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Question
Helping the step deviation method find the arithmetic mean of the distribution:
| Variable (x) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
| Frequency(f) | 20 | 43 | 75 | 67 | 72 | 45 | 39 | 9 | 8 | 6 |
Sum
Solution
Let the assumed Mean be A = 25 and h = 5.
| `x_i` | Frequencies `f_i` | Deviation `d_i = x_i - 25` | `u_i = (x_i - 25)/(5)` | `f_iu_i` |
| 5 | 20 | -20 | -4 | -80 |
| 10 | 43 | -15 | -3 | -129 |
| 15 | 75 | -10 | -2 | -150 |
| 20 | 67 | -5 | -1 | -67 |
| A = 25 | 72 | 0 | 0 | 0 |
| 30 | 45 | 5 | 1 | 45 |
| 35 | 39 | 10 | 2 | 78 |
| 40 | 9 | 15 | 3 | 27 |
| 45 | 8 | 20 | 4 | 32 |
| 50 | 6 | 25 | 5 | 30 |
| N = `sum_i = 384` | `sumf_iu_i = -214` |
We have,
N = 384, A = 25, h = 5 and `sumf_iu_i = -214`
Mean `bar"(X)" = "A" + "h" (1/"N" sumf_iu_i)`
= `25 + 5 xx ((-214)/(384))`
= 25 - 2·786
= 22·214.
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Mean of Continuous Distribution
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