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molsen · Jun 3, 2007, 04:00 PM

statistics canada reported that approx. 76% of families in a city consist of married couples. Tehre are 22 students in a grade 3 class.

what is the probability that exactly half the students have families where the parents are defined as a married couple?

I have:
n=22
p=0.76 (probability of married couples)
q=0.24 (probability of non-married couples)
____________
standard deviation=\/ 22*0.76*0.24
= 2.00

I found the z-score which was:

z= (11-16.7) / 2.00 = -2.85

and from the table of z-scores I found the answer to be 0.0183 or 1.8%

is this right, or where did I go wrong?

there is a second part to the questions that asks for the probability that at least one student does not have parents that are defined as a married couple that I am having trouble starting as well.

galactus · Jun 3, 2007, 05:01 PM

Do you have a calculator?

$C(22,11)(0.76)^{11}(0.24)^{11}=0.0052$

The method you are using is a little less accurate than the above, but sometimes when using huge numbers it isn't practical.

${\mu}=np=(22)(0.76)=16.72$

${\sigma}=\sqrt{(22)(0.76)(0.24)}=2.003$

Use the correction continuity with $P(10.5<x<11.5)$

$z_{1}=\frac{10.5-16.72}{2.003}=-3.11$

$z_{2}=\frac{11.5-16.72}{2.003}=-2.61$

$P(10.5<x<11.5)=P(-3.11<z<-2.61)=P(z<-2.61)-P(z<-3.11)=0.0045-0.0009=0.0036$

A small difference, but not that much.

For your last problem, the opposite of at least one is none. Find the probability that no student does not have parents which are defined as married, then subtract from 1.

molsen · Jun 3, 2007, 05:09 PM

0.5% is the actual answer? That seems so small to me..

I don't really understand the other method you used, I've never heard of doing it that way.

galactus · Jun 4, 2007, 09:56 AM

Hello again Molsen:

Yes, anytime you see a problem that says at least one, that is the opposite of none.
It is mostly a lot easier to figure out the probability of none and then subtract from 1 then it is to count them all up from 1 to whatever.

Example:

"If the probability of a basketball player making a free throw is .7, find the probability that the player makes at least one throw in 10 tries".

You could use the binomial and add them all up from 1 to 10, but it is easier to find the probability he makes none and subtract from 1.

$C(10,0)(.7)^{0}(.3)^{10}=.0000059049$

1-.0000059049=0.9999940951

See? You could also use:

$\sum_{k=1}^{10}C(10,k)(0.7)^{k}(0.3)^{10-k}$ and get the same thing, but the former is easier in most cases.