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

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

The number of observations is 12

Then,

Median = \( \dfrac{(\dfrac{12}{2})^{th}observation+(\dfrac{12+1}{2})^{th}}{2}\)

(\( \dfrac{12}{2}\))^{th} observation = 6^{th} = 13

(\( \dfrac{12}{2}\))+ 1)^{th} observation = 6 + 1

= 7^{th} = 14

Median = \( \dfrac{(13 + 14)}{2}\)

=\( \dfrac{27}{2}\)

= 13.5

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

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

\(\displaystyle\sum_{i=1}^{12} |X_i-M|=28\)

Mean Deviation,

\( MD(M) = \dfrac{1}{12} \displaystyle\sum_{i=1}^{12} |x_i-M|\)

= \( \dfrac{1}{12} \times 28\)

= 2.33

So, the mean deviation about the median for the given data is 2.33.

Answered by Pragya Singh | 11 months 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 |