I cant do my 10yr old sons maths homework!!!!

[davido140]

He's got a little project to do some data handling..

Easy peasy thinks me...

Firstly, read a page of his book, and write a tally of word lengths to get some data..

He's reading a book on the Battle of Britain, and ended up with this data.. (the "word length fields were defined by the teacher)

Word Length Occurence
1-3 38
4-6 34
7-9 10
10+ 0


He needs to create a bar chart... easy peasy, write some observations on the data... easy peasy... find the mean, median and mode of the data.. easy peasy... or not..

Median, easy thats 4-6, the "one in the middle" of his range of data

Mode, easy, thats 1-3, the one that occurs most often

Mean, erm, how do I add up 1-3, 4-6, 7-9 in order to divide them by 82?

Should I add, 1,2,3,4,5,6,7,8,9 and divide? that makes the average word length 0.54 letters!!!!

Is it even possible?

christ I'm thick. heeelp!

Edited by davido140 on Friday 4th June 10:55

[cs02rm0]

Odd question I'd say.

On that data, I'd just take the mean of each range of lengths and number of occurences to then total and divide by the number of words.

Edit: Think I've explained that dreadfully, but I'm saying:

(2*38 + 5*34 + 8*10) / 82 ~= 4

Edited by cs02rm0 on Friday 4th June 11:02

[Simbu]

To work it out completely accurately, you need to total up all the characters in each word, then divide by the word count. Because you are using ranges it will be less accurate.

You could go on the middle value of each range, so all words that are 1-3 in length count towards 2 in your character total, for example. This will be inaccurate though and not a true reflection on the average word length.

Editing - speeeling fail

Edited by Simbu on Friday 4th June 10:59

[LocoCoco]

You could just count how many letters total there are and then divide by the number of words.

[paul.deitch]

multiply the mid point of each range element by the count of the items in the range element

[Odie]

davido140 said:

Mean, erm, how do I add up 1-3, 4-6, 7-9 in order to divide them by 82?

I think you divide by the number of groups so divide it by 4 = 20.5

ETA - thinking about it that makes no sense, to the google mobile.

Edited by Odie on Friday 4th June 11:03

[marctwo]

I think you'll have to use the actual word-length values rather than the ranges, especially as the 10+ range has no upper bound.

[Road2Ruin]

I would count up the total number of letter and devide by the number of words giving you and average letter count for a word, it will then fit into one of the brackets.

Pete

[Justayellowbadge]

marctwo said:

I think you'll have to use the actual word-length values rather than the ranges, especially as the 10+ range has no upper bound.

It also has no values, so it's a null point.

[blinkythefish]

(2*38 + 5*34 + 8*10) / 82 = 3.98? It assumes an average for each bin which is the same as the median for each bin, but theres not muche else you can assume unless you have the number of words for each length......

Edit: Doh, just saw cs02rm0 edit.

Edited by blinkythefish on Friday 4th June 11:07

Edited by blinkythefish on Friday 4th June 11:09

[ClareC]

Given that its a school maths excericse and you have previously been asked to count the word length frequencies in data ranges.....I'm guessing they want you to work out the mean using the data ranges too.


So taking the mid-points of each data range:

Mean = (38*2+34*5+10*8)/82 = 3.975

The fact that the top category has no upper bound is irrelevant in this example as the frqwuency is 0 anyway.

[Deva Link]

marctwo said:

I think you'll have to use the actual word-length values rather than the ranges, especially as the 10+ range has no upper bound.

...that's not an observation on the data though - it would be going back a stage to the source.

[LocoCoco]

Road2Ruin said:

I would count up the total number of letter and devide by the number of words giving you and average letter count for a word, it will then fit into one of the brackets.

Pete

^^^^^

[Odie]

http://www.gcse.com/maths/median.htm


So (4+5+6+7+8+9)/6 = 6.5

edit for failure to read properly

Edited by Odie on Friday 4th June 11:09

[elster]

ClareC said:

Given that its a school maths excericse and you have previously been asked to count the word length frequencies in data ranges.....I'm guessing they want you to work out the mean using the data ranges too.


So taking the mid-points of each data range:

Mean = (38*2+34*5+10*8)/82 = 3.975

The fact that the top category has no upper bound is irrelevant in this example as the frqwuency is 0 anyway.

The lady speaks sense.

This is how you do the mean with a frequency table.

[Justayellowbadge]

Odie said:

http://www.gcse.com/maths/median.htm


So (4+5+6+7+8+9)/6 = 6.5

edit for failure to read properly

Feck me.

I hope they never let you out of the grounds unsupervised.

[cs02rm0]

Odie said:

http://www.gcse.com/maths/median.htm


So (4+5+6+7+8+9)/6 = 6.5

edit for failure to read properly

Edited by Odie on Friday 4th June 11:09

That's the improved version?

[Odie]

Justayellowbadge said:

Odie said:

http://www.gcse.com/maths/median.htm


So (4+5+6+7+8+9)/6 = 6.5

edit for failure to read properly

Feck me.

I hope they never let you out of the grounds unsupervised.

oh crap I really cant read today can I, yes please disregard my crap lol I worked out the median, probably wrongly to, the simple stuff is always my downfall..

ClareC is correct :S

Edit for spelling loads of stuff wrong, can I go back to bed and start the day again please?

Edited by Odie on Friday 4th June 11:15

[Deva Link]

ClareC said:

The fact that the top category has no upper bound is irrelevant in this example as the frqwuency is 0 anyway.

I think that's correct, but what if there were some entries at 10+? The teacher couldn't have known there wouldn't be any before setting the question as the passage examined was random.

[marctwo]

Justayellowbadge said:

marctwo said:

I think you'll have to use the actual word-length values rather than the ranges, especially as the 10+ range has no upper bound.

It also has no values, so it's a null point.

Only in this particular case though. Any other page might contain some 10+ observations and then how would you deal with it?