Question

Use the binomial expression (p+q)n to calculate a binomial distribution with n=5 and p=0.3.

@phi

Answer 1

p= 0.3 and q = 1-p = 0.7

I think they want you to expand (p+q)^n using
http://www.purplemath.com/modules/binomial.htm

and then calculate the numerical value for each term (each value of n)

Answer 2

Can you help me with that? That's where I'm really confused. @phi

Answer 3

start by writing down the binomial expansion formula
\[ (p+q)^n= \sum_{n=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)p^k\ (1-p)^{n-k}\]

Answer 4

Would I plug in the numbers I already have into that formula?

Answer 5

let me fix the index:
\[ (p+q)^n= \sum_{k=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)p^k\ (1-p)^{n-k} \]

Answer 6

there are going to be 6 terms, k=0, 1, ... , 5

Answer 7

Okay, so what would be next?

Answer 8

Let's do the first term, with k=0, n= 5, p= 0.3, q= 0.7

Answer 9

it's easier to type if we use n Choose K, rather than the paren notation
(it means the same thing)

Answer 10

\[(0.3+0.7)^5 = \sum_{k=0}^{}\]

Answer 11

Would it be something like that?

Answer 12

"sort of"

the binomial distribution has a "value" for each k
and if we plot it, it will look like a gaussian if we have a large n.

at each k, we want to get a number. If we plot a bar graph (each bar at a different k)
we would see the distribution.

They want us to figure out the values for each k.

Answer 13

How do we figure out the values for each k (from 0 to 5) ?
we use the "binomial theorem" to find each term.

Answer 14

for example, for k=0
the first term is
5 C 0 0.3^0 0.7^5

Answer 15

Okay, so next would be k=1?

Answer 16

yes, give it a try.

Answer 17

What was the second term?

Answer 18

5 C 2 0.3^2 0.7^3?

Answer 19

that is the 3rd term, with k=2

Answer 20

5 C 1 0.3^1 0.7^4?

Answer 21

yes

Answer 22

What's next?

Answer 23

do the same for k=3,4,5
also, give the numerical equivalent for each expression
and then make a table, or a plot. For example, here is a plot of the binomial distribution with p=0.3 and n= 5

Answer 24

5C3 .3^3 .7^2
5C4 .3^4 .7^1
5C5 .3^5

Answer 25

notice for such a small n (n=5), it does not look like a bell-curve.

Answer 26

What would we do next?

Answer 27

can you evaluate each of those expressions ?

Answer 28

Yes, but how would I do that?

Answer 29

Can you help me please?

Answer 30

n C k means

n! / k! (n-k)!

my calculator has a key for this operation.

Answer 31

5C0 is by definition 1
also 5C5 is 1
5C1 is 5
also
5C4 is 5

that leaves 5C2 , 5C3

Answer 32

Okay, and that means that goes into the calculator?

Answer 33

let's do the first term.
5C0 0.3^0 0.7^5
what do you get for this ?

Answer 34

I'm not sure, my calculator doesn't do that type of operation.

Answer 35

you know 5C0 (I told you up above). it is 1
you know 0.3^0

Answer 36

.16807

Answer 37

yes, so that is the value for k=0 (or x=0 if we plot this ... see my graph)

now the 2nd term
5C1 0.3^1 0.7^4

Answer 38

.36015

Answer 39

yes.

btw, google knows how to do 5 choose 2
next term, for k=2.
5 C 2 0.3^2 0.7^3

Answer 40

.3087

Answer 41

Is that correct?

Answer 42

yes. can you finish? they just want a table with those values

Answer 43

I got the rest of the values of each term, what is the next step?

Answer 44

that is the answer: a table with k and the corresponding value

Answer 45

or perhaps use x:

x Pr(x)
0 0.16807
1 0.36015
etc

Answer 46

Thank you!

Answer 47

and label it: binomial distribution for p=0.3 and n=5