Question 1167005: Cannot figure out how to figure this one out. Any help is appreciated!
Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed with a mean of 12 million dollars and a standard deviation of 1.1 million dollars.
One health care information systems company considers a quarter a "failure" if its sales level that quarter is in the bottom 20% of all quarterly sales levels.
Determine the sales level (in millions of dollars) that is the cutoff between quarters that are considered "failures" by that company and quarters that are not. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! first look up in the z-score tables for an area to the left of the z-score of .2.
the z-score table that i used can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
you will find that you have two areas, one above .2 and one below .2
specifically, .....
z-score of -.84 has an area of .20045 to the left of it.
z-score of -.85 has an area of .19766 to the left of it.
you could pick a z-score of -.84 because the area of .20 is closest to it, or you could try to get a little closer by interpolation.
the difference between .20045 and .19766 is equal to .00279.
the difference between a z-score of -.84 and -.85 is equal to .01
.2 is different from .20045 by .00045 units.
the ratio between this and .00279 is equal to .1612903226.
apply this ratio to the difference between the z-scores of -.84 and -.85 = .01 to get .1612903226 * .01 = .0016129032.
since -.84 is more than .2 and since -.85 is less than .2, you have to get somewhere between -.84 and -.85.
add .0016129032 to .84 to get a z-score of -.8416129032.
that's your z-score to the closest you can get to it manually.
now you can use the z-score formula to find the raw score.
z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation or the standard error.
in this case, s is the standard deviation.
when z = -.8416129032, the formula becomes:
-.8416129032 = (x - 12) / 1.1
multiply both sides of this equation by 1.1 and then add 12 to both sides of this equation to get x = 11.07422581.
that's your answer to the closest you can get with manual interpolation.
it's also the hard way to get your answer.
the easy way to get your answer is by using a normal distribution z-score calculator.
one that i found online is the omni normal distribution calculator.
it can be found at https://www.omnicalculator.com/statistics/normal-distribution
you clear it by hitting the round arrow button in the middle of the display under the input and results area.
for your problem, you would enter the following:
mean = 12
standard deviation = 1.1
p(x < X) = .2
the calculator will automatically fill in the rest of the fields.
here's a display of what i got after making the above entries.
it told me that my raw score is equal to 11.07407 and that my z-score is -.841756.
that compares favorably to the z-score that i derive manually of -.8416129032.
it also compares favorable to the raw score that i derived manually of 11.07422581.
it also does it with a lot less effort.
unless you are instructed to do it manually, use the calculator.
i also have a normal distribution calculator in my TI-84 Plus.
it told me that the raw score is 11.07421664.
a word about these calculators; .....
they don't all use the same algorithms, so there will be slight differences between them as to what the answer should be.
usually the differences pop up after 2 or 3 decimal places.
in most cases, your answer to 2 decimal places is close enough.
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