Class 10 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 6 - Statistics
Statistics Exercise Ex. 6.1
Solution 1
Mean of the time spent by the students for their studies is 4.36 hours.
Solution 2
Here,
Mean = 
= 550 + (-28.57)
= 521.43
∴ The mean of the toll paid by the drivers is Rs. 521.43.
Solution 3
Here, 
∴ The mean of the milk sold is 2.82 litres.
Solution 4
Here, 
= 37.5 + (-2.19)
= 35.31
= Rs (35.31 × 1000)
= Rs 35310
∴ The mean of the production of oranges is Rs. 35310.
Solution 5
Here, 
= 1250 + (-0.525 × 500)
= 1250 - 262.5
= Rs 987.5
= Rs 35310
∴ The mean of the funds collected is Rs. 987.5.
Solution 6
Here, 
= 2500 + 0.57×1000
= 2500 + 570
= Rs. 3070
∴ The mean of the weekly wages is Rs. 3070.
Statistics Exercise Ex. 6.2
Solution 1
Here, 
Now, cumulative frequency just greater than (or equal to) 500 is 650.
∴ Median class is 10 - 12.
Now, L = 10, f = 500, c.f. = 150, h = 2
=10+0.7×2
=11.4
∴ The median of the number of hours the workers work is 11.4 hours.
Solution 2
Here, 
Now, cumulative frequency just greater than (or equal to) 125 is 153.
∴ Median class is 150 - 200.
Now, L = 150, f = 90, c.f. = 63, h = 50
= 150+34.4
=184.4
≈ 184
∴ The median of the given data is 184 mangoes (approx.).
Solution 3
Here, 
Now, cumulative frequency just greater than (or equal to) 100 is 184.
∴ Median class is 74.5 - 79.5.
Now, L = 74.5, f = 85, c.f. = 99, h = 5
= 74.5+0.059
=74.559
≈ 75
∴ The median of the given data is 75 km/hr (approx.).
Solution 4
Here, 
Now, cumulative frequency just greater than (or equal to) 52.5 is 67.
∴ Median class is 50 - 60.
Now, L = 50, f = 20, c.f. = 47, h = 10
= 50.2.75
= 52.75
= 52.75 × 1000
=52750 lamps
∴ The median of the productions is 52750 bulbs (approx.).
Statistics Exercise Ex. 6.3
Solution 1
Here, the maximum frequency is 80.
∴ Modal class is 4 - 5.
Here, L = 4, h = 1, f1 = 80, f0 = 70, f2 = 60
= 4+0.33
= 4.33
∴ The mode of the fat content is 4.33%.
Solution 2
Here, the maximum frequency is 100.
∴ Modal class is 60 - 80.
Here, L = 60, h = 20, f1 = 100, f0 = 70, f2 = 80
= 60+12
= 72
∴ The mode of use of electricity is 72 units.
Solution 3
Here, the maximum frequency is 35.
∴ Modal class is 9 - 11.
Here, L = 9, h = 2, f1 = 35, f0 = 20, f2 = 18
= 9+ 0.9375
=9.9375
≈ 9.94
∴ The mode of the supply of milk is 9.94 litres (approx.).
Solution 4
Here, the maximum frequency is 50.
∴ Modal class is 9.5 - 14.5.
Here, L = 9.5, h = 5, f1 = 50, f0 = 32, f2 = 36
= 9.5+2.8125
=12.3125
≈12.31
∴ The mode of the ages of the patients is 12.31 years (approx.).
