A medium size bag of potato chips states on the bag that the

A medium size bag of potato chips states on the bag that the weight of the potato chips is 10 ounces. Suppose that in producing medium size bags of potato chips, the actual weights are a population that is distributed normal with mean µ= 10.2 ounces and standard deviation = 0.12 ounces.

9d. Suppose manufactured potato chips are bundled into 24-bag cases. You may assume each case represents a simple random sample of size 24. Consider the average weight of each 24-bag case of potato chips. Give two limits, symmetric about the mean, within which 99% of the mean weights lie.

Solution

The middle 99% correspond to the endpoints having 0.005 and 0.995 as left tailed areas.

For the left tail:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.005      
          
Then, using table or technology,          
          
z =    -2.575829304      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    10.2      
z = the critical z score =    -2.575829304      
s = standard deviation =    0.12      
n = sample size =    24      
Then          
          
x = critical value =    10.13690533   [LOWER LIMIT]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.995      
          
Then, using table or technology,          
          
z =    2.575829304      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    10.2      
z = the critical z score =    2.575829304      
s = standard deviation =    0.12      
n = sample size =    24      
Then          
          
x = critical value =    10.26309467   [UPPER LIMIT]

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Thus, the limits are 10.13690533 and 10.26309467. [ANSWER]