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# The median and standard derivative of a distribution is 20 and 4 respectively. If each item is increased by two thena) median will increase by 2.b) median will remain the same.c) standard deviation will remain samed) standard deviation will increase by 2.

Last updated date: 20th Jun 2024
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Hint: Median is the observation at the middle after you arrange the data in ascending descending order.
Standard deviation is deviation from observation by its mean.
S.D$= \sqrt {\dfrac{{\sum\limits_{x = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{N}}$
Where, $x_i$=observation
n = number of observation
$\mu$=mean of the observation
When you consider the formulas for both and find the median and standard deviation you will get two answers, one for each.

Let the observation be ${x_1},{x_2},{x_3},{x_4},{x_5}$
Median
$\mu = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5}$
S.D$= \sqrt {\dfrac{{\sum\limits_{x = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{N}}$
S.D$= \sqrt {\dfrac{{{{\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} - \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5}} \right)}^2}}}{N}}$
After increasing each of observation by 2 then, ${x_1} + 2,{x_2} + 2,{x_3} + 2,{x_4} + 2,{x_5} + 2$
New median$= {x_3} + 2$
New mean,
$\begin{array}{l}{\mu _|} = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 10}}{5}\\{\mu _|} = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} + 2\\{\mu _|} = \mu + 2\end{array}$
New S.D$= \sqrt {\dfrac{{{{\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 2 - \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 10}}{5}} \right)}^2}}}{N}}$
New S.D$= \sqrt {\dfrac{{{{\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 2 - \mu - 2} \right)}^2}}}{N}}$
New S.D = Old S.D
So, Option (a) and (c) are correct.

Note: The terms used in calculating standard deviation are deviations from the mean of the observations.
As each number/observation is increased by 2, the mean and thus the deviations from the mean remain the same.
Hence standard deviation remains the same.
On the other hand, median gives the middle term or the average of the two middle terms accordingly when the total number of terms is odd or even.
Thus, it has to increase by 2.