Statistics - SS2 Mathematics Past Questions and Answers - page 1

\(mean = \ \frac{45 + 67 + 89 + 95 + 76 + 12}{6} = \frac{384}{6} = 64\)

\(median\ \left\{ 12,\ 45,\ 67,\ 76,\ 89,\ 85 \right\} = \ \frac{67 + 76}{2} = \frac{143}{2} = 71.5\)

Range = 85 - 12 = 73

1. \(Mean,\ \overline{X} = \frac{\sum_{}^{}{Xf}}{\sum_{}^{}f} = \ \frac{3,086}{45} = 68.58 \approx 69\)

2. \(\frac{n + 1}{2} = \ \frac{45 + 1}{2} = \frac{46}{2} = 23\)

This means the median is 23^rd item in the fourth class (\(65 - 69\)), so we take the calculate the median for this grouped data as \(l_{m} + {(\frac{\frac{\sum_{}^{}f}{2} - {cf}_{cb}}{f_{m}})}^{c}\)

\[l_{m} = 64.5\]

\[\sum_{}^{}f = 45\]

\[{cf}_{cb}\ = 16\]

\[f_{m} = 10\]

\[c = 5\]

\[median = l_{m} + (\frac{\frac{\sum_{}^{}f}{2} - {cf}_{cb}}{f_{m}})c\]

\[= 64.5 + \left( \frac{\frac{45}{2} - 16}{10} \right) \times 5 = \ 67.75\]

1. The mode is the fourth class \(65 - 69\) with frequency of \(10\)

2. mean absolute deviation, \(MD = \frac{\sum_{}^{}{f|d|}}{\sum_{}^{}f} = \frac{354}{45} = 7.87\)

3. \(standard\ deviation,\ \sigma = \ \sqrt{\frac{\sum_{}^{}{f{(X - \overline{X})}^{2}}}{\sum_{}^{}f}} = \sqrt{\frac{4150}{45}} = \sqrt{92.22} = 9.603\)

4. \(variance,\ \sigma^{2} = {9.603}^{2} = 92.22\)