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Statistics and Probability

Geometry

Number Theory

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Statistics (descriptive, analysis of data, probability, xperiments) (top of page)

 

Basic Definitions . . . . (10/98)

Our data set consists of n values, or scores; x1, x2, . . . , xn.

These might be the ages xk of the n people in your class, for example.

 

The mean, xbar , of the scores is (I can't do an overscore here so imagine an x with a bar over it!)

xbar = (x1 + x2 + . . . + xn) / n .

This is the average value of the n scores.

 

Another thing to know is how spread out the scores are. This involves the variance

xvar, the average squared distance from the mean:

xvar = [ (xbar - x1)^2 + (xbar - x2)^2 + . . . + (xbar - xn)^2 ] / n

 

The standard deviation xsd is the square root of the variance; that gives the average

distance from the mean, with outliers (scores far from the mean) having the most influence.

xsd = (xvar).

 

Note: Technically these are the population variance and population standard deviation.

Sometimes the sample variance is used, where the denominator is n - 1 instead of n. The

sample standard deviation is the square root of the sample variance. The idea here is to

increase the variance a bit because in taking a sample you are less likely to include a rare outlier,

so the variance is too small with an n down there.

 

 

Example: Three burrito shops charge $2, $3, and $7 for a bean & cheese.

What are the average burrito price and the standard deviation?

 

Solution: There are n = 3 values x1 = 2, x2 = 3, x3 = 7.

So the mean (avg) is xbar = (2 + 3 + 7) / 3 = 12 / 3 = $4.

If you randomly visit the three shops, then in the long run you'll spend $4 per

burrito (even though none of the places charges $4!).

The variance is [ (2 - 4)^2 + (3 - 4)^2 + (7 - 4)^2 ] / 3 = [4 + 1 + 9] / 3 = 14 / 3

= 4.667 approx. So the standard deviation is Sqrt[14/3] = Sqrt[4.667] = $2.16 approx.

This means that in the long run, the average price difference from the mean is $2.16.

 

 

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Probability (3/04) (top of page)

The probability or 'chance' an event will happen is the fraction, decimal, or percentage

of the time you'd expect it to happen, or have observed it to happen. If there are a bunch

of simple events that have equal chance of occurring, then the prob p(A) of a compound

event A happening is p(A) = (no. of simple events in A) / (no. of possible simple events).

 

 

Example: If a coin is flipped the chance it comes out heads is 1 in 2, or 1/2, or 0.5, or 50%.

On the other hand if a baby is born, the chance it's a boy is about 52%, and the proba-

bility it comes out heads is about 96%.

 

Example: Find the probability of choosing a prime at random from an interval of numbers.

Solution: The answer depends on the set of (whole) numbers we choose from.

Between 1 and 10 the numbers 2, 3, 5, and 7 are primes; so 4 of 10, or 40% are prime.

But there are 25 primes from 1 to 100, so the probablilty of picking a prime at random

from 1 to 100 is 25/100, or 1/4. As the interval goes up, the Prime Number Theorem

states that the "density" of primes in the interval from 1 to x is about 1 / ln[x]; so the

number of primes from 1 to x is about x / ln[x]

for x = 100, 100 / ln[100] is about 21.7 (compared to the actual prime count of 25)

 

Example: What's the probability that you'll pass: a) Both your classes? b) At least one class?

 

Solution: It depends on what the chances of passing each class, and whether passing one class

has any effect on passing other class. Here's an example I used in my recent Stat class:

Say the chance of passing Algebra is 88% and the chance of passing Biology is 60%.

a) The prob of passing both can't be determined, but .if. passing one class is

independent of passing the other, then p(A and B) = p(A) * p(B).

For us, p(A and B) = (0.88)(0.60) = 0.528, or 52.8% chamce of passing both classes.

 

b) The chance of passing at least one is the sum of the indiv probs minus the prob

of both, because we don't want to count the "pass both classes" cases twice.

p(A or B) = p(A) + p(B) - p(A and B) = 0.88 + 0.60 - 0.528 = 0.952;

a 95.2% chance of passing at least one class. That means only a 4.8% chance of

failing both classes.

 

 

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Random Sampling and Surveys (7/00) (top of page)

Ask several people to "pick a random number between 1 and 100." The distribution

you'll get is far from random; lots of people will pick 50, some will pick 99, others will

try to be very random and say 37. Each of these three numbers comes up much more

than 1% of the time! People rarely say things like "80" because it doesn't sound very

offbeat. Human nature, I guess.

 

When doing a random experiment, such as polling a sample of voters, it's crucial that

these biases are not introduced. Randomizing tools like flipping coins, spinning a wheel,

or rolling dice will do the trick.

 

I've heard of surveys that say "57.1% of the voters polled favored Clinton" when it turns

out they called 7 of their friends and 4 of them were Clinton supporters. Even with a

large sample size, things like "touch-tone phones only, 95 cents per call" will only allow

people with telephones, TV's, and enough money or a strong enough opinion to call in.

I like the "no opinion" part; would you call in, wait on the line, and pay 95c, just to say

you have no opinion?

 

There are tables of random digits in stat books, and most calculators have a "Ran#"

button; that's how I select the 'lucky' people from the waiting lists that will get into my

class! Another method is to use the decimal digits of pi = 3.1415926535897932384626...

which is (believed to be) 'normal' in the sense that, in the long run, each digit of pi ()

occurs 1/10 of the time, each pair of digits occurs 1/100 of the time, etc.

