In the first proof reading of a book containing 300 pages the followin

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained: No. of misprints per pages (x):012345 No. of pages (f):1549536951Find the average number of misprints per page.

Answer»

To find : the average number of misprints per page.

Use the shortcut method to find the mean of given data.For that,Let the assumed mean be (A) = 2,The deviation of values xi from assumed mean be d_i = x_i – A.

Now to find the mean:First multiply the frequencies in column (ii) with the value of deviations in column (iii) as f_id_i.

No. of misprints per pages (xi)	No. of pages (fi)	d_i = x_i - A = x_i - 2	f_id_i
0	154	-2	-308
1	95	-1	-95
2	36	0	0
3	9	1	9
4	5	2	10
5	1	3	3
	N = 300		\(\sum\)f_id_i= -381

Now add the sum of all entries in column (iii) to obtain \(\displaystyle\sum_{i=1}^{n} f_id_i\)

and the sum of all frequencies in the column (ii) to obtain \(\displaystyle\sum_{i=1}^{n} f_id_i=N\)

So,

Average number of misprints per day = A + \(\frac{\sum f_id_i}{N}\)

where, N = total number of observations

⇒ Mean = 2 + \(\frac{-381}{300}\)

Mean = \(\frac{600-381}{300}\)

Mean = \(\frac{219}{300}\)

Mean = 0.73

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained: No. of misprints per pages (x):012345 No. of pages (f):1549536951Find the average number of misprints per page.

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