Class Interval1.5?1.92.0?2.42.5?2.93.0?2.93..

Question 1 Report

$\begin{array}{cc} Class Interval & F r e q u e n c y & Class boundaries & C l a s s M i d - p o i n t \\ 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7 \\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2 \\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2 \\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7 \\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2 \\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7 \end{array}$
Find the mode of the distribution above to find the mode of the distribution.

3.2

3.3

3.7

4.2

Answer Details

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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Representation Of Data

Measures Of Dispersion

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Class Interval1.5?1.92.0?2.42.5?2.93.0?2.93.5?3.94.0?4.44.5?4.9Frequency2214151053Class boundaries1.45?1.951.95?2.452.45?2.952.95?3.453.45?3.953.95?4.454.45...

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