# Long Questions

Find the mean deviation and its coefficient from median of the data given below:

10, 50, 60, 40, 30, 20

Arranging given data in the ascending order: 10, 20, 30, 40, 50, 60

Total number (N) = 6

Median (Md) = $(\frac{\text{N + 1}}{2})$^{th} item = 3.5^{th} item

So, Md = $\frac{\text{30 + 40}}{2}$ = 35

Now, tabulating the given data,

X | |D| = |X - Md| |
---|---|

10 | 25 |

20 | 15 |

30 | 5 |

40 | 5 |

50 | 15 |

60 | 25 |

N = 6 | ∑|D| = 90 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{15}{25}$ = 0.43

Marks obtained | 20 | 30 | 50 | 40 | 60 | 70 |
---|---|---|---|---|---|---|

No. of students | 2 | 4 | 7 | 5 | 2 | 1 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

20 | 2 | 2 |

30 | 4 | 4+2 = 6 |

40 | 5 | 5+6 = 11 |

50 | 7 | 7+11=18 |

60 | 2 | 2+18 = 20 |

70 | 1 | 1+20 = 21 |

N = 21 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 11^{th} item

Corresponding value of Cumulative frequency (c.f.) equal to 11 is 40.

So, Median = 40

Finally, Calculation of mean deviation from median,

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

20 | |20-40| = 20 | 2 | 40 |

30 | |30-40| = 10 | 4 | 40 |

40 | |40-40| = 0 | 5 | 0 |

50 | |50-40| = 10 | 7 | 70 |

60 | |60-40| = 20 | 2 | 40 |

70 | |70-40| = 30 | 1 | 30 |

N = 21 | ∑f|D| 220 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10.47}{40}$ = 0.26

Marks obtained | 5 | 10 | 15 | 20 | 25 | 30 |
---|---|---|---|---|---|---|

No. of students | 4 | 5 | 6 | 7 | 3 | 2 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

5 | 4 | 4 |

10 | 5 | 4+5 = 9 |

15 | 6 | 9+6 = 15 |

20 | 7 | 15+7 = 22 |

25 | 3 | 22+3 = 25 |

30 | 2 | 25+2 = 27 |

N = 27 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 14^{th} item

Cumulative frequency (c.f.) just greater than 14 is 15. Its corresponding value is 15.

So, Median = 15

Finally, Calculation of mean deviation from median,

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

5 | |5-15| = 10 | 4 | 40 |

10 | |10-15| = 5 | 5 | 25 |

15 | |15-15| = 0 | 6 | 0 |

20 | |20-15| = 5 | 7 | 35 |

25 | |25-15| = 10 | 3 | 30 |

30 | |30-15| = 15 | 2 | 30 |

N = 27 | ∑f|D| 160 |

Marks obtained | 20 | 30 | 40 | 50 | 60 | 70 |
---|---|---|---|---|---|---|

No. of students | 4 | 7 | 12 | 2 | 4 | 6 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

20 | 4 | 4 |

30 | 7 | 4+7 = 11 |

40 | 15 | 11+12 = 23 |

50 | 2 | 23+2 = 25 |

60 | 4 | 25+4 = 29 |

70 | 6 | 29+6 = 35 |

N = 35 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 18^{th} item

Cumulative frequency (c.f.) just greater than 18 is 23. Its corresponding value is 40.

So, Median = 40

Finally, Calculation of mean deviation from median,

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

20 | |20-40| = 20 | 4 | 80 |

30 | |30-40| = 10 | 7 | 70 |

40 | |40-40| = 0 | 12 | 0 |

50 | |50-40| = 10 | 2 | 20 |

60 | |60-40| = 20 | 4 | 80 |

70 | |70-40| = 30 | 6 | 180 |

N = 35 | ∑f|D| 430 |

Marks obtained | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
---|---|---|---|---|---|

No. of students | 5 | 4 | 5 | 4 | 2 |

Marks obtained (X) | mid-value (m) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|---|

10-20 | 15 | 5 | 5 |

20-30 | 25 | 4 | 5+4 = 9 |

30-40 | 35 | 5 | 9+5 = 14 |

40-50 | 45 | 4 | 14+4 = 18 |

50-60 | 55 | 2 | 18+2 = 20 |

N = 20 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 10^{th} item

Cumulative frequency (c.f.) which is greater than 10 is 14. Its corresponding value is 30-40

So, Median lies in class (30-40).

L = 30, f = 5, c.f. = 9 and i = 10

Now, Median = L + $\frac {i}{f}(\frac N2 - c.f.)$
= 30 + $\frac {10}{5}(10 - 9)$

= 32

Finally, Calculation of mean deviation from median,

Marks obtained (m) | |m - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

15 | |15-32| = 17 | 5 | 85 |

25 | |25-32| = 7 | 4 | 2 |

35 | |35-32| = 3 | 5 | 15 |

45 | |45-32| = 13 | 4 | 52 |

55 | |55-32| = 23 | 2 | 46 |

N = 20 | ∑f|D| = 226 |

Marks obtained | 25 | 35 | 45 | 55 | 65 |
---|---|---|---|---|---|

No. of students | 5 | 7 | 8 | 6 | 4 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

25 | 5 | 5 |

35 | 7 | 5+7 = 12 |

45 | 8 | 12+8 = 20 |

55 | 6 | 20+6 = 26 |

65 | 4 | 26+4 = 30 |

N = 30 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 15.5^{th} item

Cumulative frequency (c.f.) just greater than 15.5 is 20. Its corresponding value is 45.

