Question

Doggie Nuggets Inc. (DNI) sells large bags of dog food to warehouse clubs. DNI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 56 kilograms and a standard deviation of 0.83 kilograms. Complete parts a through d below.

DNI is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 8% of all filled bags weigh more than 58 kilograms?

Answer #1

Mean, = 56 kg

Let the adjusted standard deviation be

Let X denote the weight of a randomly selected filled bag

Given : P(X > 58) ≤ 0.08

Corresponding Z value = =

Corresponding to probability of 0.08, the critical z value = 1.405

Thus, = 1.405

-> = 2/1.405 = 1.42

So the standard deviation need to be less than or equal to 1.42 kilograms so that no more than 8% of all filled bags weigh more than 58 kilograms

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