Probability

[uselessinmath]

I am stumped.....

If P s the probability tha a person samples at random smokes, then 1-p is the probability that the person doesn't. If 40 people are samples at random, the variance of the sample will be 40p(1-p). What two probabilities p will give a variance of 6.4?

v(p)=40p-40p^2
v(p)=-40p^2+40p
v(p)=20p(-p+20)

 

[pka]

Just solve this equation: 40p(1-p)=6.4 or 40p<SUP>2</SUP>-40p+6.4=0

 

[tkhunny]

the variance of the sample will be 40p(1-p). What two probabilities p will give a variance of 6.4?

It may help to think of it as ONE probability (an it's compliment)
40*p*(1-p) = 6.4
25*p*(1-p) = 4
25*p - 25*p^2 = 4
25*p^2 - 25*p + 4 = 0

Leading to the non-unique solutions.

p = 1/5 and p = 4/5

This may seem a little odd, since the "other" solution for 'p' happens to be the corresponding value for 'q'. Don't let that bother you. Just keep track.

For p = 0.2 q = 1-p = 0.8
For p = 0.8 q = 1-p = 0.2

That's all you get for this problem. We'd need more information to know which value for 'p' is more appropriate. Since we are talking about the proportion of smokers, I sincerely hope it's p = 0.2 unless it's a population of lung cancer patients.