Question

In: Math

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly...

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.)

Solutions

Expert Solution

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


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