System Tools
Windows XP provides some basic tools to make the maintenance process. This note contains description on System Tools.
Summary
Windows XP provides some basic tools to make the maintenance process. This note contains description on System Tools.
Things to Remember
- Windows XP provides some basic tools to make the maintenance process.
- All Windows XP computers have at least one hard disk.The hard disk acts as computer’s storage area.
- Disk Cleanup inspects your hard disk and looks for the files that can be safely deleted.
MCQs
No MCQs found.
Subjective Questions
Q1:
Find the mean deviation from mean.
Daily wages (Rs) 90, 115, 100, 125, 110
Type: Long Difficulty: Easy
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Answer: <p>Calculation of mean deviation from mean</p> <table width="293"><tbody><tr><td><strong>Wages (Rs) X</strong></td> <td><strong>D = X - \(\overline X\)</strong></td> <td><strong>\(\begin {vmatrix}D\\ \end {vmatrix}\)</strong></td> </tr><tr><td>90</td> <td>-18</td> <td>18</td> </tr><tr><td>100</td> <td>-8</td> <td>8</td> </tr><tr><td>110</td> <td>2</td> <td>2</td> </tr><tr><td>115</td> <td>7</td> <td>7</td> </tr><tr><td>125</td> <td>17</td> <td>17</td> </tr><tr><td>\(\sum\)X = 540</td> <td></td> <td>\(\sum\)\(\begin {vmatrix}D\\ \end {vmatrix}\) = 52</td> </tr></tbody></table><p>Mean (\(\overline X\)) = \(\frac {\sum X}N\) = \(\frac {540}5\) = 108</p> <p>MeanDeviation from mean = \(\frac {\sum\begin {vmatrix}D\\ \end {vmatrix}}N\) = \(\frac {52}5\) = 10.4<sub>Ans</sub></p>
Q2:
Find mean deviation from median:
400, 100, 200, 300, 350, 250, 150
Type: Long Difficulty: Easy
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Answer: <p>The given data in ascending order:</p> <p>100. 150, 200, 250, 300, 350, 400</p> <p>N = 7</p> <p>\begin{align*} Median\; (M_d) &= (\frac {N + 1}2)^{th} term\\ &= (\frac {7 + 1}2)^{th}) term\\ &= (\frac 82)^{th} term\\ &= 4^{th} term\\ &= 250\\ \end{align*}</p> <p>Calculating the mean deviation from median</p> <table width="366"><tbody><tr><td><strong>x</strong></td> <td><strong>x - mdn.</strong></td> <td><strong>\(\begin{vmatrix} x - mdn\\ \end{vmatrix}\)</strong></td> </tr><tr><td>100</td> <td>100 - 250 = -150</td> <td>150</td> </tr><tr><td>150</td> <td>150 - 250 = -100</td> <td>100</td> </tr><tr><td>200</td> <td>200 - 250 = -50</td> <td>50</td> </tr><tr><td>250</td> <td>250 - 250 = 0</td> <td>0</td> </tr><tr><td>300</td> <td>300 - 250 = 50</td> <td>50</td> </tr><tr><td>350</td> <td>350 - 250 = 100</td> <td>100</td> </tr><tr><td>400</td> <td>400 - 250 = 150</td> <td>150</td> </tr><tr><td></td> <td>N = 7</td> <td>\(\sum\)\(\begin{vmatrix} x - mdn\\ \end{vmatrix}\) = 600</td> </tr></tbody></table><p>\begin{align*} Mean\; Deviation\; from\; median\; &= \frac {\sum\begin{vmatrix} x - mdn\\ \end{vmatrix}}N\\ &= \frac {600}7\\ &= 85.71\\ \end{align*}</p> <p>∴ Mean Deviation from median = 85.71<sub>Ans</sub></p>
Q3:
Calculate the mean deviation from median from the data given below: 40, 44, 54, 60, 62
Type: Long Difficulty: Easy
