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July 30, 2014

July 30, 2014

Posted by **lauren** on Sunday, February 17, 2008 at 6:51pm.

A)the probability exactly 1 patient suffers kidnay damage

B)the probabaility that less than 2 patients suffer kidney damage

c) the probability that at least 3 patients suffer kidney damage

d) the expected # of people who will suffer kidney damage

e) the standard deviation of the number of people who will suffer kidney damage

- statistics -
**Damon**, Sunday, February 17, 2008 at 7:23pmp = .01

binomial distribution, n = 30

A)

p(k) = C(30,k) p^(k) p^(30-k)

for k = 1

p(1) = [30!/(1!29!)]* .01^1 * .99^29

= 30*.01*.747

=.224

B)

That would be zero or one patient. We know p(1) = .224 so find p(0)

p(0) = (30!/0!30!) *.01^0 * .99^30

= 1 * 1 * .740

=.740

so the probability of less than 3 is .224+.740 = .964

C. 1-.964 = 0.036

D. mean = n p = 30*.01 = 0.3

E. variance = n p(1-p) =30*.01*.99 = .297

s = sqrt(.297) = .545

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