Question

The ${{8}^{th}}$ decile of the following observations is 29, 18, 15, 30, 42, 35, 34, 28, 45, 34.
$\begin{aligned} & a)44.6 \\\ & b)40.6 \\\ & c)34.15 \\\ & d)38.15 \\\ \end{aligned}$

Answer 1

Now the given observations are 29, 18, 15, 30, 42, 35, 34, 28, 45, 34. First we will arrange the numbers in ascending order. Then we know that ${{m}^{th}}$ decile of the observation is given by ${{D}_{m}}=m\times {{\left( \dfrac{n+1}{10} \right)}^{th}}$ value. Hence will calculate ${{D}_{m}}$ and the ${{m}^{th}}$ decile of the data

Complete step-by-step answer:
Now consider the set of observations 29, 18, 15, 30, 42, 35, 34, 28, 45, 34.
Arranging in ascending order we get 15, 18, 28, 29, 30, 34, 34, 35, 42, 45.
Now there are total 10 observations
Hence we have the value of n = 10. Now we know that ${{m}^{th}}$ decile of the observation is given by ${{D}_{m}}=m\times {{\left( \dfrac{n+1}{10} \right)}^{th}}$ value.
Hence to find ${{8}^{th}}$ decile let us substitute m = 8. Also we know n = 10. Hence we get
${{D}_{8}}=8\times {{\left( \dfrac{10+1}{10} \right)}^{th}}$ value.
${{D}_{8}}=8\times {{\left( 1.1 \right)}^{th}}$ value.
${{D}_{8}}={{8.8}^{th}}$ value.
Now since ${{D}_{8}}$ is in decimal we will have to calculate that 0.8 further distance
Now we know that ${{8}^{th}}$ value is 35 and ${{9}^{th}}$ value is 42
Now the distance between ${{8}^{th}}$ value and ${{9}^{th}}$ value is 42 – 35 = 7.
Hence 0.8 distance further of ${{8}^{th}}$ value will be $7\times 0.8=5.6$
Now ${{8.8}^{th}}$ value is nothing but ${{8}^{th}}$ value + 0.8 distance after ${{8}^{th}}$ value.
Hence ${{8.8}^{th}}$ value is $35+5.6=40.6$
Hence we get the ${{8}^{th}}$ decile is 40.6.

So, the correct answer is “Option b”.

Note: Note that the formula ${{D}_{m}}=m\times {{\left( \dfrac{n+1}{10} \right)}^{th}}$ does not give you ${{m}^{th}}$ decile but gives you the position of ${{m}^{th}}$ decile. Hence let us say we get ${{D}_{m}}=8$ this means the ${{m}^{th}}$ decile is ${{8}^{th}}$ number in the list. Also note that ${{D}_{m}}$ can also be non-integral value in this case we add the integral part of solution to the value of distance × decimal part. ${{\left( n+1 \right)}^{th}}-{{n}^{th}}$ Observation where n is integral part.
For example now we know that ${{8}^{th}}$ value is 28 and ${{9}^{th}}$ value is 45
Now the distance between ${{8}^{th}}$ value and ${{9}^{th}}$ value is 45 – 28 = 17.
Hence 0.8 distance further of ${{8}^{th}}$ value will be $17\times 0.8=13.6$
Now ${{8.8}^{th}}$ value is nothing but ${{8}^{th}}$ value + 0.8 distance after ${{8}^{th}}$ value.
Hence ${{8.8}^{th}}$ value is $28+13.6=41.6$