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Answer: <p>Given data is:</p> <p>40, 44, 54, 60, 62</p> <p>N = 5</p> <p>\begin{align*} Median\; (M_d) &= \frac {(N+1)^{th}}2\; term\\ &= \frac {(5+1)^{th}}2\;term\\ &= \frac {6^{th}}2\;term\\ &= 3^{rd}\; term\\ \end{align*}</p> <p>Position of 3<sup>rd</sup> term = 54</p> <p>∴ median (mdn) = 54</p> <p>Calculating mean deviation from median</p> <table width="236"><tbody><tr><td><strong>x</strong></td> <td><strong>x - median</strong></td> <td><strong>\(\begin{vmatrix}x - median\end{vmatrix}\)</strong></td> </tr><tr><td>40</td> <td>-14</td> <td>14</td> </tr><tr><td>44</td> <td>-10</td> <td>10</td> </tr><tr><td>54</td> <td>0</td> <td>0</td> </tr><tr><td>60</td> <td>6</td> <td>6</td> </tr><tr><td>62</td> <td>8</td> <td>8</td> </tr><tr><td></td> <td></td> <td>\(\sum\)\(\begin{vmatrix}x - median\end{vmatrix}\) = 38</td> </tr></tbody></table><p>\begin{align*} Mean\; deviation\; from\; median &= \frac {\sum\begin{vmatrix}x - median\end{vmatrix}}N\\ &= \frac {38}5\\ &= 7.6_{Ans}\\ \end{align*}</p>
Q4:
Compute the mean deviation from the mean and its coefficient from the data given below:
| Score (x) | 1 | 2 | 3 | 4 | 5 |
| Frequency (f) | 2 | 5 | 6 | 5 | 2 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Mean</p> <table width="353"><tbody><tr><td> <p><strong>Score</strong></p> <p><strong>(x)</strong></p> </td> <td> <p><strong>Frequency</strong></p> <p><strong>(f)</strong></p> </td> <td><strong>fx</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}x - \overline{X}\\ \end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>1</td> <td>2</td> <td>2</td> <td>2</td> <td>4</td> </tr><tr><td>2</td> <td>5</td> <td>10</td> <td>1</td> <td>5</td> </tr><tr><td>3</td> <td>6</td> <td>18</td> <td>0</td> <td>0</td> </tr><tr><td>4</td> <td>5</td> <td>20</td> <td>1</td> <td>5</td> </tr><tr><td>5</td> <td>2</td> <td>10</td> <td>2</td> <td>4</td> </tr><tr><td></td> <td>N = 20</td> <td>\(\sum\)fx = 60</td> <td></td> <td>\(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 18</td> </tr></tbody></table><p>\begin{align*} Mean\; (\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {60}{20}\\ &= 3\\ \end{align*}</p> <p>\begin{align*} Mean\; Deviation\; from\; mean\;(M.D) &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {18}{20}\\ &= 0.9_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;M.D &= \frac {M.D.}{Mean}\\ &= \frac {0.9}3\\ &= 0.3_{Ans}\\ \end{align*}</p>
Q5:
Calculate the mean deviation from mean according to the data given below:
| x | 10 | 15 | 20 | 25 | 30 |
| f | 2 | 4 | 6 | 8 | 5 |
Type: Long Difficulty: Easy
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Answer: <p>Calculating Mean Deviation from Mean</p> <table width="360"><tbody><tr><td><strong>x</strong></td> <td><strong>f</strong></td> <td><strong>fx</strong></td> <td><strong>\(\begin{vmatrix}x - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}x - \overline{X}\end{vmatrix}\)</strong></td> </tr><tr><td>10</td> <td>2</td> <td>20</td> <td>12</td> <td>24</td> </tr><tr><td>15</td> <td>4</td> <td>60</td> <td>7</td> <td>28</td> </tr><tr><td>20</td> <td>6</td> <td>120</td> <td>2</td> <td>12</td> </tr><tr><td>25</td> <td>8</td> <td>200</td> <td>3</td> <td>24</td> </tr><tr><td>30</td> <td>5</td> <td>150</td> <td>8</td> <td>40</td> </tr><tr><td></td> <td>N = 25</td> <td>\(\sum\)fx = 550</td> <td></td> <td>\(\sum\)f\(\begin{vmatrix}x - \overline{X}\end{vmatrix}\) = 128</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {550}{25}\\ &= 22\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}x - \overline{X}\end{vmatrix}}N\\ &= \frac {128}{25}\\ &= 5.12_{Ans}\\ \end{align*}</p>
Q6:
Calculate the mean deviation and its coefficient from mean of the following data.
