A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 47 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 3000 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 the whole shipment will be accepted is 0.5862.

Step-by-step explanation:

Let X = number of the aspirin tablets that doesn't meet the required specifications.

The probability of the random variable X is, P (X) = p = 0.03.

The sample of n = 47 tablets are tested from each batch.

The probability of any of the tablets being defective is independent of the others.

The probability mass function of X is,

Compute the probability that the whole shipment of 3000 tablets will be accepted as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

Thus, the probability that the whole shipment will be accepted is 0.5862.

The sample of 47 tablets is significantly small when drawn from a population of 3000 tablets. So it is difficult to make conclusion about all such shipments of aspirin tablets.

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

answer: c i think

step-by-step explanation: