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प्रश्न
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
|
Weekly wages (Rupees)
|
1000 - 2000 | 2000 - 3000 | 3000 - 4000 | 4000 - 5000 |
| No. of workers | 25 | 45 | 50 | 30 |
योग
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उत्तर
| Class
(Weekly wages rupees)
|
Class Mark (xi) |
di = xi − A = xi − 2500 |
ui = `"d"_"i"/"g"`
= `"d"_"i"/1000`
|
Frequency (Number of workers) (fi) |
Frequency × deviation (fi × ui) |
| 1000 - 2000 | 1500 | −1000 | −1 | 25 | −25 |
| 2000 - 3000 | 2500 →A | 0 | 0 | 45 | 0 |
| 3000 - 4000 | 3500 | 1000 | 1 | 50 | 50 |
| 4000 - 5000 | 4500 | 2000 | 2 | 30 | 60 |
| ∑fi = 150 | ∑fiui = 85 |
Required Mean = `(∑"f"_"i""u"_"i")/(∑"f"_"i")`
= `85/150`
= 0.57
Mean = `(barX) = "A" + bar("u")"g"`
= 2500 + 0.57 (1000)
= 2500 + 570
= Rs 3070
Hence, the mean of the weekly wages is Rs 3070.
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