the following table of grouped data represents the weight ( in pounds) of 100 computer towers. calculate the mean weight for a computer.

eddy2017

eddy2017

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Hi, this is a new twist to the types of central tendency exercises i have come across after starting studying the subject.
the following table of grouped data represents the weight ( in pounds) of 100 computer towers. calculate the mean weight for a computer.

This problem has me baffled in that I don't know what to do with the left column( weight). they give me a range of the weight, that I know, but what to do
weight # of computers
[3-5) ---------------8
{5-7)---------------25
[7-9)---------------45
[9-11)--------------18
[11-13)-------------4

I suppose I can find a mid value in every range given, right,
as in, [3-5) mid value 4, and so for and so on with the rest. but once i have done that
I might as well do it now
[3-5) -------------4
{5-7)--------------6
[7-9)-------------- 8
[9-11)--------------10
[11-13)-------------12
So, now, what do I do with the mid value, what is it useful for?.
thanks
eddy
 
BigBeachBananas said:
Are you learning about weighted average?
Click to expand...
no, but i can dig something up. wait. let's do this together. let me watch a tutorial and take some notes. hold on. I'll get back to you when I am ready with some facts. thanks, BBB
 
If that is the way the data was given to you, then the best you can do is find an ESTIMATE of the mean. You know, for example, that there are 8 computers with a weight between 3 and 5 kg. The total weight of these is somewhere between 8x3=24kg or just under 8x5=40kg or somewhere in between. You don't know. So the best you can do is assume they all weigh 4kg to give a total of 8x4=32kg.
 
Harry_the_cat said:
If that is the way the data was given to you, then the best you can do is find an ESTIMATE of the mean. You know, for example, that there are 8 computers with a weight between 3 and 5 kg. The total weight of these is somewhere between 8x3=24kg or just under 8x5=40kg or somewhere in between. You don't know. So the best you can do is assume they all weigh 4kg to give a total of 8x4=32kg.
Click to expand...
You're suggesting I multiply the mid value(the value is the fair middle/smack dab middle) times the amount of computers. it seems logical to me,
as in,
[3-5) ---------------8 = 8 * 4 =32
[5-7)---------------25 = 6 * 25=150
[7-9)---------------45 =8 * 45 =360
[9-11)--------------18 = 10*18=180
[11-13)-------------4 = 12 * 4 =48

so now, I add up all results(pounds ) and divide it into the number of computers.
is that it?
770 divided by
100 computers
the mean weight =7.7 lb approximately
 
Last edited:
eddy2017 said:
You're suggesting I multiply the mid value(the value is the fair middle/smack dab middle) times the amount of computers. it seems logical to me,
as in,
[3-5) ---------------8 = 8 * 4 =32
[5-7)---------------25 = 6 * 25=150
[7-9)---------------45 =8 * 45 =360
[9-11)--------------18 = 10*18=180
[11-13)-------------4 = 12 * 4 =48

so now, I add up all results(pounds ) and divide it into the number of computers.
is that it?
Click to expand...
Read:

www.statology.org

How to Estimate the Mean and Median of Any Histogram

This tutorial explains how to estimate the mean and median value for any histogram, including examples.
www.statology.org www.statology.org

Your data will create a histogram. The paper above gives you method of estimating the mean.
 
BigBeachBananas said:
Are you learning about weighted average?
Click to expand...
What was your idea about the weighted mean with this activity. I watch a basic tutorial. it is fairly easy. how would uyou apply it here or bette, how would you lead me to apply it here for a solution, I'm still interested.
 
eddy2017 said:
What was your idea about the weighted mean with this activity. I watch a basic tutorial. it is fairly easy. how would uyou apply it here or bette, how would you lead me to apply it here for a solution, I'm still interested.
Click to expand...
The calculation you did in post#6 is called a weighted average. We can examine it more closely.
[math]\frac{8*4 + 25*6+45*8+18*10+4*12}{100}[/math]Split up the fraction:
[math](4)*\frac{8}{100}+(6)*\frac{25}{100}+(8)*\frac{45}{100}+(10)*\frac{18}{100}+(12)*\frac{4}{100}[/math]We call the [imath]\frac{8}{100},\frac{25}{100},\frac{45}{100},\frac{18}{100},\frac{4}{100}[/imath] weights. For example, the group [3-5) and [5-7) have 8 and 25 observations, respectively. Since the group [5-7) has more observation than the [3-5) group, the [5-7) will contribute more towards the overall average, thus we assign a heavier weight to the [5-7) group compared to the [3-5) group,[imath]\frac{25}{100}>\frac{8}{100}[/imath]. The sum of the weights is always equal to 1.

A typical example is how your grades are calculated in school. Your overall grade is the weighted average of the tests and homework. Tests significantly impact your final grade than homework because schools assign a heavier weight on tests than homework.
 
eddy2017 said:
You're suggesting I multiply the mid value(the value is the fair middle/smack dab middle) times the amount of computers. it seems logical to me,
as in,
[3-5) ---------------8 = 8 * 4 =32
[5-7)---------------25 = 6 * 25=150
[7-9)---------------45 =8 * 45 =360
[9-11)--------------18 = 10*18=180
[11-13)-------------4 = 12 * 4 =48

so now, I add up all results(pounds ) and divide it into the number of computers.
is that it?
770 divided by
100 computers
the mean weight =7.7 lb approximately
Click to expand...
Yes that's the best estimate you can get if you don't have all the individual weights.
 
BigBeachBananas said:
The calculation you did in post#6 is called a weighted average. We can examine it more closely.
[math]\frac{8*4 + 25*6+45*8+18*10+4*12}{100}[/math]Split up the fraction:
[math](4)*\frac{8}{100}+(6)*\frac{25}{100}+(8)*\frac{45}{100}+(10)*\frac{18}{100}+(12)*\frac{4}{100}[/math]We call the [imath]\frac{8}{100},\frac{25}{100},\frac{45}{100},\frac{18}{100},\frac{4}{100}[/imath] weights. For example, the group [3-5) and [5-7) have 8 and 25 observations, respectively. Since the group [5-7) has more observation than the [3-5) group, the [5-7) will contribute more towards the overall average, thus we assign a heavier weight to the [5-7) group compared to the [3-5) group,[imath]\frac{25}{100}>\frac{8}{100}[/imath]. The sum of the weights is always equal to 1.

A typical example is how your grades are calculated in school. Your overall grade is the weighted average of the tests and homework. Tests significantly impact your final grade than homework because schools assign a heavier weight on tests than homework.
Click to expand...
I see. Yes, as I had told you I had watched a tutorial on weighted average. Not really difficult. But I see its use to solve the problem at hand. Thanks,BBB
 
eddy2017 said:
You're suggesting I multiply the mid value(the value is the fair middle/smack dab middle) times the amount of computers. it seems logical to me,
as in,
[3-5) ---------------8 = 8 * 4 =32
[5-7)---------------25 = 6 * 25=150
[7-9)---------------45 =8 * 45 =360
[9-11)--------------18 = 10*18=180
[11-13)-------------4 = 12 * 4 =48

so now, I add up all results(pounds ) and divide it into the number of computers.
is that it?
770 divided by
100 computers
the mean weight =7.7 lb approximately
Click to expand...
Please do not write equations like 8 = 8*4, 25 = 6*25, 45 = 8*45, 18 = 10*18 and 4=12*4.
It does not matter whether or not one understand what you mean. It is wrong and no argument from you or any else will make it correct.
 
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