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