Question

(a) The age of 20 boys in a locality is given below. Median 12 10 15 14 8 Age in Years Number of Boys un 5 3 2 6 4
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Answer 1

Step-by-step explanation:

PrepareFrequencyDistributionTable:

\begin{gathered} \begin {tabular} {|c|c|c|} < /p > < p > \cline {1-3} Age (x_{i})& Number of boys(f_{i})}&f_{i}x_{i} \\ < /p > < p > \cline {1-3} 12&5&60\\ < /p > < p > \cline {1-3} 10&3&30\\ < /p > < p > \cline {1-3} 15&2&30\\ < /p > < p > \cline {1-3}14&6&84\\ < /p > < p > \cline {1-3} 8&4&32\\ < /p > < p > \cline {1-3} -&\sum f_{i} =20&\sum f_{i}x_{i}= 236 \\ < /p > < p > \cline {1-3} < /p > < p > \end {tabular} \end{gathered}

Mean \: ( \overline {x} ) = \frac{ \sum f_{i}x_{i}}{\sum f_{i}}Mean(

x

)=

∑f

i

∑f

i

x

i

\begin{gathered} = \frac{232}{20} \\= \frac{116}{10}\\= 11.6 \end{gathered}

=

20

232

=

10

116

=11.6

Therefore.,

\red {Mean \: ( \overline {x} ) }\green {=11.6}Mean(

x

)=11.6

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Answer 2

Answer:

ans ---980--------__________________-