Chapter 6 - Statistics
Practice set 6.1
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.
Solution:
Mean (\(\overline {x}\))
= \(\large \frac {Σfixi}{Σfi}\)
= \(\large \frac {218}{50}\)
= 4.36
Ans: Mean of the time spent by students for studies is 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.
Solution:
Assumed Mean (A) = 550
\(\overline {d}\)
= \(\large \frac {Σfidi}{Σfi}\)
= \(\large \frac {–\,12000}{420}\)
= – 28.57
Mean (\(\overline {x}\))
= A + \(\overline {d}\)
= [550 + (– 28.57)]
= 550 – 28.57
= 521.43
Ans: Mean of the money collected is ₹ 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.
Solution:
Mean (\(\overline {x}\))
= \(\large \frac {Σfixi}{Σfi}\)
= \(\large \frac {141}{50}\)
= 2.82
Ans: Mean quantity of milk sold is 2.82 litres.
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.
Solution:
Assumed Mean (A) = 37.5
\(\overline {d}\)
= \(\large \frac {Σfidi}{Σfi}\)
= \(\large \frac {–\,175}{80}\)
= – 2.19
Mean (\(\overline {x}\))
= A + \(\overline {d}\)
= [37.5 + (– 2.19)]
= 37.5 – 2.19
= 35.31
∴ Mean of production Amount
= 35.31 × 1000
= ₹ 35,310
Ans: Mean of production is ₹ 35,310.
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.
Solution:
g = 500, Assumed Mean (A) = 1250
\(\overline {u}\)
= \(\large \frac {Σfiui}{Σfi}\)
= \(\large \frac {–\,63}{120}\)
= – 0.525
Mean (\(\overline {x}\))
= A + \(\overline {u}\) . g
= 1250 + 500 (– 0.525)
= 1250 – 262.5
= 987.50
Ans: Mean of the Fund collected is ₹ 987.50.
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.
Solution:
g = 1000, Assumed Mean (A) = 2500
\(\overline {u}\)
= \(\large \frac {Σfiui}{Σfi}\)
= \(\large \frac {85}{150}\)
= – 0.567
Mean (\(\overline {x}\))
= A + \(\overline {u}\) . g
= 2500 + 0.567 × 1000
= 2500 + 567
= 3067
Ans: Mean of salary is ₹ 3067.
Practice set 6.2
1. The following table shows the classification of the number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.
Solution:
Class width (h) = 2
Here,
Total frequency (N) = 1000.
∴ \(\large \frac {N}{2}\) = \(\large \frac {1000}{2}\) = 500
Cumulative frequency (less than type) which is just greater than 500 is 650.
∴ Corresponding class 10 – 12 is the median class.
∴ f = 500, c.f. = 150, L = 10 and h = 2
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 10 + \(\large [\frac {500\,–\,150}{500}]\) × 2
= 10 + \(\large [\frac {350}{500}]\) × 2
= 10 + \(\large [\frac {7}{10}]\) × 2
= 10 + 1.4
= 11.4
Ans: Median of no. of hours worked is 11.4
2. The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.
Solution:
Here,
Total frequency (N) = 250.
∴ \(\large \frac {N}{2}\) = \(\large \frac {250}{2}\) = 125
Cumulative frequency (less than type) which is just greater than 125 is 153.
∴ Corresponding class 150 – 200 is the median class.
∴ f = 90, c.f. = 63, L = 150 and h = 50
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 150 + \(\large [\frac {125\,–\,63}{90}]\) × 50
= 10 + \(\large [\frac {62}{90}]\) × 50
= 10 + 34.44
= 184.44
≈ 184
Ans: Median of no. of mangoes is 184 (approx).
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.
Solution:
Here,
Total frequency (N) = 200.
∴ \(\large \frac {N}{2}\) = \(\large \frac {200}{2}\) = 100
Cumulative frequency (less than type) which is just greater than 100 is 184.
∴ Corresponding class 74.5 – 79.5 is the median class.
∴ f = 85, c.f. = 99, L = 74.5 and h = 5
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 74.5 + \(\large [\frac {100\,–\,99}{85}]\) × 5
= 74.5 + \(\large [\frac {1}{85}]\) × 5
= 74.5 + 0.058
= 74.559
≈ 75 (approx)
Ans: Median of speed of the vehicles is 75 km/h (approx).
4. The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
Solution:
Class width (h) = 105
Here,
Total frequency (N) = 105.
