# Variance / standard deviation

#### logicandtruth

##### New member
Hi all - I wonder if you can help please.

Watching a video on youtube to help me understand about the mean, variance and standard deviation but last part of video left me confused.

The speaker said the following for the formula for standard deviation:

Consider if the variance is 200 for the first example and 2 for the second, please could someone explain in detailed simple terms why he calculated it to be 10 square roots of 2 and not 14. 14

Link to video

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#### Jameson

##### Administrator
Staff member
Hi there... just watched the video. Can you explain how you get 14? That will help me show where you are off.

#### logicandtruth

##### New member
Hi there... just watched the video. Can you explain how you get 14? That will help me show where you are off.
Hi Jameson - thanks for your response.

From timestamp 10:35 speaker starts to explain how he will calculate standard deviation using variance. As the variance in the example is 200, applying the formula used in the video you would have to find the square root of 200 which is 14. Instead he mentions the square root of 200 is equal to 10 square root's of 2 and I am not sure how/why he did this?

#### Jameson

##### Administrator
Staff member
Hi again,

Ok so $\sqrt{200}$ is not 14. Let's do it together.

$\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2}$.

Does that help?

#### logicandtruth

##### New member
Hi there,

Yes it now all makes sense as I now understand I essentially just needed to simplify the square root of an integer.

Thanks for your response, but please don't take it as a criticism but just posting an answer without explanation often does not help the intended recipient. Case in point, as someone not from a maths background motivated to improve, I went and did research where I learned the speaker was simplifying radical expressions. After reading more about this subject I also understood the rules such a
Hi again,

Ok so $\sqrt{200}$ is not 14. Let's do it together.

$\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2}$.

Does that help?
s If any factors are raised to the power of 2, they can be moved outside of the square root.

Hi there,

Yes it now all makes sense as I now understand I essentially just needed to simplify the square root of an integer.

Thanks for your response, but please don't take it as a criticism but just posting an answer without explanation often does not help the intended recipient. Case in point, as someone not from a maths background motivated to improve, I went and did research where I learned the speaker was simplifying radical expressions. After reading more about this subject I also understood the rules such as If any factors are raised to the power of 2, they can be moved outside of the square root.

Best,

#### Jameson

##### Administrator
Staff member
Hey, that's a fair point. This site is definitely of the mindset to help someone learn to do it themselves, rather than giving them the answers. It's hard to gauge sometimes what level someone is at and what is the proper amount of nudging without answering in full, but if you continue to post and ask these questions I feel confident you can both get help and get the proper balance.

#### logicandtruth

##### New member
Hey, that's a fair point. This site is definitely of the mindset to help someone learn to do it themselves, rather than giving them the answers. It's hard to gauge sometimes what level someone is at and what is the proper amount of nudging without answering in full, but if you continue to post and ask these questions I feel confident you can both get help and get the proper balance.
Thank you - I appreciate a staff member detailing to help an individual to learn for themselves. Granted, it is not an easy balance as people will have no idea as to the level of someone's ability by them just posting a question, so I understand getting the balance is not easy. I think in the future I will also try and give more context of where I am at in my learning which may prompt different tailored replies. Thanks again.