Question

Normal distributions

Answer 1

@shamil98 How was the series stuff. Do you think you got a hang of it?

Answer 2

I think I understand it , I've done series before, but there are many more series with different formulas and such.

Answer 3

When you get better at calculus, we can go into divergent and convergent series.
I wanted you to learn a bit about series so in the future we can do integrals from scratch.

Answer 4

alright, i do understand what we went over then.

Answer 5

Okay, let's start off with the concept of a distribution. When I say normal distribution, what is a distribution?

A distribution is a family of functions. They have certain inputs but there are other non input constants that need to be sorted out before you have a normal function.

Answer 6

For example \[
f(x) = mx+b
\]You could say this is the linear distribution.
\(x\) is the input.

\(m\) and \(b\) are constants that need to be sorted out before you have a normal function

Answer 7

so while \[
f(x) = 2x+6
\]Is a function \[
f(x) = mx+b
\]is a distribution.
Does that make sense?

Answer 8

Or you could think of a distribution as a function of functions.. like a meta function

Answer 9

I think so, it's a distribution because there are more than one variable that needs to be substituted or found out?..
like f(3) = 2(3) + 6

f(3) = m(3) + b
you still have too find m and b..

Answer 10

Well, for example \[
f(x,y) = x^2+y^2
\]is still a function because all of the variables are inputs, but \[
f(x,y) = ax^2+by^2
\]Is a distribution because \(a\) and \(b\) are not inputs.

Answer 11

I get it.

Answer 12

Like
ax^2 + bx + c
would that be a distribution as well?

Answer 13

Yeah

Answer 14

Now, let's talk about the simplest probability distribution,
The Bernoulli trial

Answer 15

alright

Answer 16

A Bernoulli trail is just an event that either happens or does not happen.
The probability that it happens is \(p\)
The probability that it doesn't happen is \(1-p\).

Answer 17

Clearly is must either happen or not happen because \((p)+(1-p) = 1\)

Answer 18

The next, more complicated distribution is a binomial distribution.

When you have \(n\) Bernoulli trials, the binomial distribution tells you the probability that \(k\) of the trials succeed or happen.

Answer 19

So, an example of a Bernoulli trial is a coin flip. \(p=1/2\) and \(1-p=1/2\).

Answer 20

A binomial distribution would tell us the probability we get exactly 7 heads if we flipped 10 coins.

Answer 21

The formula for a binomial distribution is: \[
f(k) = \binom nk p^k(1-p)^{n-k}
\]

Answer 22

First of all, are you familiar with this notation \[
\binom nk
\]?

Answer 23

I am not.

Answer 24

\[
\binom nk = \frac{n!}{k!(n-k)!}
\]Are you familiar with factorials, like \(5!\)

Answer 25

I've seen them before, but I do I am not familiar with those either...

Answer 26

It's an algebra 2 concept.

Answer 27

We'll have to teach you that later on, let's stick to normal distributions for now.

Answer 28

5! = 5 x 4 x 3 x 2 x 1 = 120
is it just that?..

Answer 29

Yes.

Answer 30

So \[
f(k) = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}
\]

Answer 31

Okay, here is a sample problem.

You flip a coin 5 times. What is the probability you get heads exactly 3 times?

Answer 32

Want me to do it as an example, or are you up to attempting it?

Answer 33

Do it as an example, first.. so I can learn how it's done.

Answer 34

Then afterwards i'll attempt one

Answer 35

Okay so the key here is that flipping a coin is a Bernoulli trial. Success is defined as getting heads, and the probability of that is \(p=1/2\).

We repeat this trials a total of \(n=5\) times. We are looking for success for exactly \(k=3\) times.

Answer 36

using our formula: \[
f(3) = \binom 53\left(\frac 12\right)^3\left (1-\frac 12 \right)^{5-3}
\]

Answer 37

f(3) = 5! / 3!(5-3)! 1/2^3 (1-1/2)^5-3

Answer 38

\[
\binom 53 = \frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2}=10
\]

Answer 39

f(3) = 120/12 (1/2)^3 (-1/2)^2
f(3) = 10 x 0.125 x (-0.25)
?..

