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- Variance and Standard Deviation for raw data:
- Variance and Standard Deviation for ungrouped frequency distribution:
- Variance and Standard Deviation for grouped frequency distribution :
Notes
Let `x_1, x_2, x_3, ..., x_n` be n observations and x be their mean. Then
`(x_1 - bar x)^ 2 + (x_2 - bar x) ^2 + ... + (x_n - bar x)^ 2 `
If this sum is zero, then each `(x_i - bar x)`has to be zero. This implies that there is no dispersion at all as all observations are equal to the mean `bar x` .
If \[\displaystyle\sum_{i=1}^{n} (x_i - \bar{x})^2\] is small , this indicates that the observations `x_1, x_2, x_3,...,x_n` are close to the mean x and therefore, there is a lower degree of dispersion. On the contrary, if this sum is large, there is a higher degree of dispersion of the observations from the mean `bar x` .
Related QuestionsVIEW ALL [10]
Obtain standard deviation for the following date:
| Height (in inches) | 60 – 62 | 62 – 64 | 64 – 66 | 66 – 68 | 68 – 70 |
| Number of students | 4 | 30 | 45 | 15 | 6 |
The following distribution was obtained change of origin and scale of variable X.
| di | – 4 | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |
| fi | 4 | 8 | 14 | 18 | 20 | 14 | 10 | 6 | 6 |
If it is given that mean and variance are 59.5 and 413 respectively, determine actual class intervals.
Compute variance and standard deviation for the following data:
| x | 2 | 4 | 6 | 8 | 10 |
| f | 5 | 4 | 3 | 2 | 1 |
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| x | 1 | 3 | 5 | 7 | 9 |
| Frequency | 5 | 10 | 20 | 10 | 5 |
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| Age (In years) | 10 | 11 | 12 | 13 | 14 |
| No. of students | 10 | 20 | 40 | 20 | 10 |
