Biogeochemical Cycles
This note explains about biogeochemicals and lists different biogeochemical cycles with their explanation.
Summary
This note explains about biogeochemicals and lists different biogeochemical cycles with their explanation.
Things to Remember
- That essential element which is produced by earth and is needed for living beings to continue their life is called bio-geochemical.
- The process of exchange of nitrogen gas between living and non-living world of an ecosystem is called nitrogen cycle. it has six step and they are nitrogen fixation, nitrogen assimilation, ammonification, nitrification, denitrification, and sedimentation.
MCQs
No MCQs found.
Subjective Questions
Q1:
Find the standard deviation from the data given below:
12, 25, 29, 37, 41, 45, 49
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="333"><tbody><tr><td>x</td> <td>d = x - \(\overline{X}\)</td> <td>d<sup>2</sup></td> </tr><tr><td>12</td> <td>-22</td> <td>484</td> </tr><tr><td>25</td> <td>-9</td> <td>81</td> </tr><tr><td>29</td> <td>-5</td> <td>25</td> </tr><tr><td>37</td> <td>3</td> <td>9</td> </tr><tr><td>41</td> <td>7</td> <td>49</td> </tr><tr><td>45</td> <td>11</td> <td>121</td> </tr><tr><td>49</td> <td>15</td> <td>225</td> </tr><tr><td>\(\sum x\) = 238</td> <td></td> <td>\(\sum {d^2}\) = 994</td> </tr></tbody></table><p>Here,</p> <p>\(\sum x\) = 238</p> <p>N = 7</p> <p>Mean (\(\overline{X})\) = \(\frac {\sum{x}}N\) = \(\frac {238}7\) = 34</p> <p>\begin{align*} ∴ Standard\;Deviation\;(σ) &= \sqrt {\frac {\sum{d^2}}N}\\ &= \sqrt {\frac {994}7}\\ &= \sqrt {142}\\ &= 11.92_{Ans}\\ \end{align*}</p>
Q2:
Compute the standard deviation from the following data:
11, 14, 15, 17, 18
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Given Data in ascending order is:</p> <p>11, 14, 15, 17, 18</p> <p>Let: Assumed mean (A) = 15</p> <p>Calculating Standard Deviation</p> <table width="315"><tbody><tr><td>Number (x)</td> <td>d = X - A</td> <td>d<sup>2</sup></td> </tr><tr><td>11</td> <td>-4</td> <td>16</td> </tr><tr><td>14</td> <td>-1</td> <td>1</td> </tr><tr><td>15</td> <td>0</td> <td>0</td> </tr><tr><td>17</td> <td>2</td> <td>4</td> </tr><tr><td>18</td> <td>3</td> <td>9</td> </tr><tr><td></td> <td>\(\sum d\) = 0</td> <td>\(\sum {d^2}\) = 30</td> </tr></tbody></table><p>\begin{align*}{\therefore}\;Standard \: Deviation \:(\sigma)&=\sqrt{\frac{\sum d^2}{N}- \left( \frac{\sum d}{N}\right)^2}\\&= \sqrt {\frac {30}{5}-(\frac {0}{5})^2}\\ &= \sqrt {6}\\ &= 2.45_{Ans}\end{align*}</p>
Q3:
Compute the standard deviation from the following data. Also find the coefficient of variation.