| X | 20 | 18 | 16 | 14 | 12 | 10 | 8 | 6 |
| f | 2 | 4 | 9 | 18 | 27 | 25 | 14 | 1 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of M.D. from Mean</p> <table width="408"><tbody><tr><td><strong>X</strong></td> <td><strong>f</strong></td> <td><strong>fx</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}X - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>20</td> <td>2</td> <td>40</td> <td>8</td> <td>16</td> </tr><tr><td>18</td> <td>4</td> <td>72</td> <td>6</td> <td>24</td> </tr><tr><td>16</td> <td>9</td> <td>144</td> <td>4</td> <td>36</td> </tr><tr><td>14</td> <td>18</td> <td>252</td> <td>2</td> <td>36</td> </tr><tr><td>12</td> <td>27</td> <td>324</td> <td>0</td> <td>0</td> </tr><tr><td>10</td> <td>25</td> <td>250</td> <td>2</td> <td>50</td> </tr><tr><td>8</td> <td>14</td> <td>112</td> <td>4</td> <td>56</td> </tr><tr><td>6</td> <td>1</td> <td>6</td> <td>12</td> <td>6</td> </tr><tr><td></td> <td>N = 100</td> <td>\(\sum\)fX = 1200</td> <td></td> <td>\(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 224</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fX}}N\\ &= \frac {1200}{100}\\ &= 12\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {224}{100}\\ &= 2.24_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;the\;Mean\;Deviation\;from\;Mean &= \frac {M.D.}{Mean}\\ &= \frac {2.24}{12}\\ &= 0.186_{Ans}\\ \end{align*}</p>
Q7:
Find the mean deviation from mode of the following data and its coefficient of M.D.
| Marks | 20 | 24 | 30 | 32 | 40 |
| No. of students | 12 | 28 | 40 | 26 | 14 |
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>The maximum frequencyis 40 for value of variable 30 so mode (M<sub>o</sub>) = 30</p> <table width="358"><tbody><tr><td><strong>Marks (X)</strong></td> <td><strong>No. of students (f)</strong></td> <td><strong>\(\begin{vmatrix}X - M_o \end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}X - M_o \end{vmatrix}\)</strong></td> </tr><tr><td>20</td> <td>12</td> <td>10</td> <td>120</td> </tr><tr><td>24</td> <td>28</td> <td>6</td> <td>168</td> </tr><tr><td>30</td> <td>40</td> <td>0</td> <td>0</td> </tr><tr><td>32</td> <td>26</td> <td>2</td> <td>52</td> </tr><tr><td>40</td> <td>14</td> <td>10</td> <td>140</td> </tr><tr><td></td> <td>N = 120</td> <td></td> <td>\(\sum\)f\(\begin{vmatrix}X - M_o \end{vmatrix}\) = 480</td> </tr></tbody></table><p>\begin{align*} M.D.\;from\;Mode &= \frac {\sum{f}\begin{vmatrix}X - M_o \end{vmatrix}}N\\ &= \frac {480}{120}\\ &= 4_{Ans} \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mode &= \frac {M.D.\;from\;Mode}{mode}\\ &= \frac 4{30}\\ &= 0.133_{Ans}\\ \end{align*}</p>
Q8:
Find the mean deviation from mean of the following data.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 2 | 3 | 6 | 5 | 4 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Mean</p> <table width="478"><tbody><tr><td><strong>Class interval</strong></td> <td><strong>Mid-value (m)</strong></td> <td><strong>f</strong></td> <td><strong>fm</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>0-10</td> <td>5</td> <td>2</td> <td>10</td> <td>23</td> <td>46</td> </tr><tr><td>10-20</td> <td>15</td> <td>3</td> <td>45</td> <td>13</td> <td>39</td> </tr><tr><td>20-30</td> <td>25</td> <td>6</td> <td>150</td> <td>3</td> <td>18</td> </tr><tr><td>30-40</td> <td>35</td> <td>5</td> <td>175</td> <td>7</td> <td>35</td> </tr><tr><td>40-50</td> <td>45</td> <td>4</td> <td>180</td> <td>17</td> <td>68</td> </tr><tr><td></td> <td></td> <td>N = 20</td> <td>\(\sum fm\) = 560</td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 206</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {560}{20}\\ &= 28\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &=\frac {\sum {f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {206}{20}\\ &= 10.3_{Ans}\\ \end{align*}</p>
