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प्रश्न
The weights of tea in 70 packets are shown in the following table:
| Weight | 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.
उत्तर
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 | 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+ ((sum _i f_i u_i)/( sum _i f_i)) xxh`
=`202.5 + ((-38)/70) xx1`
=202.5 – 0.542
= 201.96
Hence, the mean is 201.96 g.
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