Compute the range and sample standard deviation for strength of the concrete​ (in psi).

3920​, 4090​, 3300​, 3100​, 2940​, 3830​, 4090​, 4030

The range is

1150 psi.

sequals

nothing psi​ (Round to one decimal place as​ needed.)

Respuesta :

Answer:

[tex]Range = 1150\ psi[/tex]

[tex]Standard\ Deviation = 442.3\ psi[/tex]

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;

[tex]Range = Highest\ Strength - Least\ Strength[/tex]

From the given data;

[tex]Highest\ Strength = 4090\\ Least\ Strength = 2940[/tex]

Hence,

[tex]Range = 4090 - 2940\\\\Range = 1150\ psi[/tex]

Calculating the Standard Deviation...

Start by calculating the mean

[tex]Mean = \frac{\sum x}{n}[/tex]

Where x->3920​, 4090​, 3300​, 3100​, 2940​, 3830​, 4090​, 4030

n = 8

[tex]Mean = \frac{3920 + 4090 + 3300 + 3100+ 2940+ 3830+ 4090+4030}{8}[/tex]

[tex]Mean = \frac{29300}{8}[/tex]

[tex]Mean = 3662.5[/tex]

Subtract the mean from each observation

[tex]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[/tex]

Square the result of the above

[tex]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[/tex]

Add the above results together

[tex]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[/tex]

Divide by n

[tex]\frac{1564749.75}{8} = 195593.71875[/tex]

Take Square root of the above result to give standard deviation

[tex]SD = \sqrt{195593.71875}[/tex]

[tex]SD = 442.259786494[/tex]

[tex]SD = 442.3\ psi\ (Approximated)[/tex]

Hence,

[tex]Range = 1150\ psi[/tex]

[tex]Standard\ Deviation = 442.3\ psi[/tex]

I agree with the other person. Hope this helped you!
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