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OpenStudy (anonymous):
@Blank
OpenStudy (anonymous):
What do u need help with? :)
OpenStudy (anonymous):
In a random sample of 700 refreshment-dispensing machines, it is found that an average of 8.1 ounces is dispensed with a natural deviation of 0.75 ounce.
28. Find the standard error of the mean. Round your answer to the ten-thousandths place.
OpenStudy (anonymous):
29. Find the probability that the mean of the population will be less than 0.085 ounce from the mean of the sample. Round your answer to the tenths place.
OpenStudy (anonymous):
No. just an answer.
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OpenStudy (anonymous):
30. Find the probability that the true mean is between 8.157 and 8.185. Round your answer to the tenths place.
OpenStudy (anonymous):
And thats it
OpenStudy (anonymous):
Ok
OpenStudy (anonymous):
The answer to number 29 is a percent
OpenStudy (anonymous):
thanks. and 30?
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OpenStudy (anonymous):
it says round the answer to the tenths place @jim_thompson5910 can you help?
OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
which one?
OpenStudy (anonymous):
can you help on #30?
jimthompson5910 (jim_thompson5910):
convert those raw scores to z scores
use: z = (x-mu)/sigma
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jimthompson5910 (jim_thompson5910):
tell me what z scores you get
OpenStudy (anonymous):
turn 8.157 and 8.185 into z scores right?
jimthompson5910 (jim_thompson5910):
yeah
OpenStudy (anonymous):
sigma=700 right?
jimthompson5910 (jim_thompson5910):
n = 700 (sample size)
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to find the area to the left of z = 0.08 (I rounded z = 0.076 to 2 decimal places)
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
now what? @jim_thompson5910
jimthompson5910 (jim_thompson5910):
what area did you get?
OpenStudy (anonymous):
Wait Im confused
jimthompson5910 (jim_thompson5910):
about what
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OpenStudy (anonymous):
From where you said use the table, what did you want me to do?
jimthompson5910 (jim_thompson5910):
the table lets you find the area under the curve to the left of a given z score
jimthompson5910 (jim_thompson5910):
do you remember how to read the table?
OpenStudy (anonymous):
ok but what score do i find?
jimthompson5910 (jim_thompson5910):
z = 0.08
locate the 0.0 row, then find the 0.08 column
the value in that row/column combo is the area to the left of z = 0.08
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jimthompson5910 (jim_thompson5910):
look on page 2
OpenStudy (anonymous):
0.5319
jimthompson5910 (jim_thompson5910):
what is the area to the left of z = 0.11
jimthompson5910 (jim_thompson5910):
I'm getting z = 0.11 from rounding z = 0.113
OpenStudy (anonymous):
what area? on the graph?
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jimthompson5910 (jim_thompson5910):
the table provides the value
jimthompson5910 (jim_thompson5910):
this whole table tells you the area to the left of a given z score
OpenStudy (anonymous):
0.0?
jimthompson5910 (jim_thompson5910):
0.11 starts with 0.1
look in the 0.1 row
OpenStudy (anonymous):
ok then what
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jimthompson5910 (jim_thompson5910):
look in the column that finishes up z = 0.11
jimthompson5910 (jim_thompson5910):
0.1 is already taken care of, so the missing bit is that last 1
that's what the column 0.01 means
ie
0.11 = 0.1 + 0.01
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
0.5438
jimthompson5910 (jim_thompson5910):
now subtract the two areas
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OpenStudy (anonymous):
0.5319-0.5438=-0.0119
OR 0.5438-0.5319=0.0119
jimthompson5910 (jim_thompson5910):
the area will be positive, so 0.5438-0.5319=0.0119
jimthompson5910 (jim_thompson5910):
the area between z = 0.08 and z = 0.11 is roughly 0.0119
jimthompson5910 (jim_thompson5910):
that means the area between x = 8.157 and x = 8.185 is also roughly 0.0119
OpenStudy (anonymous):
Ok now what
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jimthompson5910 (jim_thompson5910):
the area under the curve is the same as finding the probability (think of randomly throwing a dart onto the right area)
jimthompson5910 (jim_thompson5910):
Find the probability that the true mean is between 8.157 and 8.185. Round your answer to the tenths place.
0.0119 rounds to 0.01
that's roughly a 1% chance
OpenStudy (anonymous):
And thats the answer? 1%?
jimthompson5910 (jim_thompson5910):
oh wait, not 0.01...it should be 0.0
hmm seems odd
jimthompson5910 (jim_thompson5910):
let me recheck
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OpenStudy (anonymous):
ok
jimthompson5910 (jim_thompson5910):
oh I used the wrong standard error
jimthompson5910 (jim_thompson5910):
z = (x - mu)/(sigma/sqrt(n))
z = (8.157 - 8.1)/(0.75/sqrt(700))
z = 2.01
jimthompson5910 (jim_thompson5910):
what's the area to the left of z = 2.01 ?
OpenStudy (anonymous):
there is no colomn for 2.01
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jimthompson5910 (jim_thompson5910):
combine row 2.0 with column 0.01
notice how they add to 2.01
2.0 + 0.01 = 2.01
OpenStudy (anonymous):
oh
OpenStudy (anonymous):
0.9778
jimthompson5910 (jim_thompson5910):
each row/column combo gives you pretty much every z value from z = -3.49 to z = 3.49
jimthompson5910 (jim_thompson5910):
going in increments of 0.01
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jimthompson5910 (jim_thompson5910):
correct
jimthompson5910 (jim_thompson5910):
z = (x - mu)/(sigma/sqrt(n))
z = (8.185 - 8.1)/(0.75/sqrt(700))
z = 2.9985
z = 3.00
jimthompson5910 (jim_thompson5910):
find the area to the left of z = 3.00
OpenStudy (anonymous):
0.9987
jimthompson5910 (jim_thompson5910):
subtract the two
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OpenStudy (anonymous):
0.0209
jimthompson5910 (jim_thompson5910):
That still gives 0.0, how annoying
honestly, it's better to write it to at least 2 decimal places
OpenStudy (anonymous):
ok i will do that
OpenStudy (anonymous):
Thank you for all of your help. I greatly appreciate it! :)