anonymous
  • anonymous
suppose certain coins have weights that are normally distributed with a mean of 5.767 g and a standard deviation of 0.076 g. A vending machine is configured to accept those coins with weights between 5.677 g. and 5.857 g a. If 290 different coins are inserted into the vending machine, what is the expected number of rejected coins? (round to the nearest integer
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmale
  • mathmale
Melissa, I tried to help you, but left when I received no response from you for 15-20 minutes. Can you go back to your original posting of this question?
anonymous
  • anonymous
I was trying to find the z score
anonymous
  • anonymous
I am slow :(

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anonymous
  • anonymous
I found the
mathmale
  • mathmale
yes? found the z-score? Please share y our work.
anonymous
  • anonymous
yes
anonymous
  • anonymous
5.857 - 5.767/0.076 = 0.09/0.076= 1.2
anonymous
  • anonymous
right?
anonymous
  • anonymous
then 5.677 - 5.767 / 0.076 = 0.09/0.076 = -1.2
anonymous
  • anonymous
so my z scores would be 1.2 and -1.2
mathmale
  • mathmale
At first glance it appears to be correct. So, 5.857 grams is 1.2 standard deviations above the mean. Use a table of z scores to find the AREA under the nomal curve to the left of z=1.2. Then, do the same for the left boundary: find the AREA under the normal curve to the left of z=-1.2. Subtract the smaller area from the larger. The
mathmale
  • mathmale
result is the area under the normal curve between z=-1.2 and z=1.2.
anonymous
  • anonymous
give me a second. I am very new at this. Going to do my best
anonymous
  • anonymous
having a hard time :(
anonymous
  • anonymous
Which column of numbers do I look at?
anonymous
  • anonymous
I have found 1.2 and -1.2 but then I do not know what to look for after that?
mathmale
  • mathmale
Have you found the AREA to the left of z=1.2? If so, what is that area? What is the AREA to the left of z=-1.2?
anonymous
  • anonymous
do you want me to look on the z score table?
mathmale
  • mathmale
Yes. Do you have a z-score table in front of y ou? If not, look at this one: https://www.google.com/search?q=table+of+z+scores&espv=2&tbm=isch&imgil=jqtSneBuWDdlbM%253A%253BA_vCxSVFbVWILM%253Bhttp%25253A%25252F%25252Fcosstatistics.pbworks.com%25252Fw%25252Fpage%25252F27425647%25252FLesson%25252525200311&source=iu&pf=m&fir=jqtSneBuWDdlbM%253A%252CA_vCxSVFbVWILM%252C_&biw=1360&bih=673&usg=__3DTZCKl9SIeRTcEe_T8ubTfW5LE%3D&ved=0ahUKEwifl_it5tbJAhUL9mMKHWnaDYQQyjcILA&ei=BFNsVp-FGYvsjwPptLegCA#imgrc=jqtSneBuWDdlbM%3A&usg=__3DTZCKl9SIeRTcEe_T8ubTfW5LE%3D
anonymous
  • anonymous
If so, I do not know what to look at. I am having a hard time using the chart.
mathmale
  • mathmale
You are to find 1.2 in the "z" column. Are you comfortable doing that?
anonymous
  • anonymous
I found 1.2
anonymous
  • anonymous
now what do I look for?
mathmale
  • mathmale
good. immediately to the right of the z column is a column marked 0.00. start at the top of this column and move downward to the entry next to z=1.2. What decimal fraction do you see there? Type that in here.
anonymous
  • anonymous
0.8849
mathmale
  • mathmale
beautiful.
mathmale
  • mathmale
That's the area to the left of z=1.2
anonymous
  • anonymous
0.0985
anonymous
  • anonymous
that was for -1.2
mathmale
  • mathmale
Great. Now subtract the smaller area from the larger area.
anonymous
  • anonymous
0.7864?
mathmale
  • mathmale
right. very, very good. That fraction represents the fraction of the total number of coins whose weights are within -1.2 and 1.2 standard deviations .
