Question

mean of following data is 31.4 determine the missing frequency x.
0-10 --- 5
10-20-- x
20-30--10
30-40--12
40-50--7
50-60--8​

Answer 1

$\begin {tabular} { | c | c | c | c | }</p><p>\cline { 1-4} C.I & f_{i} & x_{i} & f_{i}x_{i} \\</p><p>\cline {1-4} 0 - 10 & 5 & 5 & 25 \\</p><p>\cline {1-4} 10 - 20 & x & 15& 15x \\</p><p>\cline {1-4} 20 - 30 & 10 & 25& 250 \\</p><p>\cline {1-4} 30 - 40 & 12 & 35 & 420 \\</p><p>\cline { 1-4} 40 - 50 & 7 & 45 & 315 \\</p><p>\cline { 1-4 } 50 - 60 & 8 & 55 & 440 \\</p><p>\cline {1-4} \- & \Sigma f_{i} = 42+x & \- & \Sigma f_{i}x_{i} = 1450+15x \\</p><p>\cline {1-4} \end {tabular}$

$Mean = 31.4 \: (given)$

$\implies \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i} } = 31.4$

$\implies \frac{1450+15x}{42+x} = 31.4$

$\implies 1450 + 15x = 31.4(42+x)$

$\implies 1450 + 15x = 1318.8 + 31.4 x$

$\implies 1450 - 1318.8 = 31.4x - 15x$

$\implies 131.2 = 16.4x$

$\implies x = \frac{131.2}{16.4} = 8$

Therefore.,

$\red { Value \:of \:x } \green { = 8}$

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