Answer:
![Range = 1150\ psi](https://tex.z-dn.net/?f=Range%20%3D%201150%5C%20psi)
![Standard\ Deviation = 442.3\ psi](https://tex.z-dn.net/?f=Standard%5C%20Deviation%20%3D%20442.3%5C%20psi)
Explanation:
Given
3920, 4090, 3300, 3100, 2940, 3830, 4090, 4030
Required
- Determine the Range
- Determine the Standard Deviation
Calculating the Range...
The Range is calculated using the following formula;
![Range = Highest\ Strength - Least\ Strength](https://tex.z-dn.net/?f=Range%20%3D%20Highest%5C%20Strength%20-%20Least%5C%20Strength)
From the given data;
![Highest\ Strength = 4090\\ Least\ Strength = 2940](https://tex.z-dn.net/?f=Highest%5C%20Strength%20%3D%204090%5C%5C%20Least%5C%20Strength%20%3D%202940)
Hence,
![Range = 4090 - 2940\\\\Range = 1150\ psi](https://tex.z-dn.net/?f=Range%20%3D%204090%20-%202940%5C%5C%5C%5CRange%20%3D%201150%5C%20psi)
Calculating the Standard Deviation...
Start by calculating the mean
![Mean = \frac{\sum x}{n}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
Where x->3920, 4090, 3300, 3100, 2940, 3830, 4090, 4030
n = 8
![Mean = \frac{3920 + 4090 + 3300 + 3100+ 2940+ 3830+ 4090+4030}{8}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B3920%20%2B%204090%20%2B%203300%20%2B%203100%2B%202940%2B%203830%2B%204090%2B4030%7D%7B8%7D)
![Mean = \frac{29300}{8}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B29300%7D%7B8%7D)
![Mean = 3662.5](https://tex.z-dn.net/?f=Mean%20%3D%203662.5)
Subtract the mean from each observation
![3920 - 3662.5 = 257.5\\4090 - 3662.5 = 427.5\\3300 - 3662.5 = -362.5\\3100 - 3662.5 = -562.5\\2940 - 3662.5 = -722.5\\3830 - 3662.5 = 167.5\\4090 - 3662.5 = 427.5\\4030 - 3662.5 = 367.5](https://tex.z-dn.net/?f=3920%E2%80%8B%20-%203662.5%20%3D%20257.5%5C%5C4090%E2%80%8B%20-%203662.5%20%3D%20427.5%5C%5C3300%20-%203662.5%20%3D%20-362.5%5C%5C3100%20-%203662.5%20%3D%20-562.5%5C%5C2940%20-%203662.5%20%3D%20-722.5%5C%5C3830%20-%203662.5%20%3D%20167.5%5C%5C4090%20-%203662.5%20%3D%20427.5%5C%5C4030%20-%203662.5%20%3D%20367.5)
Square the result of the above
![257.5^2 =66,306.25\\427.5^2 =182,756\\-362.5^2 =131,406.25\\-562.5^2 =316,406.25\\-722.5^2 =522,006.25\\167.5^2 =28,056.25\\427.5^2 =182,756.25\\367.5^2 =135,056.25](https://tex.z-dn.net/?f=257.5%5E2%20%3D66%2C306.25%5C%5C427.5%5E2%20%3D182%2C756%5C%5C-362.5%5E2%20%3D131%2C406.25%5C%5C-562.5%5E2%20%3D316%2C406.25%5C%5C-722.5%5E2%20%3D522%2C006.25%5C%5C167.5%5E2%20%3D28%2C056.25%5C%5C427.5%5E2%20%3D182%2C756.25%5C%5C367.5%5E2%20%3D135%2C056.25)
Add the above results together
![66,306.25+182,756+131,406.25+316,406.25+522,006.25+28,056.25+182,756.25+135,056.25 = 1564749.75](https://tex.z-dn.net/?f=66%2C306.25%2B182%2C756%2B131%2C406.25%2B316%2C406.25%2B522%2C006.25%2B28%2C056.25%2B182%2C756.25%2B135%2C056.25%20%3D%201564749.75)
Divide by n
![\frac{1564749.75}{8} = 195593.71875](https://tex.z-dn.net/?f=%5Cfrac%7B1564749.75%7D%7B8%7D%20%3D%20195593.71875)
Take Square root of the above result to give standard deviation
![SD = \sqrt{195593.71875}](https://tex.z-dn.net/?f=SD%20%3D%20%5Csqrt%7B195593.71875%7D)
![SD = 442.259786494](https://tex.z-dn.net/?f=SD%20%3D%20442.259786494)
![SD = 442.3\ psi\ (Approximated)](https://tex.z-dn.net/?f=SD%20%3D%20442.3%5C%20psi%5C%20%28Approximated%29)
Hence,
![Range = 1150\ psi](https://tex.z-dn.net/?f=Range%20%3D%201150%5C%20psi)
![Standard\ Deviation = 442.3\ psi](https://tex.z-dn.net/?f=Standard%5C%20Deviation%20%3D%20442.3%5C%20psi)