Statistics ⇒⇒ Exercise 13.2

Question 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.

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

$${Upper Class Limit + Lower Class Limit \over 2} $$

By using Assumed Mean method.

Let the assumed mean be a = 30

Age (in years)	Number of patients Freqc. fi	Classmark xi	di = (xi-a)	(fi di )
5 - 15	6	10	10 - 30 = -20	-120
15 - 25	11	20	20 - 30 = -10	-110
25 - 35	21 (f0)	30	30 - 30 = 0	0
35 - 45	23 (f1)	40	40 - 30 = 10	230
45 - 55	14 (f2)	50	50 - 30 = 20	280
55 - 65	5	60	60 - 30 = 30	150
	$\Sigma f_i = 80 $			$\Sigma f_id_i = 430 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_id_i} \over {\Sigma f_i} }$$

$$ \bar{X} = 30 + { {430} \over {80} }$$

$$ \bar{X} = 30 + {( 5.375)}$$

$$ \bar{X} = 35.375 $$

$$ \bar{X} = 35.37 $$

Now Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 23,

So, the modal class is 35 − 45

L (the lower limit of modal class) = 35

f₁ (frequency of the modal class) = 23

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

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

h (class size) = 10

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 35 + [ { { 23 - 21} \over { 2(23) - 21 - 14} } ] ×(10) $$

$$ = 35 + [ { { 2} \over { 11} } ] ×(10) $$

$$ = 35 + [ { { 20} \over { 11} } ] $$

$$ = 35 + 1.81 $$

$$ Mode = 36.8 $$

The mode of the data shows that maximum number of patients is in the age group of 36.8, whereas the average age of all the patients is 35.37.

Question 2

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

Lifetimes (in hours)	00 − 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 :

Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 61,

So, the modal class is 60 − 80

L (the lower limit of modal class) = 60

f₁ (frequency of the modal class) = 61

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

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

h (class size) = 20

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] × h $$

$$ = 60 + [ { { 61 - 52} \over { 2(61) - 52 - 38} } ] ×(20) $$

$$ = 60 + [ { { 9} \over { (122-90)} } ] ×(20) $$

$$ = 60 + [ {( 9×20) \over { 32} } ] $$

$$ = 60 + [ {( 9×5) \over { 8} } ] $$

$$ = 60 + [ { 45 \over { 8} } ] $$

$$ = 60 + 5.625 $$

$$ Mode = 65.625 $$

Therefore, modal lifetime of electrical components is 65.625 hours

Question 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 Rs)	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

Determine the modal lifetimes of the components.

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

$${Upper Class Limit + Lower Class Limit \over 2} $$

By using Step deviation method.

Let the assumed mean be a = 2750 and Class size (h) of this data = 500

Exp. (in Rs)	Number of families fi	Class mark xi	di = (xi-a)	ui = $ d_i \over h $	(fi ui )
1000 − 1500	24 (f0)	1250	1250 - 2750 = -1500	-3	-72
1500 − 2000	40 (f1)	1750	1750 - 2750 = -1000	-2	-80
2000 − 2500	33 (f2)	2250	2250 - 2750 = -500	-1	-33
2500 − 3000	28	2750	2750 - 2750 = 0	0	0
3000 − 3500	30	3250	3250 - 2750 = 500	1	30
3500 − 4000	22	3750	3750 - 2750 = 1000	2	44
4000 − 4500	16	4250	4250 - 2750 = 1500	3	48
4500 − 5000	7	4750	4750 - 2750 = 2000	4	28
	$\Sigma f_i = 200 $				$\Sigma f_iu_i = -35 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 2750 + ({ {-35} \over {200} } )×(500)$$

$$ \bar{X} = 2750 + ({ {-175} \over {2}})$$

$$ \bar{X} = 2750 - 87.5$$

$$ \bar{X} = 2662.50 $$

Therefore, mean monthly expenditure was Rs. 2662.50

Now Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 40,

So, the modal class is 1500 − 2000

L (the lower limit of modal class) = 1500

f₁ (frequency of the modal class) = 40

f₀ (frequency of the class preceding the modal class) = 24

f₂ (frequency of the class succeeding the modal class) = 33

h (class size) = 500

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ Mode = 1500 + [ { { 40 - 24} \over { 2(40) - 24 - 33} } ] ×(500) $$

$$ = 1500 + [ { { 16 ×500} \over { 80- 57} } ] $$

$$ = 1500 + [ { { 8000} \over { 23} } ] $$

$$ = 1500 + 347.83 $$

$$ Mode = 1847.83 $$

Therefore, modal monthly expenditure was Rs 1847.83.

