A grain loader can be set to discharge grain in amounts that are normally distributed, with mean u kg and standard deviation equal to 700kg. If a company wishes to use the loader to fill containers that hold 54,440kg of grain and wants to overfill only one container in 100, at what value of u (mean) should the company set the loader?
If someone can please explain HOW to do the problem, that would help me out a lot!
If someone can please explain HOW to do the problem, that would help me out a lot!
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Yu will be looking in tables of Z for a Normal Distribution.
First standardize the standard deviation 700/54440 = 0.01286
In other words you are expressing the S.D as a fraction of the mean.
Now look up a table of Z values to be found in most text books or statistical tables
for a prob of 1/100 or 0.01 we get reading of 2.3263.
Now this is a standardized reading so in terms of grain we are talking of 2.3263 x 700 = 1628.41
so the amount that should be loaded is 54,440 - 1.628 (ignoring the decimals) = 52,812kg
Good luck with your studies
First standardize the standard deviation 700/54440 = 0.01286
In other words you are expressing the S.D as a fraction of the mean.
Now look up a table of Z values to be found in most text books or statistical tables
for a prob of 1/100 or 0.01 we get reading of 2.3263.
Now this is a standardized reading so in terms of grain we are talking of 2.3263 x 700 = 1628.41
so the amount that should be loaded is 54,440 - 1.628 (ignoring the decimals) = 52,812kg
Good luck with your studies
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If you want to assure only 1/100 or 0.01 are overfilled, then start by finding the z-score in the right tail that is associated with 0.01.
z = 2.326 or equivalently that is how many standard deviations you want.
2.326 x 700 kg = 1628 kg, now subtract that value from 54,440kg and you get 52,812 kg and that is the mean you want.
Why does that work? Because the mean + 2.326 x 700 must equal the limit of the container. And, we know that 2.326 standard deviations will cover all situations except 1%.
Hope that helps
z = 2.326 or equivalently that is how many standard deviations you want.
2.326 x 700 kg = 1628 kg, now subtract that value from 54,440kg and you get 52,812 kg and that is the mean you want.
Why does that work? Because the mean + 2.326 x 700 must equal the limit of the container. And, we know that 2.326 standard deviations will cover all situations except 1%.
Hope that helps