Exercise 13.2

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

Age(in years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
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.

Class marks(xi) as per the relation

xi = Upperclasslimit+lowerclasslimit2

Age	No.of patients(fi)	Class marks(xi)	di = xi - 275	fidi
5 - 15	6
15 - 25	11
25 - 35	21
35 - 45	23
45 - 55	14
55 - 65	5
Total

From the table,

fi = , fidi =

Mean = a + ΣfidiΣfi

x = + 43080

= 30 + =

Mean of this data is 35.38. It represents that on an average the age of patients admittied in the hospital was of age 35.38 years

Mode:

Modal class = 35 - 45

l = , f = , h = 10, f1 = , f2 = 14

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : Mode = + 23−212×23−21−14×

= 35 + 246−35 ×

= 35 + =

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

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

Lifetimes (in hours)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution :

Given data,

From the data given above, we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.

So, modal class =

f = , f1 = , f2 =

Class size(h) =

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : Mode = + 61−522×61−52−38×

= 60 + × 20

= 60 +

= 60 + =

So, the 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
1000 - 1500	24
1500 - 2000	40
2000 - 2500	33
2500 - 3000	28
3000 - 3500	30
3500 - 4000	22
4000 - 4500	16
4500 - 5000	7

Solution :

Expenditure(in ₹)	Number of families fi	fiui
1000 - 1500	24
1500 - 2000	40
2000 - 2500	33
2500 - 3000	28
3000 - 3500	30
3500 - 4000	22
4000 - 4500	16
4500 - 5000	7
	Σfi =	Σfiui =

Here, the maximum frequnecy is 4o so the modal class is - .

Therfore, l = 1500, h = 500, f = 40, f1 = 24, f2 = 33

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : Mode = + 40−2480−24−33×

= + 1623 ×

= 1500 + = Rs

Thus, the modal montly expenditure of the families is RS 1847.83.

Now, Mean monthly expenditure of the families = ΣfixiΣfi

= 532500200 =

Thus, the mean montly expenditure of the families is 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.
15 - 20	3
20 - 25	8
25 - 30	9
30 - 35	10
35 - 40	3
40 - 45	0
45 - 50	0
50 - 55	2

Solution :

Given data:

Modal class is 30 - 35, l = 30

fm = 10, f1 = 9, f2 = 3 and h = 5

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : + 10−920−9−3×

= + 58

= 30 + =

Therefore, the mode of the given data is 30.625.

Find Mean

Formula xi = uppwelimit+lowerlimit2

Class interval	Frequency(fi)	fixi
15 - 20	3
20 - 25	8
25 - 30	9
30 - 35	10
35 - 40	3
40 - 45	0
45 - 50	0
50 - 55	2
Total	Σfi =	Σ fixi =

Calculate mean,

Mean = x = ΣfixiΣfi

= 1023.535 =

Hence, the mean of the given data 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
3000 - 4000	4
4000 - 5000	18
5000 - 6000	9
6000 - 7000	7
7000 - 8000	6
8000 - 9000	3
9000 - 10000	1
10000 - 11000	1

Find the mode of the data.

Solution :

Given data:

Modal class = 4000 - 5000, l = 4000

Class width(h) = 1000, fm = 18 and f1 = 4 and f2 = 9

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : Mode = 4000 + 18−436−4−9×

Mode = + 1400023 = +

Mode = (Upto one decimal place)

Thus,the 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	0-10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	7	14	13	12	20	11	15	8

Solution :

Given data:

Modal class = 40 - 50, l = 40

Class width(h) = 10, fm = 20 and f1 = 12 and f2 = 11

Mode = l +fm−f12fm−f1−f2× h

Substitute the values : Mode = + 20−1240−12−11×

Mode = + 8017 = + =

Thus, the mode of the given data is 44.7 cars.

Chapter 13: Statistics

Glossary

Exercise 13.2

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Chapter 13: Statistics

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Glossary

Exercise 13.2