Statistics - SS2 Mathematics Past Questions and Answers - page 1

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In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\). What is the mean of the data?

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In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\), What is the median of the data?

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71.5

75.5

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In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\), What is the range of the data?

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71.5

73.5

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The data below is the scores of a set of students in a microbiology class. Solve for the relevant measures of central tendency and dispersion.

\[Score\]	\[Frequency\]
\[50 - 54\]	\[3\]
\[55 - 59\]	\[5\]
\[60 - 64\]	\[8\]
\[65 - 69\]	\[10\]
\[70 - 74\]	\[7\]
\[75 - 79\]	\[6\]
\[80 - 84\]	\[3\]
\[85 - 89\]	\[2\]
\[90 - 94\]	\[1\]

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\[Score\]	\[f\]	\[cf\]	\[X\]	\[Xf\]	\[\overline{X}\]	\[X - \overline{X} = d\]	\[d^{2}\]	\[fd^{2}\]	\[\|d\text{\|}\]	\[f\|d\|\]
\[50 - 54\]	\[3\]	\[3\]	\[52\]	156	\[69\]	\[- 17\]	\[289\]	\[867\]	\[17\]	\[51\]
\[55 - 59\]	\[5\]	\[8\]	\[57\]	\[286\]	\[69\]	\[- 12\]	\[144\]	\[720\]	\[12\]	\[60\]
\[60 - 64\]	\[8\]	\[16\]	\[62\]	\[496\]	\[69\]	\[- 7\]	\[49\]	\[392\]	\[7\]	\[56\]
\[65 - 69\]	\[10\]	\[26\]	\[67\]	\[670\]	\[69\]	\[- 2\]	\[4\]	\[40\]	\[2\]	\[20\]
\[70 - 74\]	\[7\]	\[33\]	\[72\]	\[504\]	\[69\]	\[3\]	\[9\]	\[63\]	\[3\]	\[21\]
\[75 - 79\]	\[6\]	\[39\]	\[77\]	\[462\]	\[69\]	\[8\]	\[64\]	\[384\]	\[8\]	\[48\]
\[80 - 84\]	\[3\]	\[42\]	\[82\]	\[246\]	\[69\]	\[13\]	\[169\]	\[507\]	\[13\]	\[39\]
\[85 - 89\]	\[2\]	\[44\]	\[87\]	\[174\]	\[69\]	\[18\]	\[324\]	\[648\]	\[18\]	\[36\]
\[90 - 94\]	\[1\]	\[45\]	\[92\]	\[92\]	\[69\]	\[23\]	\[529\]	\[529\]	\[23\]	\[23\]

\(Mean,\ \overline{X} = \frac{\sum_{}^{}{Xf}}{\sum_{}^{}f} = \ \frac{3,086}{45} = 68.58 \approx 69\)
\(\frac{n + 1}{2} = \ \frac{45 + 1}{2} = \frac{46}{2} = 23\)

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

\[l_{m} = 64.5\]

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

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

\[f_{m} = 10\]

\[c = 5\]

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

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

The mode is the fourth class \(65 - 69\) with frequency of \(10\)
mean absolute deviation, \(MD = \frac{\sum_{}^{}{f|d|}}{\sum_{}^{}f} = \frac{354}{45} = 7.87\)
\(standard\ deviation,\ \sigma = \ \sqrt{\frac{\sum_{}^{}{f{(X - \overline{X})}^{2}}}{\sum_{}^{}f}} = \sqrt{\frac{4150}{45}} = \sqrt{92.22} = 9.603\)
\(variance,\ \sigma^{2} = {9.603}^{2} = 92.22\)

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Statistics - SS2 Mathematics Past Questions and Answers - page 1

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