Statistics - SS2 Mathematics Past Questions and Answers - page 1
In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\). What is the mean of the data?
64
67
65
61
\(mean = \ \frac{45 + 67 + 89 + 95 + 76 + 12}{6} = \frac{384}{6} = 64\)
In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\), What is the median of the data?
71.5
76
67
75.5
\(median\ \left\{ 12,\ 45,\ 67,\ 76,\ 89,\ 85 \right\} = \ \frac{67 + 76}{2} = \frac{143}{2} = 71.5\)
In a set of data: \(45,\ 67,\ 89,\ 95,\ 76,\ 12\), What is the range of the data?
71.5
73
67
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.
| \[Score\] | \[Frequency\] |
|---|---|
| \[50 - 54\] | \[3\] |
| \[55 - 59\] | \[5\] |
| \[60 - 64\] | \[8\] |
| \[65 - 69\] | \[10\] |
| \[70 - 74\] | \[7\] |
| \[75 - 79\] | \[6\] |
| \[80 - 84\] | \[3\] |
| \[85 - 89\] | \[2\] |
| \[90 - 94\] | \[1\] |
| \[Score\] | \[f\] | \[cf\] | \[X\] | \[Xf\] | \[\overline{X}\] | \[X - \overline{X} = d\] | \[d^{2}\] | \[fd^{2}\] | \[|d\text{|}\] | \[f|d|\] |
|---|---|---|---|---|---|---|---|---|---|---|
| \[50 - 54\] | \[3\] | \[3\] | \[52\] | 156 | \[69\] | \[- 17\] | \[289\] | \[867\] | \[17\] | \[51\] |
| \[55 - 59\] | \[5\] | \[8\] | \[57\] | \[286\] | \[69\] | \[- 12\] | \[144\] | \[720\] | \[12\] | \[60\] |
| \[60 - 64\] | \[8\] | \[16\] | \[62\] | \[496\] | \[69\] | \[- 7\] | \[49\] | \[392\] | \[7\] | \[56\] |
| \[65 - 69\] | \[10\] | \[26\] | \[67\] | \[670\] | \[69\] | \[- 2\] | \[4\] | \[40\] | \[2\] | \[20\] |
| \[70 - 74\] | \[7\] | \[33\] | \[72\] | \[504\] | \[69\] | \[3\] | \[9\] | \[63\] | \[3\] | \[21\] |
| \[75 - 79\] | \[6\] | \[39\] | \[77\] | \[462\] | \[69\] | \[8\] | \[64\] | \[384\] | \[8\] | \[48\] |
| \[80 - 84\] | \[3\] | \[42\] | \[82\] | \[246\] | \[69\] | \[13\] | \[169\] | \[507\] | \[13\] | \[39\] |
| \[85 - 89\] | \[2\] | \[44\] | \[87\] | \[174\] | \[69\] | \[18\] | \[324\] | \[648\] | \[18\] | \[36\] |
| \[90 - 94\] | \[1\] | \[45\] | \[92\] | \[92\] | \[69\] | \[23\] | \[529\] | \[529\] | \[23\] | \[23\] |
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\(Mean,\ \overline{X} = \frac{\sum_{}^{}{Xf}}{\sum_{}^{}f} = \ \frac{3,086}{45} = 68.58 \approx 69\)
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\(\frac{n + 1}{2} = \ \frac{45 + 1}{2} = \frac{46}{2} = 23\)
This means the median is 23rd item in the fourth class (\(65 - 69\)), so we take the calculate the median for this grouped data as \(l_{m} + {(\frac{\frac{\sum_{}^{}f}{2} - {cf}_{cb}}{f_{m}})}^{c}\)
\[l_{m} = 64.5\]
\[\sum_{}^{}f = 45\]
\[{cf}_{cb}\ = 16\]
\[f_{m} = 10\]
\[c = 5\]
\[median = l_{m} + (\frac{\frac{\sum_{}^{}f}{2} - {cf}_{cb}}{f_{m}})c\]
\[= 64.5 + \left( \frac{\frac{45}{2} - 16}{10} \right) \times 5 = \ 67.75\]
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The mode is the fourth class \(65 - 69\) with frequency of \(10\)
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mean absolute deviation, \(MD = \frac{\sum_{}^{}{f|d|}}{\sum_{}^{}f} = \frac{354}{45} = 7.87\)
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\(standard\ deviation,\ \sigma = \ \sqrt{\frac{\sum_{}^{}{f{(X - \overline{X})}^{2}}}{\sum_{}^{}f}} = \sqrt{\frac{4150}{45}} = \sqrt{92.22} = 9.603\)
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\(variance,\ \sigma^{2} = {9.603}^{2} = 92.22\)