Posted by Sunny on Sunday, January 24, 2010 at 12:33am.
70% of the student body of a very large post-scondary institution is female. In a random sample of 12 students, what is the probability that:
a) at most half will be females
b) more than 7 will be females
c) fewer than 3 will be males
I used the Table of Binomial Probabilities:
X P(x) X P(x)
0 .0000 7 .1585
1 .0000 8 .2311
2 .0002 9 .2397
3 .0015 10 .1678
4 .0078 11 .0712
5 .0291 12 .0138
(A) P(at most 1/2 female)=
P(0 or 1 or 2 or 3 or 4 or 5 or 6)
P(x=0)+ P(x=1)+ P(x=2)+ P(x=3)+ P(x=4)+ P(x=5)+ P(x=6)
= .0000+.0000+.0002+.0015+.0078+.0291+.0792= .1178 is this right?
(B) P(more than 7 will be female)=
P(8 or 9 or 10 or 11 or 12)
P(x=8)+ P(x=9)+ P(x=10)+ P(x=11)+ P(x=12)= .2311+.2397+.1678+.0712+.0138= .7236 is this right?
(C) I don't know how to do this one :(
- Statistics - Sunny, Sunday, January 24, 2010 at 9:54am
PS I don't want the answer, for C, I just need to understand how to do the formula. Anyone?
- Statistics - Sunny, Monday, January 25, 2010 at 9:29am
Anyone? Really need help on how to do this please :)
- IS this right???????? - Sunny, Tuesday, January 26, 2010 at 6:20pm
OK last time, do I add up P(x) values from the binomial table from 3 through 12 to establish p and then subtract p value from 1 (as I would to get q value) to get answer C)? or would I totally switch the p value from women to men. I would like to keep the integrity of the question as much as possible and keep the p as women, but if I must change then I will.
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