Question 1189188
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*[tex \Large \ \ \  f_i] is the number of workers in the *[tex \Large i\text{th}] interval.  *[tex \Large q] is the number of intervals. *[tex \Large m_i] is the midpoint of the *[tex \Large i\text{th}] interval.  *[tex \Large \overline{x}] is the mean of the data.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \overline{x}\ =\ \frac{\sum_{i=0}^q\,f_i\cdot m_i}{\sum_{i=0}^q\,f_i}]


Make columns in your table for *[tex \Large f_i], *[tex \Large m_i], and *[tex \Large \overline{x}], and do the calculations for each interval and then take the appropriate sums.


Then make a column for *[tex \Large m_i\,-\,\overline{x}] and calculate each.  Then make a column for *[tex \Large \(m_i\,-\,\overline{x}\)^2] and calculate each  Then make a column for *[tex \Large f_i\(m_i\,-\,\overline{x}\)^2], calculate each, and then sum the column.


Then *[tex \Large n\ =\ \sum_{i=0}^q\,f_i]


And finally, the standard deviation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \sqrt{\frac{\sum_{i=0}^q\,f_i\(m_i\,-\,\overline{x}\)^2}{n\ -\ 1}}]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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