# Standard deviation question

In summary: If have the standard deviation of acceleration (acceleration > 0) for a car. I also have the standard deviation of deceleration (acceleration <0). With these two figures is there a way to get the a standard deviation of the acceleration (both positive and negative)?In summary, if you have the standard deviation of acceleration (acceleration > 0) for a car, you can use the standard deviation of deceleration (acceleration <0) to get the standard deviation of the acceleration (both positive and negative).
Hello,

If have the standard deviation of acceleration (acceleration > 0) for a car. I also have the standard deviation of deceleration (acceleration <0). With these two figures is there a way to get the a standard deviation of the acceleration (both positive and negative)?

Thanks

How do you define "standard deviation of acceleration (acceleration > 0)"?
Do you know the "average acceleration (acceleration > 0)" (+ same for <0)?

Hello,

If have the standard deviation of acceleration (acceleration > 0) for a car. I also have the standard deviation of deceleration (acceleration <0). With these two figures is there a way to get the a standard deviation of the acceleration (both positive and negative)?

Thanks

Maybe. What is your sample space? That is, what is the meaning of each element in the sets that are your samples?

Hi, yes I do have the average acceleration (>0) and the average deceleration (<0).

This is all the information that I have:

1. The number of acceleration observations (both acceleration and deceleration combined)
2. The SD of acceleration
3. The SD of deceleration
4. The average acceleration
5. The average deceleration

The SD is the SD of the population of the observations.

For example:
1. The number of acceleration observations (both acceleration and deceleration combined) =920
2. The SD of acceleration =0.33 m/s2
3. The SD of deceleration = 0.49 m/s2
4. The average acceleration = 0.55 m/s2
5. The average deceleration = -0.70 m/s2

Thanks for the help

If the two distributions are independent, then you will add the variances together and take the square roots of the two.

If they are dependent, you will need to include what is known as a co-variance term to get the final variance of the addition of the two random variables.

Have you come across co-variance before?

Hi chiro,

I presume the distributions are dependent? Here is sample of the raw data and the calculated SDs and means.

https://dl.dropbox.com/u/54057365/All/sample.JPG

Basically I have computed the statistics in green for a bunch of journeys. After doing this I realized that I wanted to calculate the SD of the acceleration both positive and negative. This is the statistic in red.

To save myself a lot of time and going back and doing it manually, I was wondering/hoping ithat there might be a way to get the statistic in red using the statistics in green?

Thanks for the help

Hmm... if you have the total acceleration time and the total deceleration time, you can simply add them as two datasets (correlations do not matter).

Let ta be the total (positive) acceleration time steps. Let aa be the average (positive) acceleration and σa the standard deviation in this dataset.
Define the same parameters with d for deceleration.
Let t be the total time = number of steps: t=ta+td
Your average acceleration (positive+negative, without superscript) is then given by
$$\bar{a} = \frac{a^a t^a + a^d t^d}{t^a+t^d}$$
The standard deviation can be calculated via
$$t \sigma^2 = \sum_{i,a_i>0} (a_i-\bar{a})^2 + \sum_{i,a_i<0} (a_i-\bar{a})^2$$
where the first sums run over all time steps with positive acceleration and the second runs over all with negative acceleration. N is the total number of steps. This can be simplified:
$$t \sigma^2 = \sum_{i,a_i>0} (a_i-a^a + a^a - \bar{a})^2 + \sum_{i,a_i<0} (a_i-a^d + a^d-\bar{a})^2$$
$$t \sigma^2 =\sum_{i,a_i>0} \left((a_i-a^a)^2 + (a^a - \bar{a})^2 + 2(a_i-a^a)(a^a-\bar{a})\right) + \sum_{i,a_i<0} \left((a_i-a^d)^2 + (a^d - \bar{a})^2 + 2(a_i-a^d)(a^d-\bar{a})\right)$$
In both sums, the first expression is simply the standard deviation of the individual parts. $(a^a-\bar{a})$ is constant, and the average ai is simply aa (same for d) so those terms vanish. The middle term is independent of i, so the sum just gives a factor of ta and d respectively.
$$t \sigma^2 =t^a \sigma^a + t^d \sigma^d + t^a (a^a - \bar{a})^2) + t^d (a^d - \bar{a})^2) = t^a \sigma^a + t^d \sigma^d + t\bar{a}^2 + t^a {a^a}^2 + t^d {a^d}^2 - 2t^a a^a \bar{a} - 2t^d a^d \bar{a}$$
Using the expression for $\bar{a}$, this gives
$$t \sigma^2 = t^a \sigma^a + t^d \sigma^d+ t^a {a^a}^2 + t^d {a^d}^2 - t\bar{a}^2$$
And that can be calculated, as all parts are known.If you do not now the umber of acceleration / deceleration steps: Maybe you can calculate the average acceleration based on the initial and final velocity and the total number of steps. This will allow you to calculate ta and td as well.

## What is standard deviation and why is it important?

Standard deviation is a measure of how spread out a set of data is from its mean. It is important because it allows us to understand the variability of the data and make comparisons between different datasets.

## How is standard deviation calculated?

The standard deviation is calculated by taking the square root of the variance, which is the average of the squared differences from the mean. It is represented by the symbol σ (sigma) for a population and s for a sample.

## What does a high or low standard deviation indicate?

A high standard deviation indicates that the data points are spread out over a wider range, while a low standard deviation indicates that the data points are clustered closely around the mean.

## How does standard deviation differ from mean and median?

The mean is the average of a set of data, while the median is the middle value in a set of data. Standard deviation measures the amount of variation in the data, while mean and median are measures of central tendency.

## Can standard deviation be negative?

No, standard deviation cannot be negative as it is the square root of a variance which is always positive. It represents a distance from the mean and cannot be negative.

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