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if i have a frequency table with the following data how do i work out the average or mean
word size 1,2,3,4 and Frequency of 3,11,20,29
do i add up the frequency column and divide by 4 from the word size column and help would be gratefully received.

2006-11-26 08:23:51 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

I assume you mean the average word size = total number of letters/number of words

Number of letters = Sum(word size * frequency) = 1*3+2*11+3*20+4*29 = 201

Number of words = Sum(frequency) = 3+11+20+29 = 63

Average word size = number of letters/number of words = 201/63 = 3.19 (approximately)

2006-11-26 08:29:10 · answer #1 · answered by Michel_le_Logique 4 · 0 1

Hi. Multiply the samples (3, 11, 20, 29) by the word count (1, 2, 3, 4) such that 1x3=3 + 11*2=22 + 3*20=60 + 4*29=201. Take 201/63=

2006-11-26 16:32:48 · answer #2 · answered by Cirric 7 · 2 0

x(size) f(Frequency) fx
1 x 3 = 3
2 x 11 = 22
3 x 20 = 60
4 x 29 = 116
+ ___ +___
63 201

Mean (Average) = 201/ 63 = 3.19

2006-11-26 16:31:31 · answer #3 · answered by lenpol7 7 · 2 0

If you're looking for the average word size it would be the total # of letters divided by the total # of words, or

Σs*f / Σf = (1*3 + 2*11 + 3*20 + 4*29)/(3 + 11 + 20 + 29)

You get to do the arithmetic!

2006-11-26 16:33:38 · answer #4 · answered by Steve 7 · 1 1

There are two averages you could compute with this data: average frequency of word sizes and average number of letters per word.

The others have gone for average number of letters per word, but your calculation gives average frequency of word sizes,
(3 + 11 + 20 + 29)/4 = 15.75

2006-11-26 16:48:02 · answer #5 · answered by Helmut 7 · 0 1

mean =
= (1 * 3 + 2 * 11 + 3 * 20 + 4 * 29)/(3 + 11 + 20 + 29)
= (3 + 22 + 60 + 116)/63
= 201/63
≈ 3.19

Set out table as follows

x ....... f .......... f*x
1 ...... 3 ............ 3
2 .... 11 .......... 22
3 .... 20 .......... 60
4 .... 29 ........ 116
----------------------
Σf = 63 Σfx = 201
----------------------

2006-11-26 16:30:43 · answer #6 · answered by Wal C 6 · 1 1

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