Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6 Practice Set 6.1 – 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 | 6.1 |
Practice Set – 6.1
Statistics
(1) The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Time (hrs.) | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 |
No. of students | 7 | 18 | 12 | 10 | 3 |
Solution:
Class (Time in hrs.) | Class mark xi | Frequency fi | Class mark x frequency (xifi) |
0 – 2 | 1 | 7 | 7 |
2 – 4 | 3 | 18 | 54 |
4 – 6 | 5 | 12 | 60 |
6 – 8 | 7 | 10 | 70 |
8 – 10 | 9 | 3 | 27 |
Σfi = 50 | Σxifi = 218 |
∴ Mean = Σxifi/Σfi = 218/50 = 4.36
∴ mean time spent by students for their studies
= 4.36 hrs.
(2) In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.
Toll (Rupees) | 300 – 400 | 400 – 500 | 500 – 600 | 600 – 700 | 700 – 800 |
No. of vehicles | 80 | 110 | 120 | 70 | 40 |
Solution:
Let, assumed A = 550
Class Toll | Class mark xi | di = xi – A
= xi – 550 |
Frequency fi | fidi |
300 – 400 | 350 | – 200 | 80 | -1600 |
400 – 500 | 450 | – 100 | 110 | -11000 |
500 – 600 | 550 -> A | 0 | 120 | 0 |
600 – 700 | 650 | 100 | 70 | 7000 |
700 – 800 | 750 | 200 | 40 | 8000 |
Σfi = 420 | Σfidi = -12000 |
∴ d = Σfidi/Σfi = -12000/420 = – 28.57
∴ Mean = 4 + d = 550 – 26.57 = 521.43
(3) A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
Milk Sold (Litre) | 1 – 2 | 2 – 3 | 3 – 4 | 4 – 5 | 5 – 6 |
No. of Customers | 17 | 13 | 10 | 7 | 3 |
Solution:
Class milk sold (Litre) | Class Mark xi | Frequency fi | Class Mark x frequency xifi |
1 – 2 | 1.5 | 17 | 25.5 |
2 – 3 | 2.5 | 13 | 32.5 |
3 – 4 | 3.5 | 10 | 35 |
4 – 5 | 4.5 | 7 | 31.5 |
5 – 6 | 5.5 | 3 | 16.5 |
Σfi = 50 | Σfixi = 141 |
∴ Mean = Σfixi/Σfi = 141/50 = 2.82 litre.
(4) A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.
Production (Thousand rupees) | 25 – 30 | 30 – 40 | 35 – 40 | 40 – 45 | 45 – 50 |
No. of farm owners | 20 | 25 | 15 | 10 | 10 |
Solution:
Let, Assumed mean A = 37.5
Class (1000×7) production | Class mark xi | di = xi – A
= xi – 37.5 |
Frequency fi | fidi |
25 – 30 | 27.5 | -10 | 20 | -200 |
30 – 35 | 32.5 | -5 | 25 | -125 |
35 – 40 | 37.5 -> A | 0 | 15 | 0 |
40 – 45 | 42.5 | 5 | 10 | 50 |
45 – 50 | 47.5 | 10 | 10 | 100 |
Σfi = 80 | Σfidi = -175 |
∴ d = Σfidi/Σfi = -175/80 = – 2.18
∴ Mean, x = A + d = 37.5 – 2.18 = 35.32
∴ Mean oranges produced = 35.32 × 1000
= 35320
(5) A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by ‘step deviation’ method.
Fund (Rupees) | 0 – 500 | 500 – 1000 | 1000 – 1500 | 1500 – 2000 | 2000 – 2500 |
No. of workers | 35 | 28 | 32 | 15 | 16 |
Solution:
Let, the assumed mean be, A = 1250, g = 5000
Class Fund (f) | Class Mark xi | di = xi – A
= xi – 1250 |
Ui = di/g | Frequency fi | fiui |
0 – 500 | 250 | -1000 | -2 | 35 | -70 |
500 – 1000 | 750 | -500 | -1 | 28 | -28 |
1000 – 1500 | 1250 = A | 0 | 0 | 32 | 0 |
1500 – 2000 | 1750 | 500 | 1 | 15 | 15 |
2000 – 2500 | 2250 | 1000 | 2 | 10 | 20 |
Σfi = 120 | Σfiui = -63 |
∴ u = Σfiui/Σfi = -63/120
= – 0.525
u×g = -0.525 × 500
= – 262.5
∴ Mean
x = A + g 1/u
= 12517 – 262.5 = 987.5
(6) The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.
Weekly wages (Rupees) | 1000 – 2000 | 2000 – 4000 | 3000 – 4000 | 4000 – 5000 |
No. of workers | 25 | 45 | 50 | 30 |
Solution:
Let, the assumed mean, A = 2500.
Class weekly wages (₹) | Class Mark xi | di = xi – A
= xi – 2500 |
Ui = di/g | Frequency fi | fiui |
1000 – 2000 | 1500 | – 1000 | – 0.4 | 25 | – 10 |
2000 – 3000 | 2500 -> A | 0 | 0 | 45 | 0 |
3000 – 4000 | 35000 | 1000 | 0.4 | 50 | 20 |
4000 – 5000 | 4500 | 2000 | 0.8 | 30 | 24 |
Σfi = 150 | Σfiui= 34 |
∴ u = Σfiui/Σfi
= 34/150
= 0.226
ug = 0.226×2500
= 566.66
∴ Mean x = A + ug
= 2500 + 566.66
= 3066.66
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 6 Statistics Practice Set 6.1
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