Home » (begin{array}{c|c} text{Class Interval} & Frequency & text{Class boundaries} & Class Mid-point \ hline 1.5 -…

(begin{array}{c|c} text{Class Interval} & Frequency & text{Class boundaries} & Class Mid-point \ hline 1.5 -…

(begin{array}{c|c} text{Class Interval} & Frequency & text{Class boundaries} & Class Mid-point \ hline 1.5 – 1.9 & 2 & 1.45 – 1.95 & 1.7\ 2.0 – 2.4 & 21 & 1.95 – 2.45 & 2.2\ 2.5 – 2.9 & 4 & 2.45 – 2.95 & 2.7 \ 3.0 – 2.9 & 15 & 2.95 – 3.45 & 3.2\ 3.5 – 3.9 & 10 & 3.45 – 3.95 & 3.7\ 4.0 – 4.4 & 5 & 3.95 – 4.45 & 4.2\ 4.5 – 4.9 & 3 & 4.45 – 4.95 & 4.7end{array})
Find the mode of the distribution above to find the mode of the distribution.
  • A.
    3.2
  • B.
    3.3
  • C.
    3.7
  • D.
    4.2
Correct Answer: Option B
Explanation

Mode = a + (b – a)(fm – Fb)
2Fm – Fa – Fb
= 3.0 + (frac{(3.4 – 3)(15 – 4)}{2(15) – 4 – 10})
= 3 + (frac{(6.4)(11)}{30 – 14})
= 3 + (frac{4.4}{16})
= 3 + 0.275
= 3.275
= 3.3cm

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