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November 25, 2014

November 25, 2014

Posted by **Lindy** on Friday, May 2, 2008 at 12:40pm.

- statistics -
**Damon**, Friday, May 2, 2008 at 8:22pmprobability 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

- statistics -
**Damon**, Saturday, May 3, 2008 at 1:44pmBy 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.

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