Q9:
Find the mean deviation from mean and its coefficient of the following data:
48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13, 47, 20, 53, 64, 34, 75, 66, 60, 21, 45, 57, 15, 41
Type: Long Difficulty: Easy
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Answer: <p>Calculation of M.D. from Mean</p> <table width="425"><tbody><tr><td> <p><strong>Class interval</strong></p> <p><strong>(x)</strong></p> </td> <td> <p><strong>Mid-value</strong></p> <p><strong>(m)</strong></p> </td> <td> <p><strong>frequency</strong></p> <p><strong>(f)</strong></p> </td> <td><strong>fm</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>10-20</td> <td>15</td> <td>3</td> <td>45</td> <td>29.28</td> <td>87.84</td> </tr><tr><td>20-30</td> <td>25</td> <td>5</td> <td>125</td> <td>19.28</td> <td>96.4</td> </tr><tr><td>30-40</td> <td>35</td> <td>4</td> <td>140</td> <td>9.28</td> <td>37.12</td> </tr><tr><td>40-50</td> <td>45</td> <td>5</td> <td>225</td> <td>0.72</td> <td>3.6</td> </tr><tr><td>50-60</td> <td>55</td> <td>4</td> <td>220</td> <td>10.72</td> <td>42.88</td> </tr><tr><td>60-70</td> <td>65</td> <td>4</td> <td>260</td> <td>20.72</td> <td>82.88</td> </tr><tr><td>70-80</td> <td>75</td> <td>3</td> <td>225</td> <td>30.72</td> <td>92.16</td> </tr><tr><td></td> <td></td> <td>N = 28</td> <td>\(\sum\) N = 28</td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 442.88</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {1240}{28}\\ &= 44.28\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum {f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {442.88}{28}\\ &= 15.82_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;M.D.\;from\;Mean &= \frac {M.D.}{Mean}\\ &= \frac {15.82}{44.28}\\ &= 0.36_{Ans}\\ \end{align*}</p>
Q10:
From the following data calculate the mean deviation from mean.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 2 | 3 | 6 | 5 | 4 |
Type: Long Difficulty: Easy
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Answer: <p>Let: Assumed mean (A) = 25</p> <p>Calculation of Mean Deviation from Mean</p> <table width="471"><tbody><tr><td>Class interval</td> <td>mid-value</td> <td>f</td> <td>d = \(\frac {m-25}{10}\)</td> <td>fd</td> <td>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m = \overline{X}\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}D\end{vmatrix}\)</td> </tr><tr><td>0-10</td> <td>5</td> <td>2</td> <td>-2</td> <td>-4</td> <td>23</td> <td>46</td> </tr><tr><td>10-20</td> <td>15</td> <td>3</td> <td>-1</td> <td>-3</td> <td>13</td> <td>39</td> </tr><tr><td>20-30</td> <td>25</td> <td>6</td> <td>0</td> <td>0</td> <td>3</td> <td>18</td> </tr><tr><td>30-40</td> <td>35</td> <td>5</td> <td>1</td> <td>5</td> <td>7</td> <td>35</td> </tr><tr><td>40-50</td> <td>45</td> <td>4</td> <td>0</td> <td>8</td> <td>17</td> <td>68</td> </tr><tr><td></td> <td></td> <td>N = 20</td> <td></td> <td>\(\sum {fd}\) = 6</td> <td></td> <td>\(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 206</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= A + \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N× I\\ &= 25 + \frac 6{20}× 10\\ &= 28\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {206}{20}\\ &= 10.3_{Ans}\\ \end{align*}</p>
