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Mathematics

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.

Statistics

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Answer

Mean = Sum of all observations Number of all observations \dfrac{\text{Sum of all observations }}{\text{Number of all observations }}

= 30+32+24+34+26+28+30+35+33+2510\dfrac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}

= 29710\dfrac{297}{10}

= 29.7

Hence, the mean is 29.7.

(i) x‾\overline{x} = 29.7

xxx−x‾x - \overline{x}
3029.7 - 30 = -0.3
3229.7 - 32 = -2.3
2429.7 - 24 = 5.7
3429.7 - 34 = -4.3
2629.7 - 26 = 3.7
2829.7 - 28 = 1.7
3029.7 - 30 = -0.3
3529.7 - 35 = -5.3
3329.7 - 33 = -3.3
2529.7 - 25 = 4.7

∑(x−x‾)∑(x - \overline{x}) = (-0.3) + (-2.3) + 5.7 + (-4.3) + 3.7 + 1.7 + (-0.3) + (-5.3) + (-3.3) + 4.7

= -0.3 - 2.3 + 5.7 - 4.3 + 3.7 + 1.7 - 0.3 - 5.3 - 3.3 + 4.7

= 0

Hence, the sum of the deviations of all the given observations from the mean is zero.

(ii) On arranging the given set of data in ascending order, we get :

24, 25, 26, 28, 30, 30, 32, 33, 34, 35

Number of observations, n = 10 (even)

Median = 12[the value of(n2)th+the value of(n2+1)th]\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{n}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{n}{2} + 1\Big)^{th}\Big] term

= 12[the value of(102)th+the value of(102+1)th]\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big] term

= 12[the value of(5)th+the value of(5+1)th]\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big] term

= 12[30+30]\dfrac{1}{2}\Big[30 + 30\Big]

= 12[60]\dfrac{1}{2}\Big[60\Big]

= 30

Hence, the median is 30.

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