Answer to Question #277000 in Discrete Mathematics for Himanshi

Question

The ages of 100 persons are tabulated below. Find Q1, Q3, D2, D8, P20 and P60. Above what age

we have the oldest 10% persons?

Age(in years) 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of persons 16 20 21 28 10 3 2

Note: the oldest 10% are above P90.

Ans.: P20 = 22, P60=41.071, the oldest 10% persons are of the age 55 and above.

Answer 1

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 1^st 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 1^st 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 3^rd 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 3^rd 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 2^nd 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 20^th 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 90^th 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.