Q11:
Compute the mean deviation from mean from the table given below:
| Weight in kg. | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of items | 2 | 5 | 6 | 5 | 2 |
Type: Long Difficulty: Easy
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Answer: <p>Calculating Mean Deviation from Mean</p> <table width="439"><tbody><tr><td><strong>Weight in kg.</strong></td> <td><strong>Mid-value (m)</strong></td> <td><strong>No. of men (f)</strong></td> <td><strong>fm</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>20-30</td> <td>25</td> <td>2</td> <td>50</td> <td>20</td> <td>40</td> </tr><tr><td>30-40</td> <td>35</td> <td>5</td> <td>175</td> <td>10</td> <td>50</td> </tr><tr><td>40-50</td> <td>45</td> <td>6</td> <td>270</td> <td>0</td> <td>0</td> </tr><tr><td>50-60</td> <td>55</td> <td>5</td> <td>275</td> <td>10</td> <td>50</td> </tr><tr><td>60-70</td> <td>65</td> <td>2</td> <td>130</td> <td>20</td> <td>40</td> </tr><tr><td></td> <td></td> <td>N = 20</td> <td>\(\sum {fm}\) = 900</td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 180</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {900}{20}\\ &= 45\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean\;(M.D.) &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {180}{20}\\ &= 9\;kg_{Ans}\\ \end{align*}</p>
Q12:
Find the mean deviation from mean for the following data:
| Wages (in Rs.) | 0-4 | 0-8 | 0-12 | 0-16 | 0-20 |
| No. of persons | 5 | 12 | 22 | 37 | 44 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Mean</p> <table width="439"><tbody><tr><td><strong>Wages (in Rs.)</strong></td> <td><strong>Mid-value (m)</strong></td> <td><strong>f</strong></td> <td><strong>fm</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>0-4</td> <td>2</td> <td>5</td> <td>10</td> <td>1.09</td> <td>5.45</td> </tr><tr><td>4-8</td> <td>6</td> <td>7</td> <td>42</td> <td>30.91</td> <td>216.37</td> </tr><tr><td>8-12</td> <td>10</td> <td>10</td> <td>100</td> <td>88.91</td> <td>889.1</td> </tr><tr><td>12-16</td> <td>14</td> <td>15</td> <td>210</td> <td>198.91</td> <td>2983.65</td> </tr><tr><td>16-20</td> <td>18</td> <td>7</td> <td>126</td> <td>114.91</td> <td>804.37</td> </tr><tr><td></td> <td></td> <td>N =44</td> <td>\(\sum {fm}\) =488</td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 4898.94</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {488}{44}\\ &= 11.09\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {4898.94}{44}\\ &= 113.3_{Ans}\\ \end{align*}</p>
Q13:
Calculate the mean deviation from median and its coefficient:
| Marks | 20 | 30 | 40 | 50 | 60 | 70 |
| No. of students | 4 | 7 | 12 | 2 | 4 | 6 |
Type: Long Difficulty: Easy
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Answer: <p>Calculating Mean Deviation from Median</p> <table width="455"><tbody><tr><td><strong>Marks (X)</strong></td> <td><strong>No. of students (f)</strong></td> <td><strong>c.f.</strong></td> <td><strong>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}X - mdn\end{vmatrix}\)</strong></td> <td><strong>f\(\begin{vmatrix}D\end{vmatrix}\)</strong></td> </tr><tr><td>20</td> <td>4</td> <td>4</td> <td>420</td> <td>80</td> </tr><tr><td>30</td> <td>7</td> <td>11</td> <td>10</td> <td>70</td> </tr><tr><td>40</td> <td>12</td> <td>23</td> <td>0</td> <td>0</td> </tr><tr><td>50</td> <td>2</td> <td>25</td> <td>10</td> <td>20</td> </tr><tr><td>60</td> <td>4</td> <td>29</td> <td>20</td> <td>80</td> </tr><tr><td>70</td> <td>6</td> <td>35</td> <td>30</td> <td>180</td> </tr><tr><td></td> <td>N = 35</td> <td></td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 430</td> </tr></tbody></table><p>\begin{align*} Position\;of\;Median &= (\frac {N+1}2)^{th}\;item\\ &= (\frac {35+1}2)^{th}\;item\\ &= 18^{th}\;item\\ \end{align*}</p> <p>18<sup>th</sup><sup></sup>item = 40</p> <p>∴ Mdn = 40</p> <p>\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {430}{35}\\ &= 12.28_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;Mean\;Deviation\;from\;Median &= \frac {M.D.