| X | 12 | 13 | 14 | 15 | 16 | 17 |
| f | 2 | 3 | 6 | 4 | 2 | 1 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="384"><tbody><tr><td>X</td> <td>f</td> <td>fx</td> <td>d = X - \(\overline{X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>12</td> <td>2</td> <td>24</td> <td>-2.2</td> <td>4.84</td> <td>9.68</td> </tr><tr><td>13</td> <td>3</td> <td>39</td> <td>-1.2</td> <td>1.44</td> <td>4.32</td> </tr><tr><td>14</td> <td>6</td> <td>84</td> <td>-0.2</td> <td>0.04</td> <td>0.24</td> </tr><tr><td>15</td> <td>4</td> <td>60</td> <td>0.8</td> <td>0.64</td> <td>2.56</td> </tr><tr><td>16</td> <td>2</td> <td>32</td> <td>1.8</td> <td>3.24</td> <td>6.56</td> </tr><tr><td>17</td> <td>1</td> <td>17</td> <td>2.8</td> <td>7.84</td> <td>7.84</td> </tr><tr><td></td> <td>N = 18</td> <td>\(\sum {fx}\) = 256</td> <td></td> <td></td> <td>\(\sum {fd^2}\) = 31.12</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum {fx}}N\\ &= \frac {256}{18}\\ &= 14.2\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {31.12}{18}}\\ &= \sqrt {1.72}\\ &= 1.31_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;variation &= \frac {\sigma}{\overline{X}}\;×\;100\%\\ &= \frac {1.31}{14.2}\;×\;100\%\\ &= 9.23\%_{Ans}\\ \end{align*}</p>
Q4:
Find the standard deviation, under the following data:
| Score (x) | 18 | 20 | 14 | 16 | 10 | 12 | 8 | 6 |
| Frequency (f) | 4 | 2 | 18 | 9 | 25 | 27 | 14 | 1 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="423"><tbody><tr><td>Score (x)</td> <td>Frequency (f)</td> <td>fx</td> <td>d = x - \(\overline {X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>18</td> <td>4</td> <td>72</td> <td>6</td> <td>36</td> <td>144</td> </tr><tr><td>20</td> <td>2</td> <td>40</td> <td>8</td> <td>64</td> <td>128</td> </tr><tr><td>14</td> <td>18</td> <td>252</td> <td>2</td> <td>4</td> <td>72</td> </tr><tr><td>16</td> <td>9</td> <td>144</td> <td>4</td> <td>16</td> <td>144</td> </tr><tr><td>10</td> <td>25</td> <td>250</td> <td>-2</td> <td>4</td> <td>100</td> </tr><tr><td>12</td> <td>27</td> <td>324</td> <td>0</td> <td>0</td> <td>0</td> </tr><tr><td>8</td> <td>14</td> <td>112</td> <td>-4</td> <td>16</td> <td>224</td> </tr><tr><td>6</td> <td>1</td> <td>6</td> <td>-6</td> <td>36</td> <td>36</td> </tr><tr><td></td> <td>N = 100</td> <td>\(\sum {fx}\)= 1200</td> <td></td> <td></td> <td>\(\sum {fd^2}\) = 848</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fx}}N\\ &= \frac {1200}{100}\\ &= 12\\ \end{align*}</p> <p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {848}{100}}\\ &= \sqrt {8.48}\\ &= 2.91_{Ans}\\ \end{align*}</p>
Q5:
Compute the standard deviation from the following data:
| Marks Secured | 5 | 10 | 15 | 20 | 25 | 30 |
| No. of students | 2 | 3 | 5 | 6 | 3 | 1 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="404"><tbody><tr><td> <p>Marks</p> <p>(X)</p> </td> <td> <p>No. of students</p> <p>(f)</p> </td> <td>fx</td> <td>d = x - \(\overline {X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>5</td> <td>2</td> <td>10</td> <td>-12</td> <td>144</td> <td>288</td> </tr><tr><td>10</td> <td>3</td> <td>30</td> <td>-7</td> <td>49</td> <td>147</td> </tr><tr><td>15</td> <td>5</td> <td>75</td> <td>-2</td> <td>4</td> <td>20</td> </tr><tr><td>20</td> <td>6</td> <td>120</td> <td>3</td> <td>9</td> <td>54</td> </tr><tr><td>25</td> <td>3</td> <td>75</td> <td>8</td> <td>64</td> <td>192</td> </tr><tr><td>30</td> <td>1</td> <td>30</td> <td>13</td> <td>169</td> <td>169</td> </tr><tr><td></td> <td>N = 20</td> <td>\(\sum {fx}\) = 340</td> <td></td> <td></td> <td>\(\sum {fd^2}\) =870</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline {X})\; &= \frac {\sum {fx}}N\\ &= \frac {340}{20}\\ &= 17\\ \end{align*}</p> <p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {870}{20}}\\ &= \sqrt {43.5}\\ &= 6.59_{Ans}\\ \end{align*}</p>
Q6:
Find the standard deviation of the given data:
| X | 12 | 13 | 14 | 15 | 16 | 17 |
| f | 2 | 3 | 6 | 4 | 2 | 1 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="411"><tbody><tr><td>X</td> <td>f</td> <td>fx</td> <td>fx<sup>2</sup></td> </tr><tr><td>12</td> <td>2</td> <td>24</td> <td>288</td> </tr><tr><td>13</td> <td>3</td> <td>39</td> <td>507</td> </tr><tr><td>14</td> <td>6</td> <td>84</td> <td>1176</td> </tr><tr><td>15</td> <td>4</td> <td>60</td> <td>900</td> </tr><tr><td>16</td> <td>2</td> <td>32</td> <td>512</td> </tr><tr><td>17</td> <td>1</td> <td>17</td> <td>289</td> </tr><tr><td></td> <td>N = 18</td> <td>\(\sum {fx}\) = 256</td> <td>\(\sum {fx^2}\) = 3672</td> </tr></tbody></table><p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {3672}{18} - (\frac {256}{18})^2}\\ &= \sqrt {204 - 202.27}\\ &= \sqrt {1.73}\\ &= 1.32_{Ans}\\ \end{align*}</p>
Q7:
Compute the standard deviation from the following data:
| Height | 0-8 | 8-16 | 16-24 | 24-32 | 32-40 |
| No. of plants | 6 | 7 | 10 | 8 | 9 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="447"><tbody><tr><td>Height</td> <td>No. of Plants (f)</td> <td>mid-value (m)</td> <td>fm</td> <td>d = m - \(\overline{X}\)</td> <td>fd<sup>2</sup></td> </tr><tr><td>0-8</td> <td>6</td> <td>4</td> <td>24</td> <td>-17.4</td> <td>1816.56</td> </tr><tr><td>8-16</td> <td>7</td> <td>12</td> <td>84</td> <td>-9.4</td> <td>618.52</td> </tr><tr><td>16-24</td> <td>10</td> <td>20</td> <td>200</td> <td>-1.4</td> <td>19.6</td> </tr><tr><td>24-32</td> <td>8</td> <td>28</td> <td>224</td> <td>6.6</td> <td>348.48</td> </tr><tr><td>32-40</td> <td>9</td> <td>36</td> <td>324</td> <td>14.6</td> <td>1918.44</td> </tr><tr><td></td> <td>N = 40</td> <td></td> <td>\(\sum{fm}\) = 856</td> <td></td> <td>\(\sum{fd^2}\) = 4721.6</td> </tr></tbody></table><p>\begin{align*} Mean\;({\overline {X}})\; &= \frac {\sum {fm}}N\\ &= \frac {856}{40}\\ &= 21.4\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {4721.6}{40}}\\ &= 10.87_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;Variation\;(C.V.) &= \frac {\sigma}{\overline {X}} \times {100}\%\\ &= \frac {10.87}{21.4} \times {100}\%\\ &= 50.79\%_{Ans}\\ \end{align*}</p>
Q8:
Find the coefficient of standard deviation and coefficient of variation of the following data:
59, 71, 45, 44, 35, 21, 29, 49, 42, 37, 58, 69, 55, 39, 79, 50, 65, 52, 60, 64
