1)P(shipment will be accepted)=P(X<=1)=P(X=0)+P(X=1)=46C0(0.06)0(0.94)46+46C1(0.06)1(0.94)45
=0.2285
2)
for normal distribution z score =(X-μ)/σx | |
here mean= μ= | 105 |
std deviation =σ= | 15.0000 |
probability = | P(88<X<122) | = | P(-1.13<Z<1.13)= | 0.8708-0.1292= | 0.7416 |
Question Hep * A pharmaceutical company receives large shipments of aspirin tablets. The acceptance batch if...
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% 5.2.35-T Question He A pharmaceutical company receives largo shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 52 tablets, then accept the batch there is only one or none that doesn't meet the required specifications. If one shipment of 3000 aspirin tablets actually has a 2% rate of defects, what is probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many bo rejected? The probability that...