\;from\;Median}{Median}\\ &= \frac {12.28}{40}\\ &= 0.307_{Ans}\\ \end{align*}</p>
Q14:
Find the mean deviation about the median of the following data.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No. of students | 4 | 6 | 10 | 10 | 15 | 5 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of M.D. from Median</p> <table width="474"><tbody><tr><td>Marks</td> <td>f</td> <td>c.f</td> <td>mid-value (m)</td> <td>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - md.\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}D\end{vmatrix}\)</td> </tr><tr><td>0-10</td> <td>4</td> <td>4</td> <td>5</td> <td>30</td> <td>120</td> </tr><tr><td>10-20</td> <td>6</td> <td>10</td> <td>15</td> <td>20</td> <td>120</td> </tr><tr><td>20-30</td> <td>10</td> <td>20</td> <td>25</td> <td>10</td> <td>100</td> </tr><tr><td>30-40</td> <td>10</td> <td>30</td> <td>35</td> <td>0</td> <td>0</td> </tr><tr><td>40-50</td> <td>15</td> <td>45</td> <td>45</td> <td>10</td> <td>150</td> </tr><tr><td>50-60</td> <td>5</td> <td>50</td> <td>55</td> <td>20</td> <td>100</td> </tr><tr><td></td> <td>N = 50</td> <td></td> <td></td> <td></td> <td>\(\sum {f}\)\(\begin{vmatrix}D\end{vmatrix}\) = 590</td> </tr></tbody></table><p>\begin{align*} Median &= \frac {N^{th}}2\;item\\ &= \frac {50^{th}}2\;item\\ &= 25^{th}\;term\\ \end{align*}</p> <p>Median Class = (30-40)</p> <p>Here,</p> <p>L = 30</p> <p>\(\frac N2\) = 25</p> <p>cf = 20</p> <p>f = 10</p> <p>I = 10</p> <p>\begin{align*} Median\;(mdn) &= L\;+\;\frac{\frac N2\;-\;cf}f\; ×\;I\\ &= 30\;+\;\frac {25-20}{10}\;×\;10\\ &= 30\;+\;5\\ &= 35\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {590}{50}\\ &= 11.8_{Ans}\\ \end{align*}</p>
Q15:
Find the mean deviation from median and its coefficient:
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of students | 6 | 8 | 11 | 14 | 8 | 3 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Median</p> <table width="474"><tbody><tr><td>Marks</td> <td>f</td> <td>c.f</td> <td>mid-value (m)</td> <td>\(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - md.\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}D\end{vmatrix}\)</td> </tr><tr><td>10-20</td> <td>6</td> <td>6</td> <td>15</td> <td>25</td> <td>150</td> </tr><tr><td>20-30</td> <td>8</td> <td>14</td> <td>25</td> <td>15</td> <td>120</td> </tr><tr><td>30-40</td> <td>11</td> <td>25</td> <td>35</td> <td>5</td> <td>55</td> </tr><tr><td>40-50</td> <td>14</td> <td>39</td> <td>45</td> <td>5</td> <td>70</td> </tr><tr><td>50-60</td> <td>8</td> <td>47</td> <td>55</td> <td>15</td> <td>120</td> </tr><tr><td>60-70</td> <td>3</td> <td>50</td> <td>65</td> <td>25</td> <td>75</td> </tr><tr><td></td> <td>N = 50</td> <td></td> <td></td> <td></td> <td>\(\sum {f}\)\(\begin{vmatrix}D\end{vmatrix}\) = 590</td> </tr></tbody></table><p>\begin{align*} Position\;of\;Median &= \frac {N^{th}}2\;item\\ &= \frac {50^{th}}2\;item\\ &= 25^{th}\;term\\ \end{align*}</p> <p>Median Class = (30-40)</p> <p>Here,</p> <p>L = 30</p> <p>\(\frac N2\) = 25</p> <p>cf =14</p> <p>f =11</p> <p>I = 10</p> <p></p> <p>\begin{align*} Median\;(mdn) &= L\;+\;\frac{\frac N2\;-\;cf}f\; ×\;I\\ &= 30\;+\;\frac {25-14}{11}\;×\;10\\ &= 30\;+\;\frac {11}{11}\;×\;10\\ &= 40\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {590}{50}\\ &= 11.8\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;M.D.\;from\;Median &= \frac {M.D.}{Median}\\ &= \frac {11.8}{40}\\ &= 0.295_{Ans}\\ \end{align*}</p> <p></p>
