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प्रश्न
Find the mean of the following distribution by step deviation method:
| Class interval | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 |
| Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
योग
उत्तर
| C.I. | `f` | 'X' mid values | `mu = (x - "A")/"h"` | `f.mu.` |
| 20 - 30 | 10 | 25 | -3 | -30 |
| 30 - 40 | 6 | 35 | -2 | -12 |
| 40 - 50 | 8 | 45 | -1 | -8 |
| 50 - 60 | 12 | 55 = A | 0 | 0 |
| 60 - 70 | 5 | 65 | 1 | 5 |
| 70 - 80 | 9 | 75 | 2 | 18 |
| `sumf = 50` | `sumfmu = -27` |
Here,
A = Assumed mean = 55
h = 10
∴ Mean `bar"X" = "A" + (sumfmu)/(sumf) xx "h"`
= 55 + `((-27))/(50) xx 10`
= 55 - 5·4
= 49·6.
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Mean of Continuous Distribution
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संबंधित प्रश्न
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 |
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 |
Using step-deviation method, calculate the mean marks of the following distribution
| Class Interval | 50 - 55 | 55 - 60 | 60 - 65 | 65 - 70 | 70 - 75 | 75 - 80 | 80 - 85 | 85 - 90 |
| Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
