Chapter 13 Statistics Exercise-02

EXERCISE 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years) 515 1525 2535 3545 4555 5565
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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Solution

To find the class marks (xi), the following relation is used.

xi= Upper class limit + Lower class limit 2

Taking 30 as assumed mean (a), di and fidi are calculated as follows.

Age (in years) Number of patients fi Class mark xi di=xi30 fidi
515 6 10 -20 -120
1525 11 20 -10 -110
2535 21 30 0 0
3545 23 40 10 230
4555 14 50 20 280
5565 5 60 30 150
Total 80 430

From the table, we obtain

fi=80

fidi=430

Mean, x¯=a+fidifi

=30+(43080)=30+5.375=35.37535.38

Mean of this data is 35.38 . It represents that on an average, the age of a patient admitted to hospital was 35.38 years.

It can be observed that the maximum class frequency is 23 belonging to class interval 35 - 45 .

Modal class =3545

Lower limit ( l ) of modal class =35

Frequency (f1) of modal class =23

Class size (h)=10

Frequency (f0) of class preceding the modal class =21

Frequency (f2) of class succeeding the modal class =14

Mode =

l+(f1f02f1f0f2)×h

=35+(23212(23)2114)×10

=35+[24635]×10

=35+2011

=35+1.81

=36.8

Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Lifetimes (in hours) 020 2040 4060 6080 80100 100120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

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Solution

From the data given above, it can be observed that the maximum class frequency is 61 , belonging to class interval 60 - 80.

Therefore, modal class =6080

Lower class limit (l) of modal class =60

Frequency (f1) of modal class =61

Frequency (f0) of class preceding the modal class =52

Frequency (f2) of class succeeding the modal class =38

Class size (h)=20

Mode =l+(f1f02f1f0f2)×h

=60+(61522(61)5238)(20)

=60+(912290)(20)=60+(9×2032)=60+9016=60+5.625=65.625

Therefore, modal lifetime of electrical components is 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Expenditure (in ₹) Number of families
10001500 24
15002000 40
20002500 33
25003000 28
30003500 30
35004000 22
40004500 16
45005000 7
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Solution

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 - 2000 intervals.

Therefore, modal class =15002000

Lower limit (l) of modal class =1500

Frequency (f1) of modal class =40

Frequency (f0) of class preceding modal class =24

Frequency (f2) of class succeeding modal class =33

Class size (h)=500

 Mode =l+(f1f02f1f0f2)×h=1500+(40242(40)2433)×500=1500+(168057)×500=1500+800023=1500+347.826=1847.826=1847.83

Therefore, modal monthly expenditure was Rs 1847.83.

To find the class mark, the following relation is used.

Class mark = Upper class limit + Lower class limit 2

Class size (h) of the given data =500

Taking 2750 as assumed mean (a), di,ui, and fiui are calculated as follows.

Expenditure (in Rs.) No. of families Class Mark (xi) di=xia ui=di500 fiui
10001500 24 1250 -2000 -4 -96
15002000 40 1750 -1500 -3 -120
20002500 33 2250 -1000 -2 -66
25003000 28 2750 -500 -1 -28
30003500 30 3250=a 0 0 0
35004000 22 3750 500 1 22
40004500 16 4250 1000 2 32
45005000 7 4750 1500 3 21
Total Σfi=200 Σfiui=235

From the table, we obtatain

fi=200,uifi=235 Mean monthly income =x=a+ΣfiuiΣfi×h=3250235200×500=3250587.5= Rs. 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher Number of states / U.T.
1520 3
2025 8
2530 9
3035 10
3540 3
4045 0
4550 0
5055 2
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Solution

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 - 35 .

Therefore, modal class =3035

Class size (h)=5

Lower limit (l) of modal class =30

Frequency (f1) of modal class =10

Frequency (f0) of class preceding modal class =9

Frequency (f2. ) of class succeeding modal class =3

 Mode =l+(f1f02f1f0f2)×h=30+(1092(10)93)×(5)=30+(12012)5=30+58=30.625

Mode =30.6

It represents that most of the states/U.T have a teacher-student ratio as 30.6.

To find the class marks, the following relation is used.

Class mark = Upper class limit + Lower class limit 2

Taking 32.5 as assumed mean (a), di,ui, and fiui are calculated as follows.

Number of students per teacher Number of states/U.T
(fi)
xi di=xi32.5 ui=di5 fiui
1520 3 17.5 -15 -3 -9
2025 8 22.5 -10 -2 -16
2530 9 27.5 -5 -1 -9
3035 10 32.5 0 0 0
3540 3 37.5 5 1 3
4045 0 42.5 10 2 0
4550 0 47.5 15 3 0
5055 2 52.5 20 4 8
Total 35 -23

Mean, x¯=a+(fiuifi)h

=32.5+(2335)×5=32.5237=32.53.28=29.22

Therefore, mean of the data is 29.2.

It represents that

  • The maximum number of students per teacher is 30.6
  • The average number of students per teacher is 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored Number of batsmen
30004000 4
40005000 18
50006000 9
60007000 7
70008000 6
80009000 3
900010000 1
1000011000 1

Find the mode of the data.

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Solution

From the given data, it can be observed that the maximum class frequency is 18 , belonging to class interval 40005000 .

Therefore, modal class =40005000

Lower limit (I) of modal class =4000

Frequency (f1) of modal class =18

Frequency (f0) of class preceding modal class =4

Frequency (f2. ) of class succeeding modal class =9

Class size (h)=1000

Mode =l+(f1f02f1f0f2)×h

=4000+(1842(18)49)×1000

=4000+(1400023)

=4000+608.695

=4608.695

Therefore, mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Number of cars 010 1020 2030 3040 4050 5060 6070 7080
Frequency 7 14 13 12 20 11 15 8
Show Answer

Solution

From the given data, it can be observed that the maximum class frequency is 20 , belonging to 40 - 50 class intervals.

Therefore, modal class =4050

Lower limit ( l ) of modal class =40

Frequency (f1) of modal class =20

Frequency (f0) of class preceding modal class =12

Frequency (f2) of class succeeding modal class =11

Class size =10

 Mode =l+(f1f02f1f0f2)×h=40+[20122(20)1211]×10=40+(804023)=40+8017

=40+4.7

=44.7

Therefore, mode of this data is 44.7 cars.



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