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Question
Calculate the mean of the following distribution using Assumed Mean Method
| Class Interval | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 |
| Frequency | 5 | 7 | 15 | 28 | 8 |
Sum
Solution
Assumed Mean (A) = 25
| Class Interval | Mid value (x) |
Frequency (f) |
d = x − A | fd |
| 0 − 10 | 5 | 5 | −20 | −100 |
| 10 − 20 | 15 | 7 | −10 | −70 |
| 20 − 30 | 25 | 15 | 0 | 0 |
| 30 − 40 | 35 | 28 | 10 | 280 |
| 40 − 50 | 45 | 8 | 20 | 160 |
| `sumf` = 63 | `sumf"d"` = 270 |
Arithmetic mean `(barx) = "A" + (sumf"d")/(sumf)`
= `25 + 270/63`
= 25 + 4.29
Assumed Mean = 29.29
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Method of Finding Mean for Grouped Data: Deviation Or Assumed Mean Method
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RELATED QUESTIONS
Observe the following table and find the Mean:
Assumed mean A = 300
| Class | Class mark | di = xi – A | Frequency | Frequency × Deviation |
| xi | di = xi – 300 | fi | fi di | |
| 200 – 240 | 220 | – 80 | 5 | – 400 |
| 240 – 280 | 260 | – 40 | 10 | – 400 |
| 280 – 320 | 300 `rightarrow` A | 0 | 15 | 0 |
| 320 – 360 | 340 | 40 | 12 | 480 |
| 360 – 400 | 380 | 80 | 8 | 640 |
| Total | `sumf_i` = 50 | `sumf_i d_i` = 320 |
