could somebody do question b

Printable View

- Apr 26th 2006, 06:53 PMArbiturProb Dist.. just some clarity please.
could somebody do question b

- Apr 26th 2006, 08:51 PMArbitur
changed the question

- Apr 26th 2006, 09:09 PMCaptainBlackQuote:

Originally Posted by**Arbitur**

Probability of a blue on any trial is 0.8 as we are sampling with replacement,

so:

$\displaystyle

P(N_{blue} \ge 3)=P(N_{blue} = 3)+P(N_{blue} = 4)

$

but:

$\displaystyle

P(N_{blue} = K\ \mbox{in M trials})={M \choose K}p^K(1-p)^{M-K}=$$\displaystyle \frac{M!}{K!(M-K)!}p^K(1-p)^{M-K}

$

Here $\displaystyle p=0.8$, so:

$\displaystyle

P(N_{blue} \ge 3)=\frac{4!}{3!1!}\ 0.8^3\ 0.2^1+\frac{4!}{4!0!}\ 0.8^4\ 0.2^0

$

Which since $\displaystyle 1!=0!=1$ may be reduced to:

$\displaystyle

P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4

$

RonL - Apr 26th 2006, 09:24 PMArbitur
uhm can you explain that in words?

i need to get it to a percent number

i thought it had to be something like

mean=.8*4-3.2

standard deviation=square root of 4*.8*.2

=.64

so (SOME NUMBER)-3.2/.64 =the zscore i need to get the percentage???

that is what my book is teaching me

what number do i use - Apr 26th 2006, 09:33 PMCaptainBlackQuote:

Originally Posted by**Arbitur**

should be looking at the section that deals with the Binomial distribution.

Quote:

what number do i use

$\displaystyle

P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4

$

That can be simplified to the number you want:

$\displaystyle

P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4\approx 0.819

$

RonL - Apr 26th 2006, 09:42 PMArbitur
thanks anyways but i have no idea how u get that ;/

ok im looking through my book and

because of a rule i cant use the z-score way and crap... but how do i know what way and numbers to use then

ok i get it now... but what if b used really big numbers... would it still work that easy? if not what would i do( cause those are easily added 2 gether) - Apr 26th 2006, 11:00 PMCaptainBlackQuote:

Originally Posted by**Arbitur**

binomial distribution; see this thread

RonL - Apr 27th 2006, 01:46 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

be shown by reworking this problem using the normal approximation.

Here $\displaystyle p=0.8$, and $\displaystyle N=4$, so the mean number

of blues is:

$\displaystyle

\mu_B=0.8 \times 4=3.2

$

The vaiance of the number of blues is:

$\displaystyle

\sigma_B^2=4\times 0.8 \times 0.2=0.64

$

Now we want the probability of $\displaystyle 3$ or more blues, so we want

the critical z-score for $\displaystyle 2.5$ (rounding down by $\displaystyle 0.5$

as a continuity correction):

$\displaystyle

z_{critical}=\frac{2.5-3.2}{0.8}=-0.875

$,

Look this up in a normal table we find:

$\displaystyle

P(z>-0.875)\approx 0.809

$

rather than the correct value of $\displaystyle \sim 0.819$

RonL