Question 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 :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

$${Upper Class Limit + Lower Class Limit \over 2} $$

By using Step deviation method.

Let the assumed mean be a = 37.5 and Class size (h) of this data = 5

No.of students per teacher	No.of states/U.T fi	Class mark xi	di = (xi-a)	ui = $ d_i \over h $	(fi ui )
15 − 20	3	17.5	17.5 - 37.5 = -20	-4	-12
20 − 25	8	22.5	22.5 - 37.5 = -15	-3	-24
25 − 30	9 (f0)	27.5	27.5 - 37.5 = -10	-2	-18
30 − 35	10 (f1)	32.5	32.5 - 37.5 = -5	-1	-10
35 − 40	3 (f2)	37.5	37.5 - 37.5 = 0	0	0
40 − 45	0	42.5	42.5 - 37.5 = 5	1	0
45 − 50	0	47.5	47.5 - 37.5 = 10	2	0
50 − 55	2	52.5	52.5 - 37.5 = 15	3	6
	$\Sigma f_i = 35 $				$\Sigma f_iu_i = -58 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 37.5 + ({ {-58} \over {35} } )×(5)$$

$$ \bar{X} = 37.5 + ({ {-58} \over {7}})$$

$$ \bar{X} = 37.5 - 8.285 $$

$$ \bar{X} = 29.214 $$

Therefore, mean of the data is 29.214 or 29.22

Now Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 10,

So, the modal class is 30 − 35

L (the lower limit of modal class) = 30

f₁ (frequency of the modal class) = 10

f₀ (frequency of the class preceding the modal class) = 9

f₂ (frequency of the class succeeding the modal class) = 3

h (class size) = 5

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 30 + [ { { 10 - 9} \over { 2(10) - 9 - 3} } ] ×(5) $$

$$ = 30 + [ { { 1 ×5} \over { 20- 12} } ] $$

$$ = 30 + [ { { 5} \over { 8} } ] $$

$$ = 30 + 0.625 $$

$$ Mode =30.625 $$

Therefore, Mode of the data is 30.625.

Question 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 :

Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 18,

So, the modal class is 4000 − 5000

L (the lower limit of modal class) = 4000

f₁ (frequency of the modal class) = 18

f₀ (frequency of the class preceding the modal class) = 4

f₂ (frequency of the class succeeding the modal class) = 9

h (class size) = 1000

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 4000 + [ { { 18 - 4} \over { 2(18) - 4 - 9} } ] ×(1000) $$

$$ = 4000 + [ { { 14 ×1000} \over { 36- 13} } ] $$

$$ = 4000 + [ { { 14000} \over { 23} } ] $$

$$ = 4000 + 608.695 $$

$$ Mode = 4608.695$$

Therefore, Mode of the data is 4608.695 Runs.

Question 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	Frequency
0 − 10	7
10 − 20	14
20 − 30	13
30 − 40	12
40 − 50	20
50 − 60	11
60 − 70	15
70 − 80	8

Solution :

Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 20,

So, the modal class is 40 − 50

L (the lower limit of modal class) = 40

f₁ (frequency of the modal class) = 20

f₀ (frequency of the class preceding the modal class) = 12

f₂ (frequency of the class succeeding the modal class) = 11

h (class size) = 10

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 40 + [ { { 20 - 12} \over { 2(20) - 12 - 11} } ] ×(10) $$

$$ = 40 + [ { { 8 ×10} \over { 40- 23} } ] $$

$$ = 40 + [ { { 80} \over { 17} } ] $$

$$ = 40 + 4.7 $$

$$ Mode = 44.7$$

Therefore, Mode of the data is 44.7 .