p = .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
A drug is essential to the profileration of lymphocytes can cause kidney damage in 1% of patients. If the drug is given to 30 patients, find:
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
1 answer
1 answer