When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloration, and other defects.
The whole truckload will be rejected if more than 5% of the sample is unsatisfactory.
Suppose that in fact 8% of the apples on the truck do not meet the desired standard.

What's the probability that the shipment will be accepted anyway?
a. 0.0222
b. 0.9778
c. 0.088
d. 0.912

Respuesta :

Answer:

c. 0.088

Step-by-step explanation:

Let p(s) be the proportion of apples defected in the sample. The probability that p(s)<0.05 can be calculated by calculating z statistic of 0.05:

[tex]\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }[/tex] where

  • p(s) = 0.05
  • p is the proportion of the apples in fact defected (0.08)
  • N is the sample size (150)   Then,

z(0.05)=[tex]\frac{0.05-0.08}{\sqrt{\frac{0.08*0.92)}{150} } }=-1.354[/tex]

And P(z<-1.354)≈0.0879

Therefore the probability that  a random sample of 150 among 8% defected apples can be accepted is 0.088.

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