Q16:
Find the mean deviation from median of the following data:
| Age (in year) | 10 | 12 | 15 | 16 | 17 |
| No. of people | 6 | 14 | 20 | 13 | 7 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Median</p> <table width="433"><tbody><tr><td> <p>Age (in year)</p> <p>X</p> </td> <td> <p>No. of people</p> <p>(f)</p> </td> <td> <p>Cumulative Frequency</p> <p>(c.f.)</p> </td> <td>\(\begin{vmatrix}X - M_d\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}X - M_d\end{vmatrix}\)</td> </tr><tr><td>10</td> <td>6</td> <td>6</td> <td>5</td> <td>30</td> </tr><tr><td>12</td> <td>14</td> <td>20</td> <td>3</td> <td>42</td> </tr><tr><td>15</td> <td>20</td> <td>40</td> <td>0</td> <td>0</td> </tr><tr><td>16</td> <td>13</td> <td>53</td> <td>1</td> <td>13</td> </tr><tr><td>20</td> <td>7</td> <td>60</td> <td>5</td> <td>35</td> </tr><tr><td></td> <td>N = 60</td> <td></td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}X - M_d\end{vmatrix}\) = 120</td> </tr></tbody></table><p>\begin{align*} Median\;(M_d) &= (\frac {N\;+\;1}2)^{th}\;item\\ &= (\frac {60\;+\;1}2)^{th}\;item\\ &= (30.5)^{th}\;item &= 15\\ \end{align*}</p> <p>\begin{align*} M.D.\;from\;Median &= \frac {\sum {f}\begin{vmatrix}X - M_d\end{vmatrix}}N\\ &= \frac {120}{60}\\ &= 2_{Ans}\\ \end{align*}</p>
Q17:
Find the mean deviation from the mode of the given frequency distribution.
| Age (in years) | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |
| No. of students | 4 | 6 | 8 | 5 | 2 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Mode</p> <table width="414"><tbody><tr><td>Age (in years)</td> <td>f</td> <td>Mid-value (X)</td> <td>\(\begin{vmatrix}X - M_o\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}X - M_o\end{vmatrix}\)</td> </tr><tr><td>0-4</td> <td>4</td> <td>2</td> <td>7.6</td> <td>3.04</td> </tr><tr><td>4-8</td> <td>6</td> <td>6</td> <td>3.6</td> <td>21.6</td> </tr><tr><td>8-12</td> <td>8</td> <td>10</td> <td>0.4</td> <td>3.2</td> </tr><tr><td>12-16</td> <td>5</td> <td>14</td> <td>4.4</td> <td>22</td> </tr><tr><td>16-20</td> <td>2</td> <td>18</td> <td>8.4</td> <td>16.8</td> </tr><tr><td></td> <td>N = 25</td> <td></td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}X - M_o\end{vmatrix}\) = 34</td> </tr></tbody></table><p>Model Class = The class containing maximum frequency = (8-12)</p> <p>Here,</p> <p>L = 8</p> <p>f<sub>0</sub> = 6</p> <p>f<sub>1</sub> = 8</p> <p>f<sub>2</sub> = 5</p> <p>\begin{align*} Mode\;(M_o) &= L\;+\;\frac {f_1\;-\;f_0}{2f_1\;-\;f_0\;-\;f_2}\\ &= 8 + \frac {8\;-\;6}{2\;×\;8\;-\;6\;-\;5}\\ &= 8\;+\;\frac 85\\ &= 9.6\;years\\ \end{align*}</p> <p>\begin{align*} Mean\;Deviation\;from\;Mode &= \frac {\sum {f}\begin{vmatrix}X - M_o\end{vmatrix}}N\\ &= \frac {34}{25}\\ &= 1.36\;years_{Ans}\\ \end{align*}</p>
Q18:
Find the mean from median of the following data:
| Age (in years) | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of people | 5 | 7 | 8 | 6 | 4 |
Type: Long Difficulty: Easy
