The weight of coffee in 70 jars is shown in the following table:
Weight (in grams):  200–201  201–202  202–203  203–204  204–205  205–206 
Frequency:  13  27  18  10  1  1 
Determine the variance and standard deviation of the above distribution.
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Solution
Weight (in grams)  MidValues
\[\left( x_i \right)\]

Frequency
\[\left( f_i \right)\]

\[d_i = x_i  202 . 5\]

\[d_i^2\]

\[f_i d_i\]

\[f_i d_i^2\]

200–201  200.5  13  −2  4  −26  52 
201–202  201.5  27  −1  1  −27  27 
202–203  202.5  18  0  0  0  0 
203–204  203.5  10  1  1  10  10 
204–205  204.5  1  2  4  2  4 
205–206  205.5  1  3  9  3  9 
N =
\[\sum_{} f_i = 70\]

\[\sum_{} f_i d_i =  38\]

\[\sum_{}f_i d_i^2 = 102\]

Now,
Variance,
\[\sigma^2\]
\[= \left( \frac{1}{N} \sum_{} f_i d_i^2 \right)  \left( \frac{1}{N} \sum_{} f_i d_i \right)^2 \]
\[ = \left( \frac{1}{70} \times 102 \right)  \left( \frac{1}{70} \times \left(  38 \right) \right)^2 \]
\[ = 1 . 457  0 . 295\]
\[ = 1 . 162 gm\]
\[ = \left( \frac{1}{70} \times 102 \right)  \left( \frac{1}{70} \times \left(  38 \right) \right)^2 \]
\[ = 1 . 457  0 . 295\]
\[ = 1 . 162 gm\]
Standard deviation,
\[\sigma\] = \[\sqrt{\text{ Variance} } = \sqrt{1 . 162} = 1 . 08 \text{ gm }\]
Is there an error in this question or solution?
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