# The effect of multiple samples on SD

• May 18th 2010, 05:13 AM
Peleus
The effect of multiple samples on SD
Hi all,

This is the question I am looking at...

Imagine you are building a wall from precast concrete blocks. Standard 20 cm Blocks are 19.5 cm high to allow for a 0.5 cm layer of mortar. In practice, the height of a block plus mortar row varies according to a Normal Distribution with mean 20.0 cm and standard deviation 0.25 cm. Heights of successive rows are independent. Your wall has 10 rows.

(i) What is the distribution of the height of the wall?
(ii) What is the probability that the height of the wall differs from the design height of 200 cm by more than 1 cm?)

My working is as follows.

The normal distribution should be $(\mu, \sigma)$. $\mu = 20$ still because even after 10 samples the total mean should still be the same. The SD I'm having trouble with. Is it true that the SD of the sample is $\frac{\sigma}{\sqrt{n}}$? If it is, then I guess the SD should be 0.079, giving an answer of

$N(\mu = 20, \sigma = 0.079)$ for (i)

For (ii) I believe (if (i)) is correct, that it should be a case of normalizing X then looking up the Z tables, resulting in a Z score of 201 - 200 / 0.079 = 12.658.

Obviously something that has 12.658 SD away from the mean in normal distribution is next to impossible, which makes me suspect I've done something wrong in the working.

Can anyone please check my working? Thank you.
• May 18th 2010, 05:23 AM
mr fantastic
Quote:

Originally Posted by Peleus
Hi all,

This is the question I am looking at...

Imagine you are building a wall from precast concrete blocks. Standard 20 cm Blocks are 19.5 cm high to allow for a 0.5 cm layer of mortar. In practice, the height of a block plus mortar row varies according to a Normal Distribution with mean 20.0 cm and standard deviation 0.25 cm. Heights of successive rows are independent. Your wall has 10 rows.

(i) What is the distribution of the height of the wall?
(ii) What is the probability that the height of the wall differs from the design height of 200 cm by more than 1 cm?)

My working is as follows.

The normal distribution should be $(\mu, \sigma)$. $\mu = 20$ still because even after 10 samples the total mean should still be the same. The SD I'm having trouble with. Is it true that the SD of the sample is $\frac{\sigma}{\sqrt{n}}$? If it is, then I guess the SD should be 0.079, giving an answer of

$N(\mu = 20, \sigma = 0.079)$ for (i)

For (ii) I believe (if (i)) is correct, that it should be a case of normalizing X then looking up the Z tables, resulting in a Z score of 201 - 200 / 0.079 = 12.658.

Obviously something that has 12.658 SD away from the mean in normal distribution is next to impossible, which makes me suspect I've done something wrong in the working.

Can anyone please check my working? Thank you.

Read this: Normal Sum Distribution -- from Wolfram MathWorld
• May 18th 2010, 05:27 AM
Peleus
To be honest I don't understand most of what's on the page, but is it fair to say you're referring mainly to these two formula?

http://mathworld.wolfram.com/images/...dEquation2.gif

http://mathworld.wolfram.com/images/...dEquation3.gif

Meaning the wall's distribution is (200, 2.5) ?
• May 18th 2010, 05:31 AM
mr fantastic
Quote:

Originally Posted by Peleus
To be honest I don't understand most of what's on the page, but is it fair to say you're referring mainly to these two formula?

http://mathworld.wolfram.com/images/...dEquation2.gif

http://mathworld.wolfram.com/images/...dEquation3.gif

Meaning the wall's distribution is (200, 2.5) ?

$\sigma_X = 0.25 \Rightarrow \sigma_X^2 = 0.25^2 \Rightarrow \sigma_{X_1 + ... + X_{10}}^2 = (10)(0.25^2) \Rightarrow \sigma_{X_1 + ... + X_{10}} = 0.25 \sqrt{10} = ....$
• May 18th 2010, 05:38 AM
Peleus
I really feel dumb when I post on this forum, thanks for your help.

Silly me getting mixed up between variance and SD.

So (0.25)^2 * 10 = variance of the wall.

sqrt (answer) = SD of wall.

also known as 0.25*10^0.5 = SD of wall.

So the distribution is (200, 0.791) for the wall.

Part (ii) means 1.264 SD's away from the mean, giving a chance of (1-(0.8962 - 0.1038))

meaning P(1cm or more away from mean) = 0.2076

Thanks Mr Fantastic.

(I'm reasonably confident I'm right, I'll look pretty silly if I'm wrong).