In: Math
A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 19 tablets. The entire shipment is accepted if at most 2 tablets do not meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 5.0% rate of defects, what is the probability that this whole shipment will be accepted? The probability that this whole shipment will be accepted is nothing. (Round to three decimal places as needed.)
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 19 * 0.05
= 0.95
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 19 * 0.05 * 0.95
= 0.9025
III.
standard deviation = sqrt( variance ) = sqrt(0.9025)
=0.95
a.
the probability that this whole shipment will be accepted
The entire shipment is accepted if at most 2 tablets do not meet
the required specifications
P( X < = 2) = P(X=2) + P(X=1) + P(X=0)
= ( 19 2 ) * 0.05^2 * ( 1- 0.05 ) ^17 + ( 19 1 ) * 0.05^1 * ( 1-
0.05 ) ^18 + ( 19 0 ) * 0.05^0 * ( 1- 0.05 ) ^19
= 0.933
b.
The probability that this whole shipment will be accepted is
nothing
P( X = 0 ) = ( 19 0 ) * ( 0.05^0) * ( 1 - 0.05 )^19
= 0.377