(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})
The median of the distribution above is
The median of the distribution above is
-
A.
4.0 -
B.
3.4 -
C.
3.2 -
D.
3.0
Correct Answer: Option B
Explanation
Median = (frac{a + b}{fm}) ((frac{1}{2} sum f – CF_b))
= 2.95 + (frac{0.5}{15})(2.-7)
= 2.95 + (frac{0.5}{15}) x 13
= 2.95 + 0. 43
= 3.38
= 3.4