SOLUTION: I'm having difficulty with the following question. Any help is appreciated! Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic f

Question 1117556: I'm having difficulty with the following question. Any help is appreciated!
Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill bags. Weights of the filled bags are approximately normally distributed with a mean of 50 kilograms and a standard deviation of 1.25 kilograms.

a. What is the probability that a filled bag will weigh less than 49.5 kilograms?
b. What is the probability that a randomly sampled filled bag will weight between 48.5 and 51 kilograms?
c. What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled?
d. CCI 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 2% of all filled bags weigh more than 52 kilograms?
Thank You
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Use the normal distribution and its table of z-scores
:
We are given population mean is 50 kilograms and population standard deviation is 1.25 kilograms
:
a) z-score = (49.5 - 50.0) / 1.25 = -0.4
:
lookup the associated probability with the calculated z-score
:
Probability (P) (X < 49.5) = 0.3446
:
b) P ( X > 48.5 and X < 51 ) = P (X < 51) - P (X < 48.5)
:
z-score(51) = (51 - 50) / 1.25 = 0.8
:
z-score(48.5) = (48.5 - 50.0) / 1.25 = -1.2
:
P (X < 51) - P (X < 48.5) = 0.7881 - 0.1151 = 0.673
:
c) 1 - 0.15 = 0.85
:
z-score associated with probability 0.85 is 1.04
:
1.04 = (X-50.0) / 1.25
:
X -50.0 = 1.3
:
X = 51.3 kilograms
:
d) 1 - 0.02 = 0.98
:
z-score associated with probability of 0.98 is 2.06
:
2.06 = (52.0-50.0) / standard deviation
:
standard deviation = 2 / 2.06 = 0.9708 kilograms
:
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