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The weights of tea in 70 packets are shown in the following table: Weight (in grams) 200 – 201 201 – 202 202 – 203 203 – 204 204 – 205 205 – 206 Number of packets 13 27 18 10 1 1

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Question

The weights of tea in 70 packets are shown in the following table:

Weight
(in grams)
200 – 201 201 – 202 202 – 203 203 – 204 204 – 205 205 – 206
Number of
packets
13 27 18 10 1 1

Find the mean weight of packets using step-deviation method.

Sum
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Solution

Let us choose a = 202.5, h = 1, then `d_i = x_i - 202.5 and u_i = (x_i-202.5)/1`

Using step-deviation method, the given data is shown as follows:

Weight
(in grams)
Number of
packets `(f_i)`
Class mark
`(x_i)`
`d_i = x_i - 202.5` `u_i =
(x_i−202.5)/
1`
`(f_i u_i)`
200 – 201 13 200.5 -2 -2 -26
201 – 202 27 201.5 -1 -1 -27
202 – 203 18 202.5 0 0 0
203 – 204 10 203.5 1 1 10
204 – 205 1 204.5 2 2 2
205 – 206 1 205.5 3 3 3
Total `Ʃ f_i` = 70       `Ʃ f_i u_i` = -38

The mean of the given data is given by, 

x = `a + ((sumf_i u_i)/( sumf_i)) xx h`

    = `202.5 + ((-38)/70) xx1`

    = 202.5 – 0.542
    = 201.96
Hence, the mean is 201.96 g.

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Chapter 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive - EXERCISE 18A [Page 862]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
EXERCISE 18A | Q 22. | Page 862
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