 

 

 

 



 

Number Theory (expanded 8/01) (top of page)

Prime Numbers

Prime Factorization

Number and Sum of Divisors

Supercomposite Numbers

Other Integer Functions

 

Basic Definitions

The natural numbers are the counting numbers:

1, 2, 3, 4, . . . .

The whole numbers are the natural numbers with 0 thrown in.

0, 1, 2, 3, . . .

The integers consist of the whole numbers and their negatives:

. . . , -3, -2, -1, 0, 1, 2, 3, . . .

 

A natural number bigger than 1 is called prime if there are only two numbers that go into it

evenly; itself and 1.

 

Examples: 3 is prime because only 1 and 3 go into it. But 77 is not prime because 7 and 11

go into it (as opposed to going into the 7-Eleven!). Click here for more basic info about primes and factors,

or if you like, my special dan's prime page.

 

The GCD or greatest common divisor of two or more numbers is the largest number that

goes into both (or all) of the numbers: GCD(12, 18, 28) = 2 and GCD(385, 1001) = 77.

Two numbers are relatively prime if their GCD is 1; they have no (other) common factors.

So for example 6 and 25 are rel. prime even though neither is a prime number.

 

The LCM is the least common multiple, the smallest thing that they both (or all) go into.

Here LCM(10, 14) = 70.

 

Challenge: Prove, for any numbers m and n, we have GCD(m,n) * LCM(m,n) = m*n.

 

 

Unique Prime Factorization

The Fundamental Theorem of Arithmetic states that every natural number greater than 1

can be written as a product of prime numbers, and that up to rearrangement of the factors,

this product is unique. This is called the prime factorization (or PF for short) of the number.

 

Example: 36 = 6 x 6 = 9 x 4 = 12 x 3 = 18 x 2 , but all are equal to 2 x 2 x 3 x 3.

This is the PF of 36, often written with exponents: 36 = 2^2 * 3^2. You can use these PFs

to figure out GCDs, LCMs, and the number (and sum) of divisors of n.

Not all sets of numbers have this 'UF' or unique factorization property. For example,

9 = 3 * 3 = (7 + 2 sqrt[10])(7 - 2 sqrt[10]) but none of these can be broken up further using

numbers of the form a + b sqrt[10], so the "ring of integers" Z[sqrt[10]] is not a UFD or

'unique factorization domain.'

 

 

 

Divisors (counting and adding them) (See Prob Of Week #150)

Example 1: How many numbers go into 36 evenly (with no remainder)?

There are 9 divisors of 36 : {1, 2, 3, 4, 6, 9, 12, 18, 36}.

These can be arranged into a 3 x 3 rectangle, called a divisor chart:

1	2	4
3	6	12
9	18	36

 

Notice from above that 36 = 2^2 * 3^2 ; the exponents are 2 and 2.

There is one more column than powers of 2, and one more row than powers of 3,

so there are 3 x 3 = 9 divisors. Our function notation is d(36) = 9.

 

Example 2: How many divisors does 21 million have?

21,000,000 = 3 * 7 * 10^6 = 2^6 * 3^1 * 5^6 * 7^1 ;

the exponents are 6, 1, 6, and 1 ; so the number has 7 * 2 * 7 * 2 = 196 divisors :

They are {1, 2, 3, 4, 5, 6, 7, 8, 10, . . . , 10500000, 21000000} (I left a few out.)

 

Example 3: What do the divisors in the 36-chart add up to?

See below that the actual sum is 91 ; but also notice that the column sum and row sum

that start with 1 are 1+3+9 = 13 , 1+2+4 = 7 , and that 13*7 = 91. Coincidence? Hmm . . .

Arithmetic Functions (pronounced "a-rith-MEH-tic") 8/01

 

Let's look at several integer functions using function notation, inputting only integers:

d(n) = number of (positive) divisors of n (including 1 and n)

s(n) = sum of divisors of n (incl. 1 and n; usually 'sigma(n)')

ph(n) = number of nos. less than n with no common factors with n (usu. 'phi of n').

A(n) = abundancy index of n, defined as the ratio s(n) / n .

 

For example if n = 36 we have nine divisors ; so

d(36) = #{1, 2, 3, 4, 6, 9, 12, 18, 36} = 9 ; s(36) = 1+2+3+4+6+9+12+18+36 = 91 ;

ph(36) = #{1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35} = 12 ; A(36) = 91 / 36 ~ 2.528.

 

One special case: if n = p prime, then d(p) = 2, s(p) = p+1, ph(p) = 1, A(p) = 1+(1/p).

The n is perfect if s(n) = 2n, or A(n) = 2 ; deficient if A(n) < 2 , and abundant if A(n) > 2.

For example, 36 is abundant, 35 is deficient, and 28 is perfect. Check it yourself!

 

For a good source of super-abundant numbers see dan's supercomposites page.

Challenges: Try to prove or at least verify with examples that d, s, and ph are

"multiplicative" functions, meaning that if m and n are relatively prime then

d(mn) = d(m) d(n) and so on (for s and ph).

What about the case where m and n have a common factor k? What if there

are more than two numbers? What about the A function, is it multiplicative?

 

 

 

Well, that's it for now. Check back often for new stuff!

Click below for other topics, or visit the ask dan page!

 

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