To find the mean deviation from the median, firstly let us calculate the median.

N = 50

So, \(\dfrac{ N}{2} = \dfrac{50}{2}\) = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28

So, Median = 28

By using the formula to calculate Mean,

\( x= \dfrac{\displaystyle\sum fixi}{fi}\)

= \(\dfrac{ 1350}{50}\)

= 27

Mean deviation from Median = \( \dfrac{478}{50}\) = 9.56

And, Mean deviation from Median = \( \dfrac{472}{50}\) = 9.44

The Mean Deviation from the median is 9.56 and from mean is 9.44.

Answered by Aaryan | 9 months agoThe 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 |

Compute mean deviation from mean of the following distribution:

Marks | 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 |

Find the mean deviation from the mean for the following data:

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

Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |

Find the mean deviation from the mean for the following data:

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

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

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 |