Question

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.  

Answer 1

Weight (in grams)	Mid-Values \[\left( x_i \right)\]	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, 

\[\sigma^2\]

\[= \left( \frac{1}{N} \sum_{} f_i d_i^2 \right) - \left( \frac{1}{N} \sum_{} f_i d_i \right)^2 \]
\[ = \left( \frac{1}{70} \times 102 \right) - \left( \frac{1}{70} \times \left( - 38 \right) \right)^2 \]
\[ = 1 . 457 - 0 . 295\]
\[ = 1 . 162 gm\]

Standard deviation,

\[\sigma\] = \[\sqrt{\text{ Variance} } = \sqrt{1 . 162} = 1 . 08 \text{ gm }\]