Question

Given below are the prices of shares of a company for the last 10 days. Find Q.D. $172, 164, 188, 214, 190, 237, 200, 195, 208, 230$

Answer 1

Here, we will find the Quartile deviation for the given data. We will find the first quartiles and third quartiles from the given data and by using the Lower Quartile and Upper Quartile formula. Then we will use these values and substitute it in the quartile deviation formula to find the quartile deviation for the given data set.

Formula Used:
We will use the following Formula:

1. Lower Quartile is given by the formula ${Q_1} =$ Value of $\left( {\dfrac{{n + 1}}{4}} \right)th$ observation
2. Upper Quartile is given by the formula ${Q_3} =$ Value of $3\left( {\dfrac{{n + 1}}{4}} \right)th$ observation
3. Quartile Deviation is given by the formula $Q.D. = \dfrac{{{Q_3} - {Q_1}}}{2}$ where $n$ is the number of items in the data.

Complete step by step solution:
We are given the data set as $172, 164, 188, 214, 190, 237, 200, 195, 208, 230$
Now, we will arrange the given data in the ascending order as$164, 172, 188, 190, 195, 200, 208, 214, 230, 237$
We are given with the number of items in the data $n = 10$
We will find the upper quartile and lower quartile for the given data using the formula.
Lower Quartile is given by the formula ${Q_1} =$ Value of $\left( {\dfrac{{n + 1}}{4}} \right)th$ observation
By substituting the value of $n$, we get
${Q_1} =$ Value of $\left( {\dfrac{{10 + 1}}{4}} \right)th$ observation
Simplifying the expression, we get
$\Rightarrow {Q_1} =$ Value of $\left( {2.75} \right)th$ observation
$\Rightarrow {Q_1} =$ Value of $2nd$ observation $+ 0.75$ (Value of $3rd$ observation $-$ Value of $2nd$ observation)
Substituting the respective values in the above equation, we get
$\Rightarrow {Q_1} = 172 + 0.75\left( {188 - 172} \right)$
Subtracting the terms, we get
$\Rightarrow {Q_1} = 172 + 0.75\left( {16} \right)$
By multiplying the numbers, we get
$\Rightarrow {Q_1} = 172 + 12$
Adding the terms, we get
$\Rightarrow {Q_1} = 184$
Upper Quartile is given by the formula ${Q_3} =$ Value of $3\left( {\dfrac{{n + 1}}{4}} \right)th$ observation
By substituting the value of $n$, we get
${Q_3} =$ Value of $3\left( {\dfrac{{10 + 1}}{4}} \right)th$ observation
Simplifying the expression, we get
$\Rightarrow {Q_3} =$ Value of $\left( {3 \times 2.75} \right)th$ observation
Multiplying the terms, we get
$\Rightarrow {Q_3} =$ Value of $\left( {8.25} \right)th$ observation
$\Rightarrow {Q_3} =$ Value of $8th$ observation $+ 0.25$ (Value of $9th$ observation $-$ Value of $8th$ observation)
Substituting the respective values in the above equation, we get
$\Rightarrow {Q_3} = 214 + 0.25\left( {230 - 214} \right)$
$\Rightarrow {Q_3} = 214 + 0.25\left( {16} \right)$
By multiplying the numbers, we get
$\Rightarrow {Q_3} = 214 + 4$
Adding the terms, we get
$\Rightarrow {Q_3} = 218$
Now, we will find the Quartile deviation using the upper quartiles and lower quartiles.
Quartile Deviation is given by the formula $Q.D. = \dfrac{{{Q_3} - {Q_1}}}{2}$
Now substituting ${Q_3} = 218$ and ${Q_1} = 184$ in the formula, we get
$\Rightarrow Q.D. = \dfrac{{218 - 184}}{2}$
Subtracting the terms in the numerator, we get
$\Rightarrow Q.D. = \dfrac{{34}}{2}$
Dividing the terms, we get
$\Rightarrow Q.D. = 17$

Therefore, the Quartile Deviation for the given data is 17.

Note:
We know that the Quartile Deviation is defined as the half the distance between the upper quartiles and the lower quartiles. Quartile Deviation is also called the semi inter quartile range. The lower quartile corresponds to the value which is in between the median and the lowest value in the distribution and the upper quartile corresponds to the value which is in between the median and the highest value in the distribution. The lower quartile is the $25th$ percentile of the distribution and the upper quartile is the $75th$ percentile of the distribution.