Calculate the mean weight for a computer

eddy2017

eddy2017

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hi, dear tutors and teachers:
The following table of grouped data represents the weight (in pounds) of 100 computer towers. Calculate the mean weight for a computer.
Weight (pounds) Number of Computers
[3 - 5)------------------------------8

[5 - 7)------------------------------25

[7 - 9)------------------------------45

[9 - 11)----------------------------18

[11 - 13)---------------------------4

this totally deviates from the typical mean exercise. can you help work out a solution for this?. i do not understand it .
thanks,
eddy
 
topsquark said:
Please pardon the following (unintentional) pun: Do you know what the weighted mean is?

-Dan
Click to expand...
yes, sir. i do know the definition.
instead of each data point contributing equally to the final mean some data points contribute more weight than others. that is basically it, right?
but i have never used that or seen that before.
if you want me to watch the tutorial you attached in the link i can do that.
sometimes with me it is more about the approach used to do an exercise, how would you approach this one, you don't need to tell me the answer just guide me.
thanks
 
Because it is grouped data, you don't have the actual weights. For example, you know that there are 8 computers with weights between 3-5 pounds, but you don't know their individual weights. The best you can do is assume they were 4 pounds each. To give a total of 4x8=32 pounds. Then do the same for the others to find the weighted mean.
 
Harry_the_cat said:
Because it is grouped data, you don't have the actual weights. For example, you know that there are 8 computers with weights between 3-5 pounds, but you don't know their individual weights. The best you can do is assume they were 4 pounds each. To give a total of 4x8=32 pounds. Then do the same for the others to find the weighted mean.
Click to expand...
i tink i watch the video on your link and then try to find it and report back to you.
thanks a lot. that was a good tip!
 
40 is our weighted mean.
I followed your advice. find the middle number in each one, add'em up and found them to add up to 40.
40 is the mean, then.
 
eddy2017 said:
40 is our weighted mean …
Click to expand...
40 pounds is too much for one computer, Eddy. Most of the 100 computers have an averaged weight of 6, 8, or 10 pounds, so I'd expect the average of any one computer to be about 8 pounds. (Thinking of 'central tendencies', 8 is the average of 6, 8 and 10.)

You had added the averaged weights of five computers, taking one from each group, right? (4+6+8+10+12=40) That is the weight of those five computers together. But we're asked to find the average weight of just one computer (in the set of 100 computers). Therefore, we should average the weights of all 100 computers (in other words, we'll be averaging the 100 "averaged weights").

What is the total weight of all 100 computers? Harry_the_cat showed how to find the combined weight of the first eight computers (the ones in the group that weigh 4 pounds each, on average). That group of eight computers weighs 32 pounds.

How much do the other groups weigh? Next, what do you get for the weight of all 100 computers? Use that total weight to calculate the average weight of one computer.

?
 
hi, i did not respond last night cos it was too late.
here's my work
following Harriett's tip i did the same thing with all the weights given. find the middle weight between the range in weight given
that amounted to:
32+150+360+180+48 = 770 pounds in all
I divided 770 by the number of computers (100) to find the weighted mean
that resulted in 7.7 pounds each.

if this is right let me know, and thanks a lot to both of you cos without your hints i would have never been able to solve it. it was very good.
 
The arithmetic is fine. And 7.7 pounds is very likely the answer desired. However, as a practical matter based on long experience in presenting quantitative data, I’d report it as “approximately 7.7 pounds.” The reason is that you have made a reasonable assumption that within each group the mean for that group was the group’s midpoint, but you do not know that for a fact.
 
I am not a professor, but I ran a mergers and acquisitions department for a number of years. I learned to be very careful in distinguishing between estimates and facts.
 
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