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birth weight in US are normally distributed with mean of 340g and a strandard deviation of 45g. If a hospital plans to set up observation condition for the lightest 7% of the babies, what weight is used for the cutoff separating the lightest 7% from the others?
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Hi aric
Welcome to the forum!
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birth weight in US are normally distributed with mean of 340g and a strandard deviation of 45g. If a hospital plans to set up observation condition for the lightest 7% of the babies, what weight is used for the cutoff separating the lightest 7% from the others?
What is the z-score for "7%"?
Plugging this into the equation for computing z-score from the mean, the test value, and the standard deviation, what value do you get for the test value?
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Hi aric
A good question!
Are you asking what they use - whether right or wrong?
Or are you asking what they ought to be using?
The thing about "standard distribution" is this:
It is a FORMULA that people would LIKE real measurements to APPROXIMATE
They adjust the numbers (mean and standard deviation) to "best fit" the available data.
The data fits well when near the mean. There are lots of results near there.
But it is a bodge
As far away as 7% the data are far fewer and the fit is bad - and nobody cares!
The farther we are from the mean the less we know and the fewer is the data!
A couple of standard deviations is probably as far as we usually should trust the numbers and the "normal distribution" idealisation.
What the hospital should do, politically, is ASSUME the misfits to the normal distribution are entirely "experimental errors"!!! That satisfies their bosses
What they ought to do is record ALL the birth weights and then SELECT the 7% lightest.
This will vary as the data varies from year to year and from state to state, of course.
Maybe last years results might HELP guess this year's results. Maybe not.
Last year's results in Illinois MIGHT be a better match to this year's in Michigan.
Depends on the weather and all else that maybe controls birth weights.
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Hi aric;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi aric,
Welcome to the forum.
As it would not be possible to have a normal distribution table for every set of means and standard deviations you need to begin by converting your figures into the 'standard' ones, ie. mean = zero and sd = 1
If the value you seek is 'x' then the conversion is
where mu is the mean (340) sigma is the sd (45) and phi is the function that converts this into a probability.
Use this page:
http://www.mathsisfun.com/data/standard … table.html
This interactive graph shows the probabilities as percentages in three possible ways. Click the 'up to z' button to get the 7% area you want. There is a more accurate table of values below the graph.
Once you have z, you can solve for x
Bob
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