anonymous
  • anonymous
So is that my answer?
anonymous
  • anonymous
78 coins?
anonymous
  • anonymous
or would it be 79 coins?
mathmale
  • mathmale
The experimenter takes a sample of 290 coins. To find the expected number of coins that are between the two given coin weights, multiply 290 by 0.7699 and round off your answer to the nearest whole number. Is that how you got 78 / 79?
anonymous
  • anonymous
how did you get 0.7699?
anonymous
  • anonymous
I got 0.7864
mathmale
  • mathmale
Right. I got my .7699 on my calculator. There's an arithmetic error somewhere. The important thing is that you know where this fraction comes from and what it means.
mathmale
  • mathmale
Try again: Look up z=-1.2 and copy down the fraction immediately next to it.
anonymous
  • anonymous
0.8849
anonymous
  • anonymous
0.0985
anonymous
  • anonymous
so I would take 0.8849 - 0.0985 = 0.7864
mathmale
  • mathmale
That seems to be where the error is; I get as area to the left of z=-1.2 the fraction 0.1151, whereas you've obtained 0.0985.
anonymous
  • anonymous
then take hmm wonder why?
anonymous
  • anonymous
let me look again
mathmale
  • mathmale
Wish I could see the z-score table you're using. I don't have one in front of me; my areas come from a table found on the Internet.
anonymous
  • anonymous
–1.2 0.0985 0.1003 0.1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.1151
anonymous
  • anonymous
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
mathmale
  • mathmale
Without actually seeing your z table, I can't do much problem shooting. You found z=-1.2 in your table, and then from the column immediately next to it, found 0.0985, whereas I find 0.1151.
anonymous
  • anonymous
yes. Should I use a different table?
mathmale
  • mathmale
Everything else we've done up to here has been fine. Have you a friend or fellow student with whom you could discuss this discrepancy?
mathmale
  • mathmale
It'd be worth looking for a different table, yes. The table you use MUST show negative z scores; for example, it must show z=-1.2.
anonymous
  • anonymous
mine did
anonymous
  • anonymous
http://www2.parkland.edu/businesstraining/documents/zscoretable.pdf
mathmale
  • mathmale
Melissa, I'm not sure what to tell you at this point. I've used my calculator and have come u p with the two areas 0.8849 and 0.1151, whose difference is 0.7698.
anonymous
  • anonymous
there is the link to my table
mathmale
  • mathmale
thank you so much for sharing that. Problem solved!!! Look up z=-1.2, and then look on that line in the column marked 0.000.
mathmale
  • mathmale
Your previous result came from the wrong column, the one marked 0.09.
anonymous
  • anonymous
0.1151?
anonymous
  • anonymous
oh I see
anonymous
  • anonymous
So I need to look for 0.00?
mathmale
  • mathmale
yes! So, now your results are exactly the same as the ones I've gotten from my calculator. the area under the normal curve between z=-1.2 and z=1.2 is 0.7698. Multiply that by the number (290) in the coin sample. What do you get? Don't round off yet.
anonymous
  • anonymous
my new answer with the right numbers is 0.8849 - 0.1151 = 0.7698
mathmale
  • mathmale
Same here, exactly the same. multiply that by 290.
anonymous
  • anonymous
223.242
mathmale
  • mathmale
should that be rounded up or down? why?
anonymous
  • anonymous
I have no idea!
mathmale
  • mathmale
Hint: 223.4999 would be rounded down, but 223.50001 would be rounded up. If you have 223.24, would you round up or down?
anonymous
  • anonymous
down
mathmale
  • mathmale
Thanks for sticking with me thru this long procedure. Round 223.24 down to 223 (because .24 is less than 0.50). What does y our answer, 223, signify?
anonymous
  • anonymous
number of coins?
mathmale
  • mathmale
number of coins that .... what? what is the significance?
anonymous
  • anonymous
this is so confusing sometimes
anonymous
  • anonymous
that may be rejected
mathmale
  • mathmale
actually, 223 is the number of acceptable coins; to find the number of coins that must be rejected, subtract 223 from 290 (the total number). Result?
