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# The weights of coffee in 70 jars is shown in the following table : Determine variance and standard deviation of the above distribution.

$\begin{array}{1 1}(A)\;4.15,3.049\\(B)\;25,1600\\(C)\;790,87\\(D)\;1.16,1.08\end{array}$

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A)
Toolbox:
• The formula used are $mean (\bar x)= A +\large\frac{\sum f_id_i}{\sum f_i } $$\times h • Variance = \bigg[ \large\frac{\sum f_i d_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i}{\sum f_i } \bigg)^2\bigg]$$ \times h^2$
• SD $(\sigma) =\sqrt {variance}$
Step 1:
Assumed mean $= 203 (A)$
$h=1$
Step 2:
Variance $= \bigg[ \large\frac{ \sum f_i d_i ^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i }{\sum f_i }\bigg)^2 \bigg] $$\times h^2 \qquad= \bigg[\large \frac{157.5 }{70} - \bigg(\frac{-73}{70} \bigg)^2 \bigg]^2$$ \times 1^2$
$\qquad= \large\frac{157.5 \times 70 -(-73)^2}{70 \times 70}$
$\qquad=\large\frac{11025-5329}{4900}=\frac{5696}{4900}$
$\qquad= 1.16$
Step 3:
SD $(\sigma) =\sqrt {variance}$
$\qquad=\sqrt {1.16}$
$\qquad= 1.08$
Hence D is the correct answer.