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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.)

Respuesta :

Answer:

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.

Using the binomial distribution, it is found that:

  • The probability that this whole shipment will be accepted is 0.6181 = 61.81%.
  • The company will accept 61.81% of the shipments and will reject 38.19% of the shipments, so many of the shipments will be rejected.

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For each tablet, there are only two possible outcomes. Either it is defective, or it is not. The probability of a tablet being defective is independent of any other table, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

It is the probability of exactly x successes on n repeated trials, with p probability.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]  

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  • 44 tablets are tested, thus, [tex]n = 44[/tex]
  • 3% are defective, thus, [tex]p = 0.03[/tex]
  • It will be accepted if at most 1 is defective, thus:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{44,0}.(0.03)^{0}.(0.97)^{44} = 0.2618[/tex]

[tex]P(X = 1) = C_{44,1}.(0.03)^{1}.(0.97)^{43} = 0.3563[/tex]

Thus:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.2618 + 0.3563 = 0.6181[/tex]

The probability that this whole shipment will be accepted is 0.6181 = 61.81%.

The company will accept 61.81% of the shipments and will reject 38.19% of the shipments, so many of the shipments will be rejected.

A similar problem is given at https://brainly.com/question/15557838