Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6 Practice Set 6.2 – Statistics
Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6: Statistics. Marathi or English Medium Students of Class 10 get here Statistics full Exercise Solution.
Std | Maharashtra Class 10 |
Subject | Math Part 1 Solution |
Chapter | Statistics |
Practice Set | 6.2 |
Practice Set 6.2
(1) The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.
Daily No. of hours | 8 – 10 | 10 – 12 | 12 – 14 | 14 – 16 |
Number of workers | 150 | 500 | 300 | 50 |
Solution:
Class No. of hours | Frequencies to (No of frequency) | Cumulative frequency (cf) |
8 – 10 | 150 | 150 |
10 – 12 | 500 | 650 = cf |
12 – 14 | 300 – f | 950 |
14 – 16 | 50 | 1000 |
N = 1000 |
Here = N = 1000
∴ N/2 = 500 cf = 650
f = 300
h = 10 – 8 = 2
L = 12
∴ Median = L + [(N/2 – cf)/f] A×h
= 12 + 5000 – 650/300 × 2
= 12 – 1500/150
= 12 – 1
= 11
∴ Median of number of hours they work is 11.
(2) The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.
No. of Mangoes | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 |
No. of trees | 33 | 30 | 90 | 80 | 17 |
Solution:
Class No. of mangoes | Frequency No. of trees (f) | Cumulative frequency (cf) |
50 – 100 | 33 | 33 |
100 – 150 | 30 | 63 ->cf |
150 – 200 | 90 -> f | 153 |
200 – 250 | 80 | 233 |
250 – 300 | 17 | 250 = N |
N = 250 |
Here, N = 250 ∴ N/2 = 250/2 = 125
f = 90, cf = 63, h = 200 – 150 = 50
L = 150
∴ Median = L + [(N/2 – cf)/f] × h
= 150 + [125-63/90] × 50
= 150 + 62/9 × 5
= 150 + 34.45
= 184.45
∴ 184 mangoes is the median.
(3) The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.
Average Speed of Vehicles (Km/hr) | 60 – 64 | 64 – 69 | 70 – 79 | 75 – 79 | 79 – 84 | 84 – 89 |
No. of vehicles | 10 | 34 | 55 | 85 | 10 | 6 |
Solution:
The given data is not in continuous classes so we have to break it down to continuous classes to find median.
∴ The differences in upper limit lower limit of two consecutive classes is 1.
∴ 1÷2 = 0.5, Now subtracting and adding 0.5 to down and upper limit consecutively.
∴ The following table can be observed.
Class Avg speed | Continuous classes | Frequency No. of vehicles (f) | Cumulative frequency (cf) |
60 – 64 | 59.5 – 64.5 | 10 | 10 |
64 – 69 | 64.5 – 69.5 | 34 | 44 |
70 – 74 | 69.5 – 74.5 | 55 | 99 -> cf |
75 – 79 | 74.5 – 79.5 | 85 -> f | 184 |
79 – 84 | 79.5 – 84.5 | 10 | 194 |
84 – 89 | 84.5 – 89.5 | 6 | 200 = N |
Here, N/2 = 200/2 = 100, h = 79.5 – 74.5 = 5
f = 85, cf = 99, L = 74.5
∴ Median = L + [N/2 – cf/f] × h
= 74.5 + (in – 99/85) × 5
= 74.5 + 1/85 × 8
= 74.5 + 0.58
= 74.558
= 75
∴ Median No. 1 vehicles 75
(4) The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
No. of bulbs produced (Thousand) | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 | 90 – 100 |
No. of factories | 12 | 35 | 20 | 15 | 8 | 7 | 8 |
Solution:
Class
No. of bulbs |
Frequency (f) No. of factories | Cumulative frequency (cf) |
30 – 40 | 12 | 12 |
40 – 50 | 35 | 47 |
50 – 60 | 20 | 67 ->cf |
60 – 70 | 15 -> f | 82 |
70 – 80 | 8 | 90 |
80 – 90 | 7 | 97 |
90 – 100 | 8 | 105 = N |
Here, N/2 = 105/2 = 52.5, h = 70 – 60 = 10
f = 15, cf = 67, L = 60
∴ Median = L + [N/2 – cf/f] × h
= 60 + [52.5 – 67/15] × 10
= 60 – 9.67
= 50.33
∴ Median no. of bulbs = 50.33 × 1000
= 50330 bulbs
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 6 Statistics Practice Set 6.2
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