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4. A local orchard packages apples in bags. When full, the bags weigh 5 pounds
each and contain a whole number of apples. The weights are normally distributed
with a mean of 5 pounds and a standard deviation of 0.25 pound. An inspector
weighs each bag and rejects all bags that weigh less than 4.75 pounds. What
percentage of bags will the inspector reject? *
Your answer

Respuesta :

JeanaShupp

Answer: The percentage of bags will the inspector reject = 15.87%

Step-by-step explanation:

Given, The weights of apple bags are normally distributed  with a mean of 5 pounds and a standard deviation of 0.25 pound.

i.e. [tex]\mu=5[/tex] and [tex]\sigma=0.25[/tex]

Let X be the weight of any random apple bag.

Since an inspector  weighs each bag and rejects all bags that weigh less than 4.75 pounds.

Then, the probability of bags will be rejected = probability that bags weigh less than 4.75 pounds.

[tex]=P(X<4.75)\\\\=P(\dfrac{X-\mu}{\sigma}<\dfrac{4.75-5}{0.25})\\\\=P(z<-1)\ \ \ [z=\dfrac{X-\mu}{\sigma}][/tex]

[tex]=1-P(z<1)\\\\=1-0.8413\ \ \ [\text{By z-value table}]\\\\=0.1587=15.87\%[/tex]

Hence, the percentage of bags will the inspector reject = 15.87%

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