Statistics Practice Test 1

A = Assumed; Mean = 153.5
Mean $=\mathit{A}+\frac{\mathit{\Sigma }\mathit{f}\mathit{d}}{\mathit{\Sigma }\mathit{f}}=153.5+\frac{56}{50}$ = 153.3 + 1.12 = 154.62
Since, $\frac{\mathit{n}}{2}=\frac{50}{2}=25$
∴ Median group is 155 - 156.
$\therefore \mathit{l}=115,\mathit{h}=2,\mathit{c}=20,\mathit{f}=15$
Hence, Median $=\mathit{l}+\frac{\frac{\mathit{n}}{2}-\mathit{c}}{\mathit{f}}\times \mathit{h}$ $=155+\frac{25-20}{15}\times 2=155.66$
Since maximum frequency is 15, then the model group is 155 - 156.
$\therefore {\mathit{f}}_0=8,{\mathit{f}}_1=15$, ${\mathit{f}}_2=9,\mathit{h}=2,\mathit{l}=155$
Mode $=\mathit{l}+\frac{{\mathit{f}}_1+{\mathit{f}}_0}{2{\mathit{f}}_1-{\mathit{f}}_0-{\mathit{f}}_2}\times \mathit{h}$
$=155+\frac{15-8}{30-8-9}\times 2=156.07$
When rectangles are drawn and mean is the least and median lies between mode and mean, then distribution is negatively skewed.