anonymous
  • anonymous
do I take that number and subtract it from the total number of coins?
anonymous
  • anonymous
YAY!!
anonymous
  • anonymous
I got 67
mathmale
  • mathmale
yes. 67 coins would be rejected; 223 would be accepted. congrats! We have not discussed every detail involved in this problem, but I think you've gotten a good first exposure to the tasks at hand.
mathmale
  • mathmale
Have to move on. Thanks for your perseverance. Would be happy to work with you again in the future. Bye!
anonymous
  • anonymous
thank you!!
mathmale
  • mathmale
You're welcome!
anonymous
  • anonymous
I got it wrong?
anonymous
  • anonymous
it said the correct answer was 69
mathmale
  • mathmale
We are so close that we can assume our approaches have been correct. Different results, in a problem such as this one, are most likely due to round-off error.
mathmale
  • mathmale
I don't think you did anything wrong.
mathmale
  • mathmale
Personally I think you should move on to answering other questions, since your result was so close.
anonymous
  • anonymous
It wants to know if 290 different coins are inserted into the vending machine, what is the probability that the mean falls between the limits of 5.677 and 5.857
mathmale
  • mathmale
Melissa: Wait. I multiplied 290 by 0.7698 and got 223.242. Subtracting this from 290, I got 66.758. Round that off, please.l
mathmale
  • mathmale
We have already answered the question you've re-typed: that probability is the area under the nomral curve between z=-1.2 and z=1.2 and is 0.7698, which rounds up to 0.77. that's a probability.
anonymous
  • anonymous
I got the same thing
mathmale
  • mathmale
Good. Then that's our answer. The prob. that the coin weights will be between z=-1.2 and z=1.2 is 0.77.
anonymous
  • anonymous
it said that it was 0.9998
anonymous
  • anonymous
I am missing something.
mathmale
  • mathmale
that response makes no sense to me at all. What was the question? copy and paste it here.
anonymous
  • anonymous
if 290 different coins are inserted into the vending machine, what is the probability that the mean falls between the limits of 5.677 and 5.857
mathmale
  • mathmale
Now I see. Your 290 is a SAMPLE, not a population, and so the standard deviation is not the same as it would be for the population. Does any of this sound familiar?
anonymous
  • anonymous
yes. I am becoming familiar with sample vs population.
mathmale
  • mathmale
The sample standard deviation is \[ssd=\frac{ \mu }{ \sqrt{n} }\]
mathmale
  • mathmale
does that look familiar?
anonymous
  • anonymous
OMG!
anonymous
  • anonymous
NO lol
anonymous
  • anonymous
looks like Japanese to me!
mathmale
  • mathmale
;) You already have mu, the population mean, and you know the sample size is 290. find the ssd.
mathmale
  • mathmale
So, you're not Japanese? ;)
anonymous
  • anonymous
american girl here
anonymous
  • anonymous
cant read japanese or statistics lol
anonymous
  • anonymous
how do I have the population mean?
anonymous
  • anonymous
This stuff makes me feel so stupid!
mathmale
  • mathmale
I used to hate stats, but once I learned it in some depth, grew to like the subject.
mathmale
  • mathmale
Look at the original problem statement. It gives y ou the population mean and population std. dev.
mathmale
  • mathmale
We go thru the same thing again: calculate z scores and then calculate the area between 2 z scores.
mathmale
  • mathmale
the only difference is that the std. dev. we use is \[\frac{ \mu }{ \sqrt{n} }\]
mathmale
  • mathmale
Find the new z score if the population mean is 5.767 and the std. dev. is the sample std. dev, discussed immediately above. You'll get 2 areas, one of which is the area to the left of your larger z score, and the other is the area to the left of your smaller z score. Subtract. The result is the probability that you were looking for.
mathmale
  • mathmale
I've been on OpenStudy for hours already and would like to get off now. However, you can continue to type messages about this problem to me here; either someone else or I will pick it up later on.

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