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