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The frequency table for the data given above is as shown below.

class boundary f cf

10-20 16 16

20-30 20 36

30-40 21 57

40-50 28 85

50-60 10 95

60-70 3 98

70-80 2 100

Here, $n=100.$

$i)$

$Q_1$ is the $1^{st}$ quartile defined by the formula,

$Q_1=l+({n\over4}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $Q_1$

$cf$ is the cumulative frequency of the class preceding the class containing $Q_1$

$c$ is the width of the class in which $Q_1$ lies

$f$ is the frequency of the class containing $Q_1$

To determine the 1st quartile, we first find its position in order to find its class boundary as follows,

$Q_1=({n\over4})^{th} value=({100\over 4})^{th}=25^{th}value$

Therefore, the class boundary for the 1st quartile is 20-30

Thus,

$Q_1=20+(25-16)\times({10\over20})=20+4.5=24.5$

$ii)$

$Q_3$ is the third quartile defined by the formula,

$Q_3=l+({3\over4}\times n-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $Q_3$

$cf$ is the cumulative frequency of the class preceding the class containing $Q_3$

$c$ is the width of the class containing $Q_3$

$f$ is the frequency of the class containing $Q_3$

To determine the 3rd quartile, we first its position in order to find its class boundary as follows,

$Q_1=({3\over4}\times n)^{th} value=({3\over 4}\times 100)^{th}=75^{th}value$

Therefore, the class boundary for the 3rd quartile is 40-50

Thus,

$Q_3=40+(75-57)\times({10\over28})=40+6.4286=46.4286(4dp)$

Therefore, $Q_3=46.4286$

$iii)$

To find the 2nd decile; $D_2$, we use the formula stated below,

$D_2=l+({(2\times n)\over10}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $D_2$

$cf$ is the cumulative frequency of the class preceding the class containing $D_2$

$c$ is the width of the class in which $D_2$ lies

$f$ is the frequency of the class containing $D_2$

We first determine the class containing $D_2$,

$D_2=({(2\times100)\over10})^{th} value=20^{th}value$

Therefore, the class where $D_2$ can be found is 20-30

Hence,

$D_2=20+(20-16)\times({10\over20})=20+2=22$

Therefore, $D_2=22$

$iv)$

To find $D_8,$ we apply the formula below,

$D_8=l+({(8\times n)\over10}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $D_8$

$cf$ is the cumulative frequency of the class preceding the class containing $D_8$

$c$ is the width of the class in which $D_8$ lies

$f$ is the frequency of the class containing $D_8$

The position of $D_8$ is,

$D_8=({(8\times100)\over10})^{th}value.=80^{th}value$

The class in which $D_8$ lies is 40-50

Thus,

$D_8=40+(80-57)\times({10\over28})=40+8.2143=48.2143(4dp)$

Therefore, $D_8=48.2143$

$v)$

To find the 20th percentile given as $P_{20}$, we apply the following formula,

$P_{20}=l+({(20\times n)\over 100}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $P_{20}$

$cf$ is the cumulative frequency of the class preceding the class containing $P_{20}$

$c$ is the width of the class in which $P_{20}$ lies

$f$ is the frequency of the class containing $P_{20}$

Let us determine the class where $P_{20}$ lies by first finding its position,

$P_{20}={(20\times100)\over 100}=20^{th} position$

Now, $P_{20}$ lies in the class 20-30

Thus,

$P_{20}=20+(20-16)\times({10\over 20})=20+2=22$

Therefore, $P_{20}=22$

$vi)$

To find the $60^{th}$ percentile given as, $P_{60}$, the following the formula is used,

$P_{60}=l+({(60\times n)\over 100}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $P_{60}$

$cf$ is the cumulative frequency of the class preceding the class containing $P_{60}$

$c$ is the width of the class in which $P_{60}$ lies

$f$ is the frequency of the class containing $P_{60}$

To find the class where $P_{60}$ lies, we first determine its position as follows,

$P_{60}={(60\times100)\over100}=60^{th}position$

Thus, $P_{60}$ is in the class 40-50

Therefore,

$P_{60}=40+(60-57)\times({10\over28})=40+1.0714=41.0714$

Therefore, $P_{60}=41.0714(4dp)$

vii)

To find the age where the oldest 10% persons are above, we determine the 90th percentile given as, $P_{90}$. We shall apply the formula below,

$P_{90}=l+({(90\times n)\over 100}-cf)\times({c\over f})$ where,

$l$ is the lower class boundary of the class containing $P_{90}$

$cf$ is the cumulative frequency of the class preceding the class containing $P_{90}$

$c$ is the width of the class containing $P_{90}$

$f$ is the frequency of the class containing $P_{90}$

To find the class where $P_{90}$ lies, we first determine its position as follows,

$P_{90}={(90\times100)\over100}=90^{th}position$

Thus, $P_{90}$ is in the class 50-60

Therefore,

$P_{90}=50+(90-85)\times({10\over10})=50+5=55$

Hence, the oldest 10% persons are of the age 55 and above.