SOLUTION: Boxes of a certain cereal say they hold 15 oz. To make sure they​ do, the manufacturer fills the box to a mean weight of 15.3 ​oz, with a standard deviation of 0.15 oz.
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-> SOLUTION: Boxes of a certain cereal say they hold 15 oz. To make sure they​ do, the manufacturer fills the box to a mean weight of 15.3 ​oz, with a standard deviation of 0.15 oz.
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Question 1125900: Boxes of a certain cereal say they hold 15 oz. To make sure they do, the manufacturer fills the box to a mean weight of 15.3 oz, with a standard deviation of 0.15 oz. If the weights have a normal curve, what percentage of boxes actually contain 15 oz or more? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mean is 15.3
standard deviation is .15
you want to know what percentage of boxes actually contain 15 ounces or more.
the following graph shows that it will be 97.72%.
if you want to find the z-score, you would do the following.
z = (x-m)/s
z is the z-score
m is the mean
x is the raw score
s is the standard deviation.
formula becomes z = (15 - 15.3) / .15 = -2
look up in the z-score table for the area to the left of a z-score of -2 to find that the area to the left of the z-score is .0228.
the area to the right of that z-score is therefore 1 - .0228 = .9772
the z-score calculator i used will show you that as well, as shown below.
it appears i'm having trouble uploading the graphs.
rest assured that the probability is 97.72%.
i'll have to do some troubleshooting and will update the solution when i'm successful.
in the meantime, look up the table and you'll find that the area to the left of a z-score of -2 is .0228 which means that the area to the right of that z-score score is .9772.
