1.

The lengths (in cm) of 10 rods in a shop are given below:40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2(i) Find the mean deviation from the median.(ii) Find the mean deviation from the mean also.

Answer»

(i) Find the mean deviation from the median

Let us arrange the data in ascending order,

15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0

We know that,

MD = 1/n ∑ni=1|di|

Where, |di| = |xi – M|

The number of observations are Even then Median = (40+52.3)/2 = 46.15

Median = 46.15

Number of observations, ‘n’ = 10

xi|di| = |xi – 46.15|
40.06.15
52.36.15
55.29.05
72.926.75
52.86.65
79.032.85
32.513.65
15.230.95
27.919.25
30.215.95
Total167.4

MD = 1/n ∑ni=1|di|

= 1/10 × 167.4

= 16.74

∴ The Mean Deviation is 16.74.

(ii) Find the mean deviation from the mean also.

We know that,

MD = 1/n ∑ni=1|di|

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10

= 458/10

= 45.8

Number of observations, ‘n’ = 10

xi

|di| = |xi – 45.8|
40.05.8
52.36.5
55.29.4
72.927.1
52.87
79.033.2
32.513.3
15.230.6
27.917.9
30.215.6
Total166.4

MD = 1/n ∑ni=1|di|

= 1/10 × 166.4

= 16.64

∴ The Mean Deviation is 16.64


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