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Sagot :
Given,
The frequency of length of fishes is,
Value frequency
5 3
6 1
7 1
8 2
9 3
`10 4
11 6
12 1
Required
The mean length of the fish.
The mean is calculated as,
[tex]\begin{gathered} Mean\text{ =}\frac{\sum^xf}{\sum^f} \\ =\frac{5\times3+6\times1+7\times1+8\times2+9\times3+10\times4+11\times6+12\times1}{21} \\ =\frac{15+6+7+16+27+40+66+12}{21} \\ =\frac{189}{21} \\ =9 \end{gathered}[/tex]Hence, the mean length is 9.
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