So, Median = 45

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

25 | |25-45| = 20 | 5 | 100 |

35 | |35-45| = 10 | 7 | 70 |

45 | |45-45| = 0 | 8 | 0 |

55 | |55-45| = 10 | 6 | 60 |

65 | |65-45| = 20 | 4 | 80 |

N = 30 | ∑f|D| 310 |

Marks obtained | 55 | 65 | 75 | 85 | 95 |
---|---|---|---|---|---|

No. of students | 4 | 5 | 6 | 3 | 2 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

55 | 4 | 4 |

65 | 5 | 4+5 = 9 |

75 | 6 | 9+6 = 15 |

85 | 3 | 15+3 = 18 |

95 | 2 | 18+2 = 20 |

N = 20 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 10.5^{th} item

Cumulative frequency (c.f.) just greater than 10.5 is 15. Its corresponding value is 75.

So, Median = 75

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

55 | |55-75| = 20 | 4 | 80 |

65 | |65-75| = 10 | 5 | 50 |

75 | |75-75| = 0 | 6 | 0 |

85 | |85-75| = 10 | 3 | 30 |

95 | |95-75| = 20 | 2 | 40 |

N = 20 | ∑f|D| 200 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{75}$ = 0.133

Marks obtained | 10 | 20 | 30 | 40 | 50 |
---|---|---|---|---|---|

No. of students | 2 | 3 | 6 | 5 | 4 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

10 | 2 | 2 |

20 | 3 | 2+3 = 5 |

30 | 6 | 5+6 = 11 |

40 | 5 | 11+5 = 16 |

50 | 4 | 16+4 = 20 |

N = 20 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item

= value of 10.5^{th} item

Cumulative frequency (c.f.) just greater than 10.5 is 11. Its corresponding value is 30.

So, Median = 30

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

10 | |10-30| = 20 | 2 | 40 |

20 | |20-30| = 10 | 3 | 30 |

30 | |30-30| = 0 | 6 | 0 |

40 | |40-30| = 10 | 5 | 50 |

50 | |50-30| = 20 | 4 | 80 |

N = 20 | ∑f|D| 200 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{30}$ = 0.33

Marks obtained | 5 | 10 | 15 | 20 | 25 |
---|---|---|---|---|---|

No. of students | 3 | 4 | 6 | 2 | 1 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

5 | 3 | 3 |

10 | 4 | 3+4 = 7 |

15 | 6 | 7+6 = 13 |

20 | 2 | 13+2 = 15 |

25 | 1 | 15+1 = 16 |

N = 16 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 8.5^{th} item

So, Median = 15

Finally, Calculation of mean deviation from median,

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

5 | |5-15| = 10 | 3 | 30 |

10 | |10-15| = 5 | 4 | 20 |

15 | |15-15| = 0 | 6 | 0 |

20 | |20-15| = 5 | 2 | 10 |

25 | |25-15| = 10 | 1 | 10 |

N = 16 | ∑f|D| 70 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{4.375}{15}$ = 0.29

Marks obtained | 10 | 30 | 50 | 20 | 40 | 60 |
---|---|---|---|---|---|---|

No. of students | 2 | 10 | 8 | 5 | 12 | 3 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

10 | 2 | 2 |

20 | 5 | 2+5 = 7 |

30 | 10 | 7+10 = 17 |

40 | 12 | 17+12 = 29 |

50 | 8 | 29+8 = 37 |

60 | 3 | 37+3 = 40 |

N = 40 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 20.5^{th} item

Cumulative frequency (c.f.) just greater than 20.5 is 29. Its corresponding value is 40.

So, Median = 40

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

10 | |10-40| = 30 | 2 | 60 |

20 | |20-40| = 20 | 5 | 100 |

30 | |30-40| = 10 | 10 | 100 |

40 | |40-40| = 0 | 12 | 0 |

50 | |50-40| = 10 | 8 | 80 |

60 | |60-40| = 20 | 3 | 60 |

N = 40 | ∑f|D| 400 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{40}$ = 0.25

Marks obtained | 200 | 400 | 300 | 350 | 250 |
---|---|---|---|---|---|

No. of students | 4 | 7 | 6 | 1 | 3 |

Marks obtained (X) | Number of students (f) | Cumulative frequency (c.f.) |
---|---|---|

200 | 4 | 4 |

250 | 3 | 4+3 = 7 |

300 | 6 | 7+6 = 13 |

350 | 1 | 13+1 = 14 |

400 | 7 | 14+7 = 21 |

N = 21 |

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 11^{th} item

Cumulative frequency (c.f.) just greater than 11 is 13. Its corresponding value is 300.

So, Median = 300

Marks obtained (X) | |X - Median| = |D| | Number of students (f) | f|D| |
---|---|---|---|

200 | |200-300| = 100 | 4 | 400 |

250 | |250-300| = 50 | 3 | 150 |

300 | |300-300| = 0 | 6 | 0 |

350 | |350-300| = 50 | 1 | 50 |

400 | |400-300| = 100 | 7 | 700 |

N = 21 | ∑f|D| 1300 |

Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{61.9}{300}$ = 0.206

Marks obtained | 20 | 10 | 30 | 50 | 40 |
---|---|---|---|---|---|

No. of students | 3 | 4 | 5 | 2 | 1 |