Question 1: Compute the mean deviation from the median of the following distribution:

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency | 5 | 10 | 20 | 5 | 10 |

Answer:

*Calculation of Mean Deviation from Median *

Question 2: Find the mean deviation from the mean from the following data:

(i)

Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

Frequency | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |

(ii)

Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |

Frequency | 9 | 13 | 16 | 26 | 30 | 12 |

(iii)

Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 6 | 8 | 14 | 16 | 4 | 2 |

Answer:

(i)

*Calculation of Mean Deviation from Mean *

From the above table we get

(ii)

*Calculation of Mean Deviation from Mean *

From the above table we get

(iii)

*Calculation of Mean Deviation from Mean *

From the above table we get

Question 3: Compute the mean deviation from the mean of the following distribution:

Class | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

No. of Students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |

Answer:

*Calculation of Mean Deviation from Mean *

From the above table we get

Question 4: The age distribution of 100 life insurance policy holders is as follows:

Age ( on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |

No. of Persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |

Calculate the mean deviation from the median age.

Answer:

To make the function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

*Calculation of Mean Deviation from Median *

Question 5: Find the mean deviation from the mean and from median of the following distribution:

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of Students | 5 | 8 | 15 | 16 | 6 |

Answer:

*Calculation of Mean Deviation from Median *

*Calculation of Mean Deviation from Mean *

From the above table we get

Question 6: Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |

No. of Persons | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |

Answer:

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains the same, while the function becomes continuous.

*Calculation of Mean Deviation from Median *

Thus the mean deviation from the median age is 7.35 years

Question 7: Calculate the mean deviation about the mean for the following frequency distribution:

Class Interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |

Frequency | 4 | 6 | 8 | 5 | 2 |

Answer:

In order to avoid tedious calculations of computing mean , let us compute by step-deviation method. The formula for the same is :

Let us take the assumed mean and and form the following table.

*Calculation of Mean Deviation from Mean *

From the above table we get

Question 8: Calculate the mean deviation from the median of the following data:

Class Interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |

Frequency | 4 | 5 | 3 | 6 | 2 |

Answer:

*Calculation of Mean Deviation from Median *