Solution to The ages of 100 persons are tabulated below. Find Q1, Q3, D2, D8, P20 and … - Sikademy
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Archangel Macsika

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.

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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.

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