Approximately 4% of people are colorblind. Consider a class of 460. What is the probability that 15 or fewer students are colorblind?

Question

anonymous · Answer

do you want a fraction?

perl · Answer

This problem can be solved using a binomial distribution.

anonymous · Answer

oh i obviously dont belong here then ....;-;

carry on young chap

perl · Answer

$$
\Large \Pr(X \le k) = \sum_{i=0}^{k} {n\choose i}p^i(1-p)^{n-i}
\ 	herefore \
\Large \Pr(X \le 15) = \sum_{i=0}^{15} {460\choose i}0.4^i(1-0.4)^{460-i}

$$

anonymous · Answer

wow you guys are smart o,o

this is like next gen math 4 me

thatonegirl_ · Answer

What o.o @perl

thatonegirl_ · Answer

I know it's a binomial distribution and first i find the mean which wold be 18.4 and the standard deviation is 4.2, right?

perl · Answer

you would do that if you are going to approximate the binomial with a normal standard curve. but the problem did not say to do this

perl · Answer

want to do it that way?

thatonegirl_ · Answer

Well when n is sufficiently large enough can't you consider it to be normal?

thatonegirl_ · Answer

That's the only thing I've learned Dx

perl · Answer

yes you can, it just didnt say to use normal approximation explicitly . 
we could solve this exactly using a calculator

thatonegirl_ · Answer

Oh so you don't need to find the z value?

perl · Answer

normal curve binomial approximation was useful before the age of calculators, since solving 450 choose 15 is difficult by hand

perl · Answer

well lets do it both ways, and we can compare the results

thatonegirl_ · Answer

kk

perl · Answer

the exact answer using a calculator is : http://www.wolframalpha.com/input/?i=sum+%28i%3D0..15%29++%28460+choose+i%29*0.04^i%281-0.04%29^%28460-i%29

perl · Answer

there is a typo above in the formula it should be $$

\Large \Pr(X \le 15) = \sum_{i=0}^{15} {460\choose i}\color{red}{0.04}^i(1-\color{red}{0.04})^{460-i}

$$

thatonegirl_ · Answer

Oh okay.. Is that a binomial setting just with that first thing?

perl · Answer

right, the exact answer is 
`0.251193749212680812548679303635273122372737486042499189442302...`
now lets compare this to the normal approximation

perl · Answer

mean = n*p
st. dev = sqrt( n * p * (1-p)) 

mean = 460 * .04 = 18.4
st.dev = sqrt( 460 * .04 * .96) = 4.202856

First find the z score, for X = 15, since we want less than it. 
also make a continuity correction because we want P ( X <= 15)
Z = ( 15.5 - 18.4) / (  4.202856)

thatonegirl_ · Answer

.209?

perl · Answer

Z = -.6900

thatonegirl_ · Answer

ohh i thought u mean on the chart.

thatonegirl_ · Answer

.2451?

perl · Answer

oh ok lets check the chart . but do you agree with with that z score?

P( X <= 15) = P( Z <= -.6900) = 0.245094

perl · Answer

yes i got the same

thatonegirl_ · Answer

yeah

perl · Answer

and thats pretty close to the exact answer

thatonegirl_ · Answer

so the answer is 24.51%?

thatonegirl_ · Answer

oh cool

perl · Answer

yes, using the normal approximation

thatonegirl_ · Answer

How would i do probability of more than 12? with the same data

thatonegirl_ · Answer

12 or more*

thatonegirl_ · Answer

I got a z-score of -1.52 but you would use 1.52 because it's greater than, right?

perl · Answer

right

thatonegirl_ · Answer

I end up getting 93.57% which is very off

perl · Answer

so if you are using a table, once you find the area to the left of the z score you can subtract from 1 to get the area to the right of that z score

thatonegirl_ · Answer

So if i did -1.52, just find that value and subtract it from 1?

perl · Answer

i think we need to make another continuity correction

perl · Answer

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