**Question 1. Calculate the mean deviation from the median of the following frequency distribution:**

Heights in inches | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |

No. Of Students | 15 | 20 | 32 | 35 | 35 | 22 | 20 | 10 | 8 |

**Solution:**

Median is the middle term of the observation in ascending order,

So, Median = 61

Let us assume,

x

_{i}=Heights in inchesf

_{i}= Number of students

x _{i}f _{i}Cumulative Frequency |d

_{i}| = |x_{i}– M|= |x

_{i}– 61|f _{i}|d_{i}|58 15 15 3 45 59 20 35 2 40 60 32 67 1 32 61 35 102 0 0 62 35 137 1 35 63 22 159 2 44 64 20 179 3 60 65 10 189 4 40 66 8 197 5 40 N = 197 Total = 336 N=197

= 1/197 × 336

= 1.70

Therefore, the mean deviation is 1.70.

**Question 2.** **The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:**

Number of calls | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Frequency | 14 | 21 | 25 | 43 | 51 | 40 | 39 | 12 |

**Compute the mean deviation about the median.**

**Solution:**

Median is the middle term of the observation in ascending order,

We know, Median is the even term, (3+5)/2 = 4

So, Median = 8

Let us assume

x

_{i}=Number of callsf

_{i}= FrequencyN = 245

x _{i}f _{i}Cumulative Frequency |d

_{i}| = |x_{i}– M|= |x

_{i}– 61|f _{i}|d_{i}|0 14 14 4 56 1 21 35 3 63 2 25 60 2 50 3 43 103 1 43 4 51 154 0 0 5 40 194 1 40 6 39 233 2 78 7 12 245 3 36 Total = 245 Total = 366 = 1/245 × 336

= 1.49

Therefore, mean deviation is 1.49.

**Question 3. Calculate the mean deviation about the median of the following frequency distribution:**

x_{i} | 5 | 7 | 9 | 11 | 13 | 15 | 17 |

f_{i} | 2 | 4 | 6 | 8 | 10 | 12 | 8 |

**Solution:**

Calculating the median,

We know, Number of observations, N = 50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 13.

x _{i}f _{i}Cumulative Frequency |d

_{i}| = |x_{i}– M|= |x

_{i}– 61|f _{i}|d_{i}|5 2 2 8 16 7 4 6 6 24 9 6 12 4 24 11 8 20 2 16 13 10 30 0 0 15 12 42 2 24 17 8 50 4 32 Total = 50 Total = 136 = 1/50 × 136

= 2.72

Therefore, the mean deviation is 2.72.

**Question 4.** **Find the mean deviation from the mean for the following data:**

**(i)**

x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |

f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |

**(ii)**

x_{i} | 5 | 10 | 15 | 20 | 25 |

f_{i} | 7 | 4 | 6 | 3 | 5 |

**(iii)**

x_{i} | 10 | 30 | 50 | 70 | 90 |

f_{i} | 4 | 24 | 28 | 16 | 8 |

**Solution:**

(i)We know,

x _{i}f _{i}Cumulative Frequency (x _{i}f_{i})|d _{i}| = |x_{i}– Mean|f _{i}|d_{i}|5 8 40 4 32 7 6 42 2 12 9 2 18 0 0 10 2 20 1 2 12 2 24 3 6 15 6 90 6 36 Total = 26 Total = 234 Total = 88 Now, Mean = 234/26

= 9

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii)We know,

x _{i}f _{i}Cumulative Frequency (x _{i}f_{i})|d _{i}| = |x_{i}– Mean|f _{i}|d_{i}|5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 Total = 25 Total = 350 Total = 158 Mean = 350/25

= 14

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii)We know,= 4000/80

= 50

x _{i}f _{i}Cumulative Frequency (x _{i}f_{i})|d _{i}| = |x_{i}– Mean|f _{i}|d_{i}|10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 Total = 80 Total = 4000 Total = 1280 = 1280/80

= 16

∴ The mean deviation is 16

**Question 5. Find the mean deviation from the median for the following data :**

**(i)**

x_{i} | 15 | 21 | 27 | 30 |

f_{i} | 3 | 5 | 6 | 7 |

**(ii)**

x_{i} | 74 | 89 | 42 | 54 | 91 | 94 | 35 |

f_{i} | 20 | 12 | 2 | 4 | 5 | 3 | 4 |

**(iii)**

Marks obtained | 10 | 11 | 12 | 14 | 15 |

No. of students | 2 | 3 | 8 | 3 | 4 |

**Solution:**

(i)We know,Number of observations, N = 21

Median = (21)/2 = 10.5

Therefore, the median corresponding to 10.5 is 27

x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 Total = 21 Total = 46 Total = 108 = 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii)We know,Number of observations, N =50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 74.

x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|74 20 4 39 156 89 12 6 32 64 42 2 10 20 80 54 4 30 0 0 91 5 42 15 180 94 3 47 17 85 35 4 50 20 60 Total = 50 Total = 189 Total = 625 = 1/50 × 625

= 12.5

Therefore, the mean deviation is 12.5

(iii)We know,Number of observations, N =20

Median = (20)/2 = 10

So, the median corresponding to 10 is 12.

x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|10 2 2 2 4 11 3 5 1 3 12 8 13 0 0 14 3 16 2 6 15 4 20 3 12 Total = 20 Total = 25 = 1/20 × 25

= 1.25

∴ The mean deviation is 1.25