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Let: Assumed Mean (A) = 45</p> <p>Calculation of the Standard Deviation</p> <table width="525"><tbody><tr><td>Class Interval (x)</td> <td>Mid-value (m)</td> <td>Frequency (f)</td> <td>d = m - A</td> <td>d<sup>2</sup></td> <td>fd</td> <td>fd<sup>2</sup></td> </tr><tr><td>20-30</td> <td>25</td> <td>2</td> <td>-20</td> <td>400</td> <td>-40</td> <td>800</td> </tr><tr><td>30-40</td> <td>35</td> <td>3</td> <td>-10</td> <td>100</td> <td>-30</td> <td>300</td> </tr><tr><td>40-50</td> <td>45</td> <td>4</td> <td>0</td> <td>0</td> <td>0</td> <td>0</td> </tr><tr><td>50-60</td> <td>55</td> <td>5</td> <td>10</td> <td>100</td> <td>50</td> <td>500</td> </tr><tr><td>60-70</td> <td>65</td> <td>4</td> <td>20</td> <td>400</td> <td>80</td> <td>1600</td> </tr><tr><td>70-80</td> <td>75</td> <td>2</td> <td>30</td> <td>900</td> <td>60</td> <td>1800</td> </tr><tr><td></td> <td></td> <td>N = 20</td> <td></td> <td></td> <td>\(\sum{fd}\) = 120</td> <td>\(\sum{fd^2}\) = 5000</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= A + \frac {\sum {fd}}N\\ &= 45 + \frac {120}{20}\\ &= 45 + 6\\ &= 51\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {5000}{20} - (\frac {120}{20})^2}\\ &= \sqrt {250 - 36}\\ &= \sqrt {214}\\ &= 14.63_{Ans}\\ \end{align*}</p> <p>\begin{align*} Coefficient\;of\;Variation\;(C.V.) &= \frac {\sigma}{\overline {X}} \times {100}\%\\ &= \frac {14.63}{51} \times {100}\%\\ &= 29\%_{Ans}\\ \end{align*}</p>
Q9:
Calculate the standard deviation from the following data:
| Marks obtained | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No. of students | 4 | 6 | 10 | 3 | 2 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="485"><tbody><tr><td>Marks obtained</td> <td>Mid-value</td> <td>No. of students (f)</td> <td>fx</td> <td>d= X - \(\overline{X}\)</td> <td>fd</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>10-20</td> <td>15</td> <td>4</td> <td>60</td> <td>-17.5</td> <td>-70</td> <td>306.25</td> <td>1225</td> </tr><tr><td>20-30</td> <td>25</td> <td>6</td> <td>150</td> <td>-7.5</td> <td>-45</td> <td>56.25</td> <td>337.5</td> </tr><tr><td>30-40</td> <td>35</td> <td>10</td> <td>350</td> <td>2.5</td> <td>25</td> <td>6.25</td> <td>62.5</td> </tr><tr><td>40-50</td> <td>45</td> <td>3</td> <td>135</td> <td>12.5</td> <td>37.5</td> <td>156.25</td> <td>468.75</td> </tr><tr><td>50-60</td> <td>55</td> <td>2</td> <td>110</td> <td>22.5</td> <td>45</td> <td>506.25</td> <td>1012.5</td> </tr><tr><td></td> <td></td> <td>N = 25</td> <td>\(\sum{fx}\) = 805</td> <td></td> <td>\(\sum{fd}\) = 7.5</td> <td></td> <td>\(\sum{fd^2}\) = 3106.25</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}{N}\\ &= \frac {805}{25}\\ &= 32.5\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {3106.25}{25} - (\frac {-7.5}{25})^2}\\ &= \sqrt {124.25 - 0.09}\\ &= \sqrt {124.16}\\ &= 11.14_{Ans}\\ \end{align*}</p>
Q10:
Calculate the standard deviation from the table given below:
| Daily Sales | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 4 | 10 | 12 | 8 | 6 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Let: Assumed Mean (A) = 35</p> <p>Calculating Standard Deviation</p> <table width="491"><tbody><tr><td>Daily Sales</td> <td>Mid-value(m)</td> <td>Frequency (f)</td> <td>d = \(\frac {x - A}{10}\)</td> <td>d<sup>2</sup></td> <td>fd</td> <td>fd<sup>2</sup></td> </tr><tr><td>10-20</td> <td>15</td> <td>4</td> <td>-2</td> <td>4</td> <td>-8</td> <td>16</td> </tr><tr><td>20-30</td> <td>25</td> <td>10</td> <td>-1</td> <td>1</td> <td>-10</td> <td>10</td> </tr><tr><td>30-40</td> <td>35</td> <td>12</td> <td>0</td> <td>0</td> <td>0</td> <td>0</td> </tr><tr><td>40-50</td> <td>45</td> <td>8</td> <td>1</td> <td>1</td> <td>8</td> <td>8</td> </tr><tr><td>50-60</td> <td>55</td> <td>6</td> <td>2</td> <td>4</td> <td>12</td> <td>24</td> </tr><tr><td></td> <td></td> <td>N = 40</td> <td></td> <td></td> <td>\(\sum {fd}\) = 2</td> <td>\(\sum{fd^2}\) = 58</td> </tr></tbody></table><p>\begin{align*}\therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\;\times i\\ &= \sqrt {\frac {58}{40} - (\frac {2}{40})^2}\;\times {10}\\ &= \sqrt {1.45 - 0.0025}\;\times {10}\\ &= \sqrt {1.4475}\;\times {10}\\ &= 1.203\;\times{10}\\&= 12.03_{Ans}\\ \end{align*}</p>
Q11:
Compute the standard deviation of the following:
| Class Interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 | 20-24 |
| Frequency | 7 | 7 | 10 | 15 | 7 | 6 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="507"><tbody><tr><td>Class Interval</td> <td>Mid-value (m)</td> <td>Frequency (f)</td> <td>fm</td> <td>fm<sup>2</sup></td> </tr><tr><td>0-4</td> <td>2</td> <td>7</td> <td>14</td> <td>28</td> </tr><tr><td>4-8</td> <td>6</td> <td>7</td> <td>42</td> <td>252</td> </tr><tr><td>8-12</td> <td>10</td> <td>10</td> <td>100</td> <td>1000</td> </tr><tr><td>12-16</td> <td>14</td> <td>15</td> <td>210</td> <td>2940</td> </tr><tr><td>16-20</td> <td>18</td> <td>7</td> <td>126</td> <td>2268</td> </tr><tr><td>20-24</td> <td>22</td> <td>6</td> <td>132</td> <td>2904</td> </tr><tr><td></td> <td></td> <td>N = 52</td> <td>\(\sum{fm}\) = 624</td> <td>\(\sum{fm^2}\) = 9392</td> </tr></tbody></table><p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fm^2}}N - (\frac {\sum {fm}}N)^2}\\ &= \sqrt {\frac {9392}{52} - (\frac {624}{52})^2}\\ &= \sqrt {180.61 - 144}\\ &= \sqrt {36.61}\\ &= 6.05_{Ans}\\ \end{align*}</p>
Q12:
Compute the standard deviation from the following data:
| Marks secured | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No. of students | 8 | 12 | 15 | 9 | 6 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="420"><tbody><tr><td>Marks secured</td> <td>Mid-value (m)</td> <td>No. of Students (f)</td> <td>fm</td> <td>d = m - \(\overline{X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>10-20</td> <td>15</td> <td>8</td> <td>120</td> <td>-18.6</td> <td>345.96</td> <td>2767.68</td> </tr><tr><td>20-30</td> <td>25</td> <td>12</td> <td>300</td> <td>8.6</td> <td>73.96</td> <td>887.52</td> </tr><tr><td>30-40</td> <td>35</td> <td>15</td> <td>525</td> <td>1.4</td> <td>1.96</td> <td>29.4</td> </tr><tr><td>40-50</td> <td>45</td> <td>9</td> <td>405</td> <td>11.4</td> <td>129.96</td> <td>1169.64</td> </tr><tr><td>50-60</td> <td>55</td> <td>6</td> <td>330</td> <td>21.4</td> <td>457.96</td> <td>2747.76</td> </tr><tr><td></td> <td></td> <td>N = 50</td> <td>\(\sum {fm}\) = 1680</td> <td></td> <td></td> <td>\(\sum {fd^2}\) = 7602</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {1680}{50}\\ &= 33.6\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {7602}{50}}\\ &= \sqrt {152.04}\\ &= 12.33_{Ans}\\ \end{align*}</p>
Q13:
Compute the standard deviation from the following data:
