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The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure (in Rs)100 − 150150 − 200200 − 250250 − 300300 − 350 - Mathematics

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Question

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Sum
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Solution

To find the class mark (xi) for each interval, the following relation is used:

Let the assumed mean, a = 225

Class size, h = 50

= `u_i = (x_i - a)/h`

= `(x_i - 225)/50`

Taking 225 as the assumed mean (a), di, ui, fiui are calculated as follows.

Daily expenditure (in Rs) fi xi di = x− 225 `u_i=(x_i-225)/50` fiui
100 − 150 4 125 − 100 − 2 − 8
150 − 200 5 175 − 50 − 1 − 5
200 − 250 12 225 0 0 0
250 − 300 2 275 50 1 2
300 − 350 2 325 100 2 4
Total 25       -7

From the table, we obtain:

`sumf_i = 25`

`sumf_iu_i = -7`

`"Mean " barx = a+ ((sumf_1u_i)/(sumf_i))xh`

= `225 + ((-7)/25)xx(50)`

= 225 − 14

= 211

Therefore, mean daily expenditure on food is Rs 211.

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Chapter 14: Statistics - Exercise 14.1 [Page 271]

APPEARS IN

NCERT Mathematics [English] Class 10
Chapter 14 Statistics
Exercise 14.1 | Q 6 | Page 271
RD Sharma Mathematics [English] Class 10
Chapter 15 Statistics
Exercise 15.3 | Q 22 | Page 24
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