
Calculate Quartile Deviation. for the following data
X 24 25 26 27 28 29 30 F 6 5 3 2 4 7 3
| X | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| F | 6 | 5 | 3 | 2 | 4 | 7 | 3 |
Answer
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Hint: We need to find the cumulative frequency for the given data. Then using the cumulative frequency we need to find the value of $Q_1$ & $Q_3$., using which we can find the value of Quartile Deviation.
Formula.
$Q_1$= Value of $ {(\dfrac{{N + 1}}{4})^{th}} $ observation, where N is Cumulative Frequency
$Q_3$= Value of $ {[3(\dfrac{{N + 1}}{4})]^{th}} $ observation, where N is Cumulative Frequency
$ Quartile\; Deviation = \dfrac{{{Q_3} - {Q_1}}}{2} $
Complete step-by-step answer:
We have to find Quartile Deviation for the given set of data. For that we need to find Cumulative Frequency (N) and also $Q_1$ & $Q_3$.
So, at first let us find the Cumulative Frequency (N)
Hence, the value of N for the given set of values is 30.
Now, we need to find the value of $Q_1$
Where, $Q_1$= Value of $ {(\dfrac{{N + 1}}{4})^{th}} $ observation
$Q_1$= Value of $ {(\dfrac{{30 + 1}}{4})^{th}} $ observation
$Q_1$= Value of $ {(7.75)^{th}} $ observation
Cumulative frequency which is just greater than (or equal to) 7.75 is 11.
Therefore, $Q_1$= 25
Similarly, we need to find the value of $Q_3$
Where, $Q_3$= Value of $ {[3(\dfrac{{N + 1}}{4})]^{th}} $ observation
$Q_3$= Value of $ {[3(\dfrac{{30 + 1}}{4})]^{th}} $ observation
$Q_3$= Value of $ {(23.25)^{th}} $ observation
Cumulative frequency which is just greater than (or equal to) 23.25 is 27.
Therefore, $Q_3$= 29
Now, using the derived value of $Q_1$ and $Q_3$ we need to find the value of Quartile Deviation using the below mentioned formula.
$\Rightarrow Quartile\;Deviation = \dfrac{{{Q_3} - {Q_1}}}{2} $
$\Rightarrow Quartile\;Deviation = \dfrac{{29 - 25}}{2} $
$\Rightarrow Quartile\;Deviation = \dfrac{4}{2} $
$\Rightarrow Quartile\;Deviation = 2 $
Hence, the value of “Quartile Deviation” for the given set is 2.
So, the correct answer is “2”.
Note: Generally, $Q_1$ is the median (the middle) of the lower half of the data, and $Q_3$ is the median (the middle) of the upper half of the data. But for calculating the actual value and also to save time, we need to remember the above formulas without which we won’t be solved or proceed further.
Formula.
$Q_1$= Value of $ {(\dfrac{{N + 1}}{4})^{th}} $ observation, where N is Cumulative Frequency
$Q_3$= Value of $ {[3(\dfrac{{N + 1}}{4})]^{th}} $ observation, where N is Cumulative Frequency
$ Quartile\; Deviation = \dfrac{{{Q_3} - {Q_1}}}{2} $
Complete step-by-step answer:
We have to find Quartile Deviation for the given set of data. For that we need to find Cumulative Frequency (N) and also $Q_1$ & $Q_3$.
So, at first let us find the Cumulative Frequency (N)
| X | F | C.F. |
| 24 | 6 | 6 |
| 25 | 5 | 11 |
| 26 | 3 | 14 |
| 27 | 2 | 16 |
| 28 | 4 | 20 |
| 29 | 7 | 27 |
| 30 | 3 | 30 |
Hence, the value of N for the given set of values is 30.
Now, we need to find the value of $Q_1$
Where, $Q_1$= Value of $ {(\dfrac{{N + 1}}{4})^{th}} $ observation
$Q_1$= Value of $ {(\dfrac{{30 + 1}}{4})^{th}} $ observation
$Q_1$= Value of $ {(7.75)^{th}} $ observation
Cumulative frequency which is just greater than (or equal to) 7.75 is 11.
Therefore, $Q_1$= 25
Similarly, we need to find the value of $Q_3$
Where, $Q_3$= Value of $ {[3(\dfrac{{N + 1}}{4})]^{th}} $ observation
$Q_3$= Value of $ {[3(\dfrac{{30 + 1}}{4})]^{th}} $ observation
$Q_3$= Value of $ {(23.25)^{th}} $ observation
Cumulative frequency which is just greater than (or equal to) 23.25 is 27.
Therefore, $Q_3$= 29
Now, using the derived value of $Q_1$ and $Q_3$ we need to find the value of Quartile Deviation using the below mentioned formula.
$\Rightarrow Quartile\;Deviation = \dfrac{{{Q_3} - {Q_1}}}{2} $
$\Rightarrow Quartile\;Deviation = \dfrac{{29 - 25}}{2} $
$\Rightarrow Quartile\;Deviation = \dfrac{4}{2} $
$\Rightarrow Quartile\;Deviation = 2 $
Hence, the value of “Quartile Deviation” for the given set is 2.
So, the correct answer is “2”.
Note: Generally, $Q_1$ is the median (the middle) of the lower half of the data, and $Q_3$ is the median (the middle) of the upper half of the data. But for calculating the actual value and also to save time, we need to remember the above formulas without which we won’t be solved or proceed further.
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