The variance of a set of data is 196. Then the standard deviation of the data is. A. ± 14 B. 14 C. 96
B
In statistics, we use measures to describe the spread or dispersion of a dataset. Two fundamental measures of spread are variance and standard deviation. They are closely related, and understanding their relationship is key to solving this type of problem.
Variance measures how far each number in the set is from the mean, and thus from every other number in the set. It is the average of the squared differences from the Mean.
Standard deviation is the square root of the variance. It measures the typical distance of the data points from the mean. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.
The standard deviation is simply the positive square root of the variance. If $\sigma^2$ represents the variance and $\sigma$ represents the standard deviation, the relationship is:
$$ \sigma = \sqrt{\sigma^2} $$
We take the positive square root because the standard deviation represents a distance or a measure of spread, which cannot be negative. While the square root of a positive number technically has both a positive and a negative value (e.g., $\sqrt{4} = \pm 2$), in the context of standard deviation, we only consider the positive value.
The question provides the variance of a set of data as 196. We need to find the standard deviation of the data.
Given:
We need to find the standard deviation ($\sigma$).
Using the relationship $\sigma = \sqrt{\sigma^2}$, we can calculate the standard deviation:
$$ \sigma = \sqrt{196} $$
To find $\sqrt{196}$, we need to find a number that, when multiplied by itself, equals 196.
So, the square root of 196 is 14.
$$ \sigma = 14 $$
The standard deviation must be a non-negative value, as it represents a measure of spread or distance. Therefore, the standard deviation is 14.
Let's look at the given options based on our calculation:
Our calculated standard deviation is 14, which corresponds to option B.
| Measure | Value Given | Relationship | Calculated Value |
|---|---|---|---|
| Variance ($\sigma^2$) | 196 | $\sigma = \sqrt{\sigma^2}$ | - |
| Standard Deviation ($\sigma$) | - | $\sigma = \sqrt{196}$ | 14 |
| Measure | Description | Relationship to Others |
|---|---|---|
| Range | Difference between the highest and lowest values. | Simplest measure of spread. |
| Variance ($\sigma^2$) | Average of squared differences from the mean. | Square of the standard deviation. |
| Standard Deviation ($\sigma$) | Square root of the variance. Typical distance from the mean. | Square root of the variance. |
Understanding standard deviation and variance is crucial in many fields, including finance, quality control, and scientific research.
Calculate the mean from the following table.
Scores | Frequencies |
0-10 | 2 |
10-20 | 4 |
20-30 | 12 |
30-40 | 21 |
40-50 | 6 |
50-60 | 3 |
60-70 | 2 |
Find the standard deviation of the following data (rounded off to two decimal places).
5, 3, 4, 7
If the standard deviation of a population is 5, what will be its variance?
A. 10
B. 15
C. 25
D. 12.5
The variance of a set of data is 144. Then the standard deviation of the data is:
A. ±12
B. 12
C. 44
D. 72
The mean of a distribution is 24 and the standard deviation is 6. What is the value of variance coefficient?
A. 50%
B. 25%
C. 100%
D. 75%