begin array c c Class Interval amp 3 - 5 amp 6 ... - JAMB Mathematics 2011 Question
\(\begin{array}{c|c}
Class Interval & 3 - 5 & 6 - 8 & 9 - 11 \ \hline Frequency & 2 & 2 & 2 \end{array}\). Find the standard deviation of the above distribution.
Class Interval & 3 - 5 & 6 - 8 & 9 - 11 \ \hline Frequency & 2 & 2 & 2 \end{array}\). Find the standard deviation of the above distribution.
A
√5
B
√6
C
√7
D
√2
correct option: b
\(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \ x & 4 & 7 & 10 \ f & 2 & 2 & 2 \ f - x & 8 & 14 & 20 \ |x - \bar{x}|^2 & 9 & 0 & 9 \ |x - \bar{x}|^2 & 18 0 & 18 \end{array}\)
\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)
= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)
= \(\frac{42}{6}\)
\(\bar{x}\) = 7
S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\)
= \(\sqrt\frac{18 + 0 + 18}{6}\)
= \(\sqrt\frac{36}{6}\)
= \(\sqrt {6}\)
\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)
= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)
= \(\frac{42}{6}\)
\(\bar{x}\) = 7
S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\)
= \(\sqrt\frac{18 + 0 + 18}{6}\)
= \(\sqrt\frac{36}{6}\)
= \(\sqrt {6}\)
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