The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be:
- 81
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- 36
- 25
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The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be:
The following information relates to a sample of size 60 $\sum\text{x}^2=18000$ and $\sum\text{x}=960,$ then the variance is:
Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is:
Mean deviation for n observations x1, x2, ...... , xn from their mean x is given by:
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})$
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
Let x1, x2, ..., xn be n observations and $\bar{\text{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
$\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(\bar{\text{x}})^{-2}$
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