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Answer: <p>Calculation of Mean Deviation from Median</p> <table width="466"><tbody><tr><td>Age (in years)</td> <td>Mid-value (X)</td> <td>Frequency (f)</td> <td>c.f.</td> <td>\(\begin{vmatrix}X - M_d\end{vmatrix}\)</td> <td>f\(\begin{vmatrix}X - M_d\end{vmatrix}\)</td> </tr><tr><td>20-30</td> <td>25</td> <td>5</td> <td>5</td> <td>18.75</td> <td>93.75</td> </tr><tr><td>30-40</td> <td>35</td> <td>7</td> <td>12</td> <td>8.75</td> <td>61.25</td> </tr><tr><td>40-50</td> <td>45</td> <td>8</td> <td>20</td> <td>1.25</td> <td>10</td> </tr><tr><td>50-60</td> <td>55</td> <td>6</td> <td>26</td> <td>11.25</td> <td>67.5</td> </tr><tr><td>60-70</td> <td>65</td> <td>4</td> <td>30</td> <td>21.25</td> <td>85</td> </tr><tr><td></td> <td></td> <td>N = 30</td> <td></td> <td></td> <td>\(\sum f\)\(\begin{vmatrix}X - M_d\end{vmatrix}\) = 317.5</td> </tr></tbody></table><p>\begin{align*} Class\;of\;Median &= (\frac N2)^{th}\;item\\ &= (\frac {30}2)^{th}\;item\\ &= 15^{th}\;item\\ \end{align*}</p> <p>15<sup>th</sup> item represents (40-50) class</p> <p>Here,</p> <p>L = 40</p> <p>\(\frac N2\) = 15</p> <p>c.f. = 12</p> <p>f = 8</p> <p>I = 10</p> <p>\begin{align*} Median\;(M_d) &= L\;+\;\frac {\frac N2 - c.f.}f\;×\;I\\ &= 40\;+\;\frac {15\;-\;12}8\;×\;10\\ &= 40\;+\;\frac {30}8\\ &= 43.75\\ \end{align*}</p> <p>\begin{align*} M.D.\;from\;Median &= \frac {\sum {f}\begin{vmatrix}X - M_d\end{vmatrix}}N\\ &= \frac {317.5}{30}\\ &= 10.58years_{Ans}\\ \end{align*}</p>
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System Tools
What are System Tools?
Do some basic maintenance from time to time to keep Windows XP running smoothly. Windows XP provides you with some basic tools to make the maintenance process. You can access these tools via the System Tools menu.
Disk Defragmenter
All Windows XP computers have at least one hard disk. The hard disk acts as your computer's storage area. Almost everything installed on your computer- applications, files, folders, and operating system- is stored here.
With general use, your hard disk can become fragmented. This means that parts of the same disk file become scattered over different areas of the disk. A fragmented hard disk slows down your computer and hinders its performance.
To keep your hard disk running smoothly, you must routinely defragment or defrag the hard disk. The Disk Defragmenter tool can help you do this.
To use Disk Defragmenter
- Choose Start.-----All Programs------Accessories----------System Tools-----Disk Defragmenter.
- The Disk Defragmenter opens.
To open Microsoft Management Console
- Open Disk Defragmenter.
- Open the Action menu.
- Click Help.
A microsoft Management Console opens and explains how to run Disk Defragmenter.
Disk Cleanup
Disk Cleanup is another tool that helps to keep Windows XP operating as it should. It inspects your hard disk and looks for the files that can be safely deleted. Deleting unnecessary files free up valuable disk space.
To Run Disk Cleanup
- Choose Start--------All Programs-------Accessories-------System Tools--------Disk cleanup.
- The Disk Cleanup window opens. Click the Disk Cleanup tab if it's not showing.
- Disk Cleanup lists several potential files that can be deleted and the amount of disk space you'll gain by emptying each one.
- To learn more about a file category, click it and click the View Files button. A description appears.
- Click the check boxes next to the categories you want to delete and click OK.
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Lesson
Operating System
Subject
Computer
Grade
Grade 9
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