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%