| Marks obtained | 0-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No. of students | 2 | 8 | 16 | 10 | 4 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="440"><tbody><tr><td>Marks Secured</td> <td>Mid-value (m)</td> <td>No. of students (f)</td> <td>fm</td> <td>d= m - \(\overline {X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>0-20</td> <td>10</td> <td>2</td> <td>20</td> <td>-43</td> <td>1849</td> <td>3698</td> </tr><tr><td>20-40</td> <td>30</td> <td>8</td> <td>240</td> <td>-23</td> <td>529</td> <td>4232</td> </tr><tr><td>40-60</td> <td>50</td> <td>16</td> <td>800</td> <td>-3</td> <td>9</td> <td>144</td> </tr><tr><td>60-80</td> <td>70</td> <td>10</td> <td>700</td> <td>17</td> <td>289</td> <td>2890</td> </tr><tr><td>80-100</td> <td>90</td> <td>4</td> <td>360</td> <td>37</td> <td>1369</td> <td>5476</td> </tr><tr><td></td> <td></td> <td>N = 40</td> <td>\(\sum {fm}\) = 2120</td> <td></td> <td></td> <td>\(\sum{fd^2}\) = 16440</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {2120}{40}\\ &= 53\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {16440}{40}}\\ &= \sqrt {411}\\ &= 20.27_{Ans}\\ \end{align*}</p>
Q14:
Compute the standard deviation from the following data:
| Weight in kg. | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of boys | 2 | 5 | 6 | 3 | 2 | 2 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="430"><tbody><tr><td>Weight in kg.</td> <td>No. of boys (f)</td> <td>Mid-value (m)</td> <td>fm</td> <td>d = m - \(\overline{X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>10-20</td> <td>2</td> <td>15</td> <td>30</td> <td>-22</td> <td>484</td> <td>968</td> </tr><tr><td>20-30</td> <td>5</td> <td>25</td> <td>125</td> <td>-12</td> <td>144</td> <td>720</td> </tr><tr><td>30-40</td> <td>6</td> <td>35</td> <td>210</td> <td>-2</td> <td>4</td> <td>24</td> </tr><tr><td>40-50</td> <td>3</td> <td>45</td> <td>135</td> <td>8</td> <td>64</td> <td>192</td> </tr><tr><td>50-60</td> <td>2</td> <td>55</td> <td>110</td> <td>18</td> <td>324</td> <td>648</td> </tr><tr><td>60-70</td> <td>2</td> <td>65</td> <td>130</td> <td>28</td> <td>784</td> <td>1568</td> </tr><tr><td></td> <td>N = 20</td> <td></td> <td>\(\sum{fm}\) = 740</td> <td></td> <td></td> <td>\(\sum{fd^2}\) = 4120</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {740}{20}\\ &= 37\\ \end{align*}</p> <p>\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\sum{fd^2}N}\\ &= \sqrt {\frac {4120}{20}}\\ &= \sqrt {206}\\ &= 14.35_{Ans}\\ \end{align*}</p>
Q15:
Compute the standard deviation from the following data:
| Daily Sales | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Number of shops | 2 | 9 | 10 | 7 | 1 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="461"><tbody><tr><td>Daily Sales</td> <td>Mid-value (x)</td> <td>No. of shops (f)</td> <td>fx</td> <td>d = x - \(\overline{X}\)</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>0-10</td> <td>5</td> <td>2</td> <td>10</td> <td>-18.6</td> <td>345.96</td> <td>691.92</td> </tr><tr><td>10-20</td> <td>15</td> <td>9</td> <td>135</td> <td>-8.6</td> <td>73.96</td> <td>665.64</td> </tr><tr><td>20-30</td> <td>25</td> <td>10</td> <td>250</td> <td>1.4</td> <td>1.96</td> <td>19.6</td> </tr><tr><td>30-40</td> <td>35</td> <td>7</td> <td>245</td> <td>11.4</td> <td>129.96</td> <td>909.72</td> </tr><tr><td>40-50</td> <td>45</td> <td>1</td> <td>45</td> <td>21.4</td> <td>457.96</td> <td>457.96</td> </tr><tr><td></td> <td></td> <td>N = 29</td> <td>\(\sum {fx}\) = 685</td> <td></td> <td></td> <td>\(\sum {fd^2}\) = 2744.84</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {685}{29}\\ &= 23.6\\ \end{align*}</p> <p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {2744.84}{29}}\\ &= \sqrt {94.65}\\ &= 9.73_{Ans}\\ \end{align*}</p>
