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प्रश्न
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
पर्याय
`sum_(i = 1)^n (x_i - barx)`
`1/n sum_(i = 1)^n |x_i - barx|`
`sum_(i = 1)^n (x_i - barx)^2`
`1/n sum_(i = 1)^n (x_i - barx)^2`
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उत्तर
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by `1/n sum_(i = 1)^n |x_i - barx|`.
Explanation:
M.D. = `1/n sum_(i = 1)^n |x_i - barx|`
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संबंधित प्रश्न
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38, 70, 48, 40, 42, 55, 63, 46, 54, 44
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36, 72, 46, 42, 60, 45, 53, 46, 51, 49
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| fi | 7 | 4 | 6 | 3 | 5 |
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| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |
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| fi | 3 | 5 | 6 | 7 | 8 |
Find the mean deviation about the mean for the data.
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| 0-100 | 4 |
| 100-200 | 8 |
| 200-300 | 9 |
| 300-400 | 10 |
| 400-500 | 7 |
| 500-600 | 5 |
| 600-700 | 4 |
| 700-800 | 3 |
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| 16 - 20 | 5 |
| 21 - 25 | 6 |
| 26 - 30 | 12 |
| 31 - 35 | 14 |
| 36 - 40 | 26 |
| 41 - 45 | 12 |
| 46 - 50 | 16 |
| 51 - 55 | 9 |
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38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
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(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
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300 4000 4200 4400 4600 4800 5800 |
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\[\bar{ X } \] + M.D, where M.D. is the mean deviation from the mean.
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\[\bar { X } \] − M.D. and
\[\bar { X } \] + M.D, where M.D. is the mean deviation from the mean.
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| xi | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
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| fi | 3 | 5 | 6 | 7 | 8 |
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| Frequency | 4 | 6 | 8 | 5 | 2 |
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