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Question
The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram. by the Step Deviation Method.
| Weights in grams | No. of apples |
| 80 - 55 | 5 |
| 85 - 90 | 8 |
| 90 - 95 | 10 |
| 95 - 100 | 12 |
| 100 - 105 | 8 |
| 105 - 110 | 4 |
| 110 - 115 | 3 |
Sum
Solution
| Weight in grams. | No. of apples | `bb(x_i)` | `bb(x_i - "A")` | `bb(u_i = (x_i - "A")/(5))` | `bb(f_iu_i)` |
| 80 - 85 | 5 | 82.5 | −15 | −3 | −15 |
| 85 - 90 | 8 | 87.5 | −10 | −2 | −16 |
| 90 - 95 | 10 | 92.5 | −5 | −1 | −10 |
| 95 - 100 | 12 | (97.5)A | 0 | 0 | 0 |
| 100 - 105 | 8 | 102.5 | 5 | 1 | 8 |
| 105 - 110 | 4 | 107.5 | 10 | 2 | 8 |
| 110 - 115 | 3 | 112.5 | 15 | 3 | 9 |
| `sumf_i = 50` | `sumf_iu_i = -16` |
A = 97.5, `sumf_i = 50, sumf_iu_i = -16, "h" = 5`.
∴ Mean `bar"X" = "A" + (sumf_iu_i)/(sumf_i) xx "h"`
= 97.5 + `(-16)/(50) xx 5`
= 95·9.
= 96 g
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Mean of Continuous Distribution
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