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W12 - Examples

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Summations

Binomial expectation and variance

Suppose we have repeated Bernoulli trials with .

The sum is a binomial variable: .

We know and .

The summation rule for expectation:

The summation rule for variance:

Multinomial covariances

Each trial of an experiment has possible outcomes labeled with probabilities of occurrence . The experiment is run times.

Let count the number of occurrences of outcome . So .

Find .

Solution

Notice that is also a binomial variable with success probability . (‘Success’ is an outcome of either or .)

The variance of a binomial is known to be for whatever relevant and .

So we compute by solving:

Hats in the air

All sailors throws their hats in the air, and catch a random hat when they fall down.

How many sailors do you expect will catch the hat they own? What is the variance of this number?

Solution Strangely, the answers are both 1, regardless of the number of sailors. Here is the reasoning:

(1) Let be an indicator of sailor catching their own hat. So when sailor catches their own hat, and otherwise. Thus is Bernoulli with success probability .

Then counts the total number of hats caught by original owners.

(2) Note that .

Therefore:

(3) Similarly:

We need and .

(4) Use . Observe that . Therefore:

(5) Now for covariance:

We need to compute .

Notice that when and both catch their own hats, and 0 otherwise.

We have:

Therefore:

(6) Putting everything together back in (1):

Months with a birthday

Suppose study groups of 10 are formed from a large population.

For a typical study group, how many months out of the year contain a birthday of a member of the group? (Assume the 12 months have equal duration.)

Solution

Let be 1 if month contains a birthday, and 0 otherwise.

So we seek . This equals .

The answer will be because all terms are equal.

For a given :

The complement event:

Therefore:

Pascal expectation and variance

Let .

Let be independent random variables, where:

  • counts the trials until the first success
  • counts the trials after the first success until the second success
  • counts the trials after the success until the success

Observe that .

Notice that for every . Therefore:

Using the summation rule, conclude:

Central Limit Theorem

Test scores distribution

Explain what is wrong with the claim that test scores should be normally distributed when a large number of students take a test.

Can you imagine a scenario with a good argument that test scores would be normally distributed?

(Hint: think about the composition of a single test instead of the number of students taking the test.)

Height follows a bell curve

The height of female American basketball players follows a bell curve. Why?

Binomial estimation: 10,000 flips

Flip a fair coin 10,000 times. Write for the number of heads.

Estimate the probability that .

Solution

Check the rule of thumb: and , so and the approximation is effective.

Now, calculate needed quantities:

Set up CDF:

Compute desired probability:

Summing 1000 dice

About 1,000 dice are rolled.

Estimate the probability that the total sum of rolled numbers is more than 3,600.

Solution

Let be the number rolled on the die.

Let , so counts the total sum of rolled numbers.

We seek .

Now, calculate needed quantities:

Set up CDF:

Compute desired probability:

Nutrition study

A nutrition review board will endorse a diet if it has any positive effect in at least 65% of those tested in a certain study with 100 participants.

Suppose the diet is bogus, but 50% of participants display some positive effect by pure chance.

What is the probability that it will be endorsed?

Answer

Continuity correction of absurd normal approximation

Let denote the number of sixes rolled after rolls of a fair die. Estimate .

Solution

We have , and and .

The usual approximation, since is continuous, gives an estimate of 0, which is useless.

Now using the continuity correction:

The exact solution is 0.0318, so this estimate is quite good: the error is 1.9%.