Question

The mean deviation of the numbers 1, 2, 3, 4 and 5 is
(a) 0
(b) 1.2
(c) 2
(d) 1.4

Answer 1

Hint:In this question, we have been given 5 numbers and we have to find their mean deviation. Therefore, we should understand the definition for mean deviation, calculate the absolute distances of the numbers from their average and then take the mean of those deviations to obtain the answer to this question.

Complete step-by-step answer:
Mean:A mean is the simple mathematical average of a set of two or more numbers.
Mean deviation: The mean of the absolute values of the numerical differences between the numbers of a set and their mean.
In this question, the given numbers are 1, 2, 3, 4 and 5.
The average of n numbers ${{a}_{1}},{{a}_{2}}...{{a}_{n}}$ is defined to be
\[\bar{a}=\dfrac{{{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}}{n}................(1.1)\]
Now, the mean deviation (dev) of the n numbers ${{a}_{1}},{{a}_{2}}...{{a}_{n}}$ is defined to be
$dev=\dfrac{\left| {{a}_{1}}-\bar{a} \right|+\left| {{a}_{2}}-\bar{a} \right|+...+\left| {{a}_{n}}-\bar{a} \right|}{n}............(1.2)$
Where for any number x, $\left| x \right|$denotes the modulus of x and $\bar{a}$ is as defined in equation (1.1).
Therefore, in this case, as the numbers are given to be 1, 2, 3, 4 and 5, we can take ${{a}_{1}}=1,{{a}_{2}}=2,{{a}_{3}}=3,{{a}_{4}}=4,{{a}_{5}}=5$ and n=5 in equation (1.1) to obtain their average $\bar{a}$ to be
\[\bar{a}=\dfrac{1+2+3+4+5}{5}=3................(1.3)\]
Now, using equation (1.3) in equation (1.2), we obtain the standard deviation of these numbers to be
$dev=\dfrac{\left| 1-3 \right|+\left| 2-3 \right|+\left| 3-3 \right|+\left| 4-3 \right|+\left| 5-3 \right|}{5}=\dfrac{2+1+0+1+2}{5}=\dfrac{6}{5}=1.2............(1.4)$
Therefore, from equation (1.2), we obtain the standard deviation of 1, 2, 3, 4 and 5 to be 1.2 which matches option (b). Hence (b) is the correct answer to this question.

Note: We should note that in equation (1.2), we have to take the modulus of the difference between the numbers and the average because otherwise the mean deviation will always come out to be zero.