probability that 13 respond
p(13)= C(20,13) (.4^13) (.6^7)
= 77520 * 6.71 * 10^-6 * 2.8 * 10^-2
= 1.46 * 10^-2
and for 14
p(14) = C(20,14)(.4^14)(.6^6)
38760 * 2.68 * 10^-6 * 4.67 *10^-2
=.485 * 10^-2
etc for binomial distribution
p(15) = 15504 * .4^15 * .6^5
= .129 * 10^-2
p(16) = 4845 *.4^16 *.6^4
= .027 *10^-2
p(17) = 1140 * .4^17 * .6^3
=.0042 *10^-2
p(18) = 190 * .4^18 * .6*2
=.00047 *10^-2
forget p(19) and p(20), getting too small
add them up
2.1*10^-2 = .02
It is known that 60% of people do not respond to mailed questionnaires. In a sample of 20 questionnaires mailed, what is the probability that more than 12 people will respond?
2 answers
By the way, n = 20 is high enough so we could use the continuous version, the normal distribution, to approximate and see if we are way off base.
mean no response = 20 * .6 = 12
sigma = sqrt (20*.6*.4) = 2.2
To get more than 12 responding on the continuous distribution we want to be between 12 and 13 responding or between 7 and 8 not responding, call it 7.5
(7.5-12)/2.2 = -2 sigmas from mean
which is .023 in table of F(z) versus z on normal function table.
mean no response = 20 * .6 = 12
sigma = sqrt (20*.6*.4) = 2.2
To get more than 12 responding on the continuous distribution we want to be between 12 and 13 responding or between 7 and 8 not responding, call it 7.5
(7.5-12)/2.2 = -2 sigmas from mean
which is .023 in table of F(z) versus z on normal function table.