A manufacturing company measures the weight of boxes before shipping them to the

Question

A manufacturing company measures the weight of boxes before shipping them to the customers. The box weights are normally distributed with a population mean and a population standard deviation of 90 lbs. and 24 lbs., respectively. Use this information to answer questions 8 to 10.

Question 8:

What is the probability that a randomly chosen box weighs more than 100 lbs.?     (Round to the nearest decimal number with 4 decimal places.)

Question 9:

What is the probability that a random sample of 25 boxes would weigh, on average, more than 100 lbs.?                              (Round to the nearest decimal number with 4 decimal places.)

Question 10:

If the average weight of a random sample of 25 boxes was 100 lbs., what could we conclude? (Hint: refer to your answer to Question 9.)

About 34% of all samples should have an average weight of at least 100 lbs., so there is nothing wrong.

Less than 2% of all samples should have an average weight of at least 100 lbs., so it would be unlikely to observe a sample average weight of 100 lbs. if the original population parameters were true.

We wouldn't expect to see an average weight of 100 lbs. for a sample of 25 boxes. So the original population mean parameter might be wrong.

The sample data suggest that box weights are not normally distributed.

About 34% of all samples should have an average weight of at least 100 lbs., so there is nothing wrong.

Less than 2% of all samples should have an average weight of at least 100 lbs., so it would be unlikely to observe a sample average weight of 100 lbs. if the original population parameters were true.

We wouldn't expect to see an average weight of 100 lbs. for a sample of 25 boxes. So the original population mean parameter might be wrong.

The sample data suggest that box weights are not normally distributed.

Explanation / Answer

a.
Mean ( u ) =90
Standard Deviation ( sd )=24
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 100) = (100-90)/24
= 10/24 = 0.4167
= P ( Z >0.417) From Standard Normal Table
= 0.3385                  
b.
P(X > 100) = (100-90)/24/ Sqrt ( 25 )
= 10/4.8= 2.0833
= P ( Z >2.0833) From Standard Normal Table
= 0.0186                  
c.
Less than 2% of all samples should have an average weight of at least 100 lbs., so it would be unlikely to observe a sample average weight of 100 lbs. if the original population parameters were true.

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