Statistics Exercise Ex. 6.4
Solution 1
For the given data, the histogram will be as follows:
Solution 2
The histogram is as follows:
Solution 3
The histogram is as follows:
Solution 4
The histogram is as follows:
Statistics Exercise Ex. 6.6
Solution 1
Total number of persons = 80 + 60 + 35 + 25 = 200
Measure of
central angle (θ) = 
Solution 2
Total marks obtained = 50 + 70 + 80 + 90 + 60 + 50 = 400
Measure of
central angle (θ) = 
Solution 3
Total number of trees planted = 40 + 50 + 75 + 50 + 70 + 75 = 360
Measure of
central angle (θ) = 
Solution 4
Total percentage = 30 + 15 + 25 + 20 + 10 = 100%
Measure of
central angle (θ) = 
Solution 5
Measure of
central angle (θ) = 
(1)
Central angle for construction field = 72o
∴ Number of workers in construction = 2000
(2)
Central angle for construction field = 36o
∴ Number of workers in administration = 1000
(3)
Central angle for construction field = 90o
∴ Number of workers in construction = 2500
Now, 
∴ 25% of workers are working in the production field.
Solution 6
Measure
of central angle (θ) = 
(1)
Central angle for shares = 60o
(2)
Central angle for deposit in bank = 90o
(3)
Difference in central angle for immovable property and mutual fund = 120° - 60° = 60°
∴ Rs. 2000 more is invested in immovable property than in mutual fund.
(4) Central angle for post = 30°
∴ Rs. 1000 invested in post.
Statistics Exercise Ex. 6.5
Solution 1
(1) The class 60 - 70 has the maximum number of students.
(2) The classes 20 - 30 and 90 - 100 have frequency zero.
(3) The class mark of the class having 50 students is 55.
(4) The lower and upper class limits of the class having class mark 85 are 80 and 90 respectively.
(5) There are 15 students in the class 80 - 90.
Solution 2
The frequency polygon will be as follows:
Solution 3
The frequency polygon is as follows:
Statistics Exercise Problem Set 6
Solution 1(1)
Solution 1(2)
Solution 1(3)
Cumulative frequencies in a grouped frequency table are useful to find median.
Solution 1(4)
Solution 1(5)
The cumulative frequencies are as follows:
Here, N = 50
Therefore, N/2 = 25
The cumulative frequency just greater than (or equal to) 25 is 43.
Therefore, the median lies in the group 16 - 18.
Solution 1(6)
The class marks will be:
2, 5, 8, 11
The coordinates of the points will be:
(2, 7), (5, 8), (8, 6), (11, 4)
The coordinates of the point to show number of students in the class 4 - 6 are (5, 8).
Solution 2
∴ The mean of the income of the farmers is Rs. 52,500.
Solution 3
∴ The mean of the loans given by the bank is Rs. 65,400.
Solution 4
∴ The mean of the weekly wages of the workers is Rs. 4250.
Solution 5
∴ The mean of the amount of aid given to families is Rs. 72400.
Solution 6
Cumulative frequency which is just greater than (or equal to) 125 is 180.
∴ The median class is 220 - 230.
Now, L = 220, f = 80, cf = 100, h = 10
= 220 + 3.125
≈ 223.13
∴ The median of the distances is 223.13 km (approx.).
Solution 7
Cumulative frequency which is just greater than (or equal to) 200 is 240.
∴ The median class is 20 - 40.
Now, L = 20, f = 100, cf = 140, h = 20
= 20 + 12
= 32
∴ The median of the prices of different articles is Rs. 32.
Solution 8
Here, the maximum frequency is 60.
∴ The modal class is 250 - 500.
Now, L = 250, f0 = 10, f1 = 60, f2 = 25, h = 250
= 250 + 147.06
= 397.06
∴ The mode of the demand of sweet is 397.06 grams.
Solution 9
The histogram is as follows:
Solution 10
The frequency polygon will be as follows:
Solution 11
The frequency polygon is as follows:
Solution 12
The frequency polygon is as follows:
Solution 13
The frequency polygon is as follows:
Solution 14
(1)
(2)
Central angle for two wheelers (θ) = 
∴ Total number of vehicles is 3000.
Solution 15
Solution 16
(1)
Central angle for cricket 
∴ 225 students like cricket.
(2)
Central angle for cricket 
∴ 175 students like football.
(3)
Central angle for cricket 
∴ 200 students like other games.
Solution 17
Total number of women = 180
Solution 18
Total number of trees planted = 120