Q16:
Compute the standard deviation from the data below:
| Marks Obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of boys | 4 | 7 | 14 | 16 | 22 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="467"><tbody><tr><td>Marks Obtained</td> <td>No. of Students (f)</td> <td>Mid-value (m)</td> <td>fm</td> <td>fm<sup>2</sup></td> </tr><tr><td>0-10</td> <td>4</td> <td>5</td> <td>20</td> <td>100</td> </tr><tr><td>10-20</td> <td>7 - 4 = 3</td> <td>15</td> <td>45</td> <td>675</td> </tr><tr><td>20-30</td> <td>14 - 7 = 7</td> <td>25</td> <td>175</td> <td>4375</td> </tr><tr><td>30-40</td> <td>16 - 14 = 2</td> <td>35</td> <td>70</td> <td>2450</td> </tr><tr><td>40-50</td> <td>22 - 16 = 6</td> <td>45</td> <td>270</td> <td>12150</td> </tr><tr><td></td> <td>N = 22</td> <td></td> <td>\(\sum{fm}\) = 580</td> <td>\(\sum{fm^2}\) = 19750</td> </tr></tbody></table><p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fm^2}}N - (\frac {\sum{fm}}N)^2}\\ &= \sqrt {\frac {19750}{22} - (\frac {580}{22})^2}\\ &= \sqrt {897.72 - 695.04}\\ &= \sqrt {202.69}\\ &= 14.24_{Ans}\\ \end{align*}</p>
Q17:
Calculate the standard deviation from the table given below:
| Marks Obtained | 5 | 15 | 25 | 35 | 45 | 55 |
| No. of Students | 5 | 8 | 10 | 15 | 8 | 4 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="458"><tbody><tr><td>Marks Obtained (x)</td> <td>No. of Students (f)</td> <td>fx</td> <td>d</td> <td>d<sup>2</sup></td> <td>fd<sup>2</sup></td> </tr><tr><td>5</td> <td>5</td> <td>25</td> <td>-25</td> <td>625</td> <td>3125</td> </tr><tr><td>15</td> <td>8</td> <td>120</td> <td>-15</td> <td>225</td> <td>1800</td> </tr><tr><td>25</td> <td>10</td> <td>250</td> <td>-5</td> <td>25</td> <td>250</td> </tr><tr><td>35</td> <td>15</td> <td>525</td> <td>5</td> <td>25</td> <td>375</td> </tr><tr><td>45</td> <td>8</td> <td>360</td> <td>15</td> <td>225</td> <td>1800</td> </tr><tr><td>55</td> <td>4</td> <td>220</td> <td>25</td> <td>625</td> <td>2500</td> </tr><tr><td></td> <td>N = 50</td> <td>\(\sum{fx}\) = 1500</td> <td></td> <td></td> <td>\(\sum {fd^2}\) = 9850</td> </tr></tbody></table><p>\begin{align*} Mean\;(\overline{X}) &= \frac {\sum {fx}}N\\ &= \frac {1500}{50}\\ &= 30\\ \end{align*}</p> <p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {9850}{50}}\\ &= \sqrt {197}\\ &= 14.04_{Ans}\\ \end{align*}</p> <p></p>
Q18:
Compute the standard deviation from the following data:
| Marks Obtained | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Numbers of Students | 4 | 6 | 10 | 3 | 2 |
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Calculating Standard Deviation</p> <table width="482"><tbody><tr><td>Marks Obtained</td> <td>Mid-value (m)</td> <td>d = m - 35</td> <td>f</td> <td>fd</td> <td>fd<sup>2</sup></td> </tr><tr><td>10-20</td> <td>15</td> <td>-20</td> <td>4</td> <td>-80</td> <td>1600</td> </tr><tr><td>20-30</td> <td>25</td> <td>-10</td> <td>6</td> <td>-60</td> <td>600</td> </tr><tr><td>30-40</td> <td>35</td> <td>0</td> <td>10</td> <td>0</td> <td>0</td> </tr><tr><td>40-50</td> <td>45</td> <td>10</td> <td>3</td> <td>30</td> <td>300</td> </tr><tr><td>50-60</td> <td>55</td> <td>20</td> <td>2</td> <td>40</td> <td>800</td> </tr><tr><td></td> <td></td> <td></td> <td>N = 25</td> <td>\(\sum{fd}\) = -70</td> <td>\(\sum{fd^2}\) = 3300</td> </tr></tbody></table><p>\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N - (\frac {\sum{fd}}N)^2}\\ &= \sqrt {\frac {3300}{25} - (\frac {-70}{25})^2}\\ &= \sqrt {132 - 7.84}\\ &= \sqrt {124.16}\\ &= 11.14_{Ans}\\ \end{align*}</p>
