A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 44 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3 % rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

The probability that this whole shipment will be accepted is

nothing.

(Round to four decimal places as needed.)

The company will accept

nothing % of the shipments and will reject

nothing % of the shipments, so

▼

almost all of the shipments will be accepted.

many of the shipments will be rejected.

(Round to two decimal places as needed.)

Question

Mathematics

Guest

28 March, 12:27

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 44 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3 % rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

The probability that this whole shipment will be accepted is

nothing.

(Round to four decimal places as needed.)

The company will accept

nothing % of the shipments and will reject

nothing % of the shipments, so

▼

almost all of the shipments will be accepted.

many of the shipments will be rejected.

(Round to two decimal places as needed.)

+3

Answers (1)

Know the Answer?

Howard Cannon · Answer 1

prob for accepting = 61.8%

Step-by-step explanation:

Here let X be the no of defectives in the sample of 44 tablets.

Each part is independent of the other to be defective and also there are only two outcomes, defective or non defective.

Then X is binomial.

Prob for one success = p=3% = 0.03 (given)

q = 1-p = 0.97

For the shipment to be accepted we must get 0 or 1 defective item in the sample of 44 items.

i. e. x can be either 0 or 1 only.

n = 44.

P (X=r) = 44Cr (0.03) ^r (0.97) ^100-r

From the above we calculate P (X=0) and P (X=1)

Prob for shipment to be accepted = P (x=0) + P (X=1)

= 0.2618+0.3562

= 0.6180

i. e. 61.8% have chance to be accepted and 38.2% to be rejected.