = 1 + 1 = 2, so the sum is the wider curve.

xy

Var(X − Y) = 1 + 1 = 2 as well — independent spreads add even when you subtract.

means add: alwaysif X and Y are dependent, the width changesvariances add: only if independent

Means always add. Variances add only when X and Y are independent.

Now you

Var(X − Y)what does this come to?

X and Y are independent with Var(X) = 6 and Var(Y) = 4. What is Var(X − Y)?

Var(X − Y)what does this come to?

X and Y are independent with Var(X) = 8 and Var(Y) = 7. What is Var(X − Y)?

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