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प्रश्न
Find the mean of the following data: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25
(i) Show that the sum of the deviations of all the given observations from the mean is zero.
(ii) Find the median of the given data.
उत्तर
Let `barx` be the mean of n number of observation x1, x2, x3, ...., xn
Mean = `[ x_1 + x_2 + x_3 + ....... + x_n ]/n`
Therefore,
Mean of given data = `[ 30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25 ]/10`
= `297/10`
= 29.7
(i) Let us tabulate the observations and their deviations from the mean
Observation xi |
Deviation `x_i - barx` |
30 | 0.3 |
32 | 2.3 |
24 | - 5.7 |
34 | 4.3 |
26 | - 3.7 |
28 | - 1.7 |
30 | 0.3 |
35 | 5.3 |
33 | 3.3 |
25 | - 4.7 |
Total | 0 |
From the table, it is clear that the sum of the deviations from the mean is Zero.
(ii) Consider the given data :
30, 32, 24, 34, 26, 28, 30, 35, 33, 25.
Let us rewrite the above data in ascending order
24, 25, 26, 28, 30, 30, 32, 33, 34, 35.
There are 10 observations, which is even.
Therefore,
Median = `1/2[ (n/2)^"th" "term" + ( n/2 + 1 )^"th" "term" ]`
= `1/2[ (10/2)^"th" "term" + ( 10/2 + 1)^"th" "term" ]`
= `1/2[ (5)^"th" "term" + ( 10/2 + 1)^"th" "term" ]`
= `1/2[ 5^"th" "term" + ( 5 + 1 )^"th" "term" ]`
= `1/2[ 5^"th" "term" + 6^"th" "term" ]`
= `1/2[ 30 + 30 ]`
= `1/2[ 60 ]`
= 30.
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