Videos
Statistics: Standard deviation | Descriptive statistics | Probability and Statistics | Khan Academy
Standard Deviation : What is it and how to work it out : ExamSolutions
Standard Deviation
Biogeochemical Cycles
Biogeochemical:
That essential element which is produced by earth and is needed for living beings to continue their life is called biogeochemical.
Biogeochemical cycle:
The exchange of biogeochemical between the living and non-living world of biosphere is called biogeochemical cycle.
Some biogeochemical cycles:
- Water cycle:
- Oxygen cycle:
- Carbon cycle
- Nitrogen cycle:
The process of exchange of nitrogen gas between living and non-living world of an ecosystem is called nitrogen cycle. Nitrogen cycle completes in following six steps
- Nitrogen fixation: The process of conversion of atmospheric nitrogen gas into usable nitrogenous compounds is called nitrogen fixation. Nitrogen fixation in nature is going on with following processes:
- Biological nitrogen fixation: The roots nodes of leguminous plants contain symbiotic bacteria, which converts atmospheric nitrogen gas into usable nitrogenous compounds, which can be used by plants.
- Atmospheric fixation: Under the high electric current produced during thundering the atmospheric nitrogen gas combines with oxygen to give oxides of nitrogen. Those oxides combine with rain water and come onto soil and produce many usable nitrogenous compounds.
- Industrial fixation: Sometimes to increase the nitrogen content of soil we prepare the nitrogenous fertilizer in industry and add into the soil.
- Nitrogen assimilation: The process of entering nitrogenous compound into living world is called nitrogen assimilation. Those nitrogenous compounds are first used by plants and they prepare plant protein and after eating plants protein produces animal protein.
- Ammonification: The protein assimilated in the body by living beings will not remain locked permanently but it will be digested and after digesting some ammonium salts like urea will be produced. This process of producing ammonium salts after digestion of proteins is called ammonification.
- Nitrification: The ammonium salts that are produced after digesting protein will be excreted out of the body of living being. Those ammonium salts will be converted to simple nitrates by the action of nitrifying bacteria in soil.
- Denitrification: The process of conversion of soil nitrates to pure atmospheric nitrogen with the help of denitrifying bacteria is called denitrification.
- Sedimentation: Some parts of nitrogenous compound of embedded into rock and remain on its surface for long time. This process is called sedimentation.
- Nitrogen fixation: The process of conversion of atmospheric nitrogen gas into usable nitrogenous compounds is called nitrogen fixation. Nitrogen fixation in nature is going on with following processes:
Lesson
Eco-System
Subject
Science
Grade
Grade 10
Recent Notes
No recent notes.
Related Notes
No related notes.