Answer 40

(0.25)* forgot mb

Answer 41

That negative sign is wrong though

Answer 42

\(1-1/2=1/2\)

Answer 43

Oh, sorry, i thought it was 1/2 -1, read it wrong..
so f(3) =0.3125 ?

Answer 44

Yep, also written as \(5/16\)

Answer 45

Alright, I believe I understand it.

Answer 46

Here is a tip. When I was your age I got really used to changing fractions into decimals and changing square roots into decimals.

Resist the urge to do that. Keep things as fractions and as radicals. That keeps things exact for as long as necessary.

Answer 47

I was using a calculator..
but in fraction form
10 x 1/8 x 1/4
10/32
5/16

Answer 48

Now for a test problem.
Suppose you roll \(8\) dice. What is the probability that \(5\) of the dice land on \(2\).

Answer 49

Think you can do this one?

Answer 50

First step is to identify the bernoulli trial.

Answer 51

5/8 of the dice land on 2?..

Answer 52

k = 2?..

Answer 53

Yes. Also the other 3 dice don't land on 2 obviously

Answer 54

No, \(k\neq 2\)

Answer 55

First of all, what is the Bernoulli trial here?

Answer 56

p = 5/8
right?

Answer 57

I'm a bit lost..

Answer 58

The Bernoulli trial here is "a die lands on 2"

Answer 59

How many times is the trial done? How many times are we saying it succeeds?

Answer 60

What is the probability this trial succeeds just once?

Answer 61

A die has \(6\) sides. And \(2\) is on only one of those sides.

Answer 62

1/6 chances..
1/6 x 5/8 then?

Answer 63

Hint: It is a binomial distribution.

Answer 64

You have identified \(p=1/6\) but what about \(n\) and \(k\)?

Answer 65

would k become 1/6 x 5/8 then?
1/8 that 2 is the number on the dice?

Answer 66

\(n\) and \(k\) must be natural numbers (not fractions) for it to be a binomial distribution.

Answer 67

Suppose you roll \(8\) dice. What is the probability that \(5\) of the dice land on \(2\).

Answer 68

\(k=5\)

Answer 69

would n be 2 then?

Answer 70

\(n=8, k=5, p=1/6\)

Answer 71

oh.

Answer 72

Does that not make sense?

Answer 73

Not really.

Answer 74

I understand that 5 and 8 is from 5/8 correct?..

Answer 75

Okay think of each die as a B trial. Success is if it lands on 2.

Since we roll 8 dice, that is 8 total trials. Wanting 5 die on 2 means we want 5 successes.

Answer 76

Oh! That makes sense.

Answer 77

Okay the calculation isn't important here.

Answer 78

http://www.wolframalpha.com/input/?i=%288+choose+5%29%281%2F6%29%5E5+%281-1%2F6%29%5E%288-5%29
However it is really low probability. Less than half of a percent. You can do this 1000 times and only get it to happen 4 times.

Answer 79

Okay

Answer 80

Let's try one more.

Answer 81

alright

Answer 82

Suppose that a college class is 65% female and 35% male.
Then 6 people are randomly put into a study group.
What is the probability that the group has 4 girls.

Answer 83

k = 4
n = 6

Answer 84

right?

Answer 85

Yes

Answer 86

Hm, what would p be then..
65/100
?

Answer 87

Yes.

Answer 88

so..
6! /4! (6-4)! (65/100)^4 (1-65/100)^2

720/ 48 (17.85%) (0.35)^2
15(17.85/100)(0.1225)

Answer 89

52479/160000
is the probability, if i did that correctly.

Answer 90

http://www.wolframalpha.com/input/?i=%286+choose+4%29%2865%2F100%29%5E4+%281-65%2F100%29%5E%286-4%29

Answer 91

It's about 32 percent.

Answer 92

Yeah, I did correctly then.

Answer 93

These problems are not tough when you say exactly 4 girls or exactly 5 dice

Answer 94

What if I said "at least 4 girls" for example

Answer 95

You would have to do k=4, k=5, and k=6, then add those probabilities up.

Answer 96

So basically k >= 4

Answer 97

with the limit 4 <= k <= 6

Answer 98

So, the formula for a for the number of successes that are less than or equal to \(k\) is given by: \[
S_k = \sum_{i=0}^k\binom nk p^k(1-p)^{n-k}
\]