First we have to arrange the given observations into ascending order,

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations is 10

Then,

Median = \(\dfrac{(\dfrac{10}{2})^{th}observation+(\dfrac{10}{2})^{th}observation}{2} \)

(\( \dfrac{10}{2}\))^{th} observation = 5^{th} = 46

(\( \dfrac{10}{2}\))+ 1)^{th} observation = 5 + 1

= 6^{th} = 49

Median = \(\dfrac{(46 + 49)}{2}\)

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

\( \displaystyle\sum_{i=1}^{10} f_i |X_i-M|=70\)

Mean Deviation,

\( MD(M) = \dfrac{1}{10} \displaystyle\sum_{i=1}^{10} |X_i-M|\)

= \( \dfrac{1}{10}\times 70\)

= 7

So, the mean deviation about the median for the given data is 7.

Answered by Abhisek | 1 year agoFind the mean deviation from the mean and from a median of the following distribution:

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

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

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 |

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 |