The distribution of actual weights of 8 oz. wedges of cheddar cheese produced by a certain company is normal with mean 8.1 ounces and standard deviation 0.1 ounces.
There is only a 5% chance that the average weight of the sample of 5 of the cheese wedges will be below:?
I know the answer is 8.03 oz. but how do I get there?
There is only a 5% chance that the average weight of the sample of 5 of the cheese wedges will be below:?
I know the answer is 8.03 oz. but how do I get there?
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You have µ = 8.1 ounces and σ = 0.1 ounces. For samples of size 5, the mean will still be 8.1 ounces, but the standard deviation will be
s = σ/√5 = 0.1/2.236067977 = 0.04472136
The z score for 5% is -1.6449, so we have
z = ( x - µ)/s
-1.6449 = (x - 8.1)/0.04472136
x - 8.1 = (-1.6449)(0.04472136) = -0.073562165
x = 8.1 - 0.073562165 = 8.026437835
That rounds to x = 8.03
s = σ/√5 = 0.1/2.236067977 = 0.04472136
The z score for 5% is -1.6449, so we have
z = ( x - µ)/s
-1.6449 = (x - 8.1)/0.04472136
x - 8.1 = (-1.6449)(0.04472136) = -0.073562165
x = 8.1 - 0.073562165 = 8.026437835
That rounds to x = 8.03
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The z score is obtained from a table for the standard normal distribution.
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