∴ \(\large \frac {N}{2}\) = \(\large \frac {105}{2}\) = 52.5
Cumulative frequency (less than type) which is just greater than 52.5 is 67.
∴ Corresponding class 50 – 60 is the median class.
∴ f = 20, c. f. = 47, L = 50 and h = 10
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 50 + \(\large [\frac {52.5\,–\,47}{20}]\) × 10
= 50 + \(\large [\frac {5.5}{20}]\) × 10
= 50 + 2.75
= 52.75
Median production
= 52.75 × 1000
= 52750
Ans: Median of production of bulbs is 52750.
Practice set 6.3
1. The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
Solution:
f₁ = Maximum frequency = 80.
The corresponding class 4 – 5 is modal class.
f₀ = 70, f₂ = 60, L = 4 and h = 1
Mode
= L + \(\large [\frac {f₁\,–\,f₀}{2f₁\,–\,f₀\,–\,f₂}]\) × h
= 4 + \(\large [\frac {80\,–\,70}{2(80)\,–\,70\,–\,60}]\) × 1
= 4 + \(\large [\frac {10}{160\,–\,130}]\)
= 4 + \(\large [\frac {10}{30}]\)
= 4 + 0.33
= 4.33
Ans: Mode of weight of fat in milk is 4.33 %.
2. Electricity used by some families is shown in the following table. Find the mode for use of electricity.
Solution:
f₁ = Maximum frequency = 100.
The corresponding class 60 – 80 is modal class.
f₀ = 70, f₂ = 80, L = 60 and h = 20
Mode
= L + \(\large [\frac {f₁\,–\,f₀}{2f₁\,–\,f₀\,–\,f₂}]\) × h
= 60 + \(\large [\frac {100\,–\,70}{2(100)\,–\,70\,–\,80}]\) × 20
= 60 + \(\large [\frac {30}{200\,–\,150}]\) × 20
= 60 + \(\large [\frac {30}{50}]\) × 20
= 60 + 12
= 72
Ans: Mode of no. of units consumed is 72.
3. Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Solution:
f₁ = Maximum frequency = 35.
The corresponding class 9 – 11 is the modal class.
f₀ = 20, f₂ = 18, L = 9 and h = 2
Mode
= L + \(\large [\frac {f₁\,–\,f₀}{2f₁\,–\,f₀\,–\,f₂}]\) × h
= 9 + \(\large [\frac {35\,–\,20}{2(35)\,–\,20\,–\,18}]\) × 2
= 9 + \(\large [\frac {15}{70\,–\,38}]\) × 2
= 9 + \(\large [\frac {15}{32}]\) × 2
= 9 + \(\large [\frac {15}{16}]\)
= 9 + 0.94
= 9.94
Ans: Modal of quantity of milk consumed by hotels is 9.94 litres.
4. The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
Solution:
f₁ = Maximum frequency = 50.
The corresponding class 9.5 – 14.5 is the modal class.
f₀ = 32, f₂ = 36, L = 9.5 and h = 5
Mode
= L + \(\large [\frac {f₁\,–\,f₀}{2f₁\,–\,f₀\,–\,f₂}]\) × h
= 9.5 + \(\large [\frac {50\,–\,32}{2(50)\,–\,32\,–\,36}]\) × 5
= 9.5 + \(\large [\frac {18}{100\,–\,68}]\) × 5
= 9.5 + \(\large [\frac {18}{32}]\) × 5
= 9.5 + 2.81
= 9.5 + 0.94
= 12.31 years.
Ans: Mode of age of patients is is 12.31 years.
Practice set 6.4
1. Draw a histogram of the following data.
Solution:
2. The table below shows the yield of jowar per acre. Show the data by histogram.
Solution:
3. In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.
Solution:
4. Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
Solution:
Practice set 6.5
1. Observe the following frequency polygon and write the answers of the questions below it.
(1) Which class has the maximum number of students?
Ans: Maximum number of students are present in the group 60 – 70.
(2) Write the classes having zero frequency.
Ans: Zero students are present in 20 – 30 and 90 – 100 groups.
(3) What is the class-mark of the class, having a frequency of 50 students?
Ans: 50 students are present in the group 50 – 60 whose class mark is 55.
(4) Write the lower and upper class limits of the class whose class mark is 85.
Ans: 85 is the class mark of the group 80 – 90.
(5) How many students are in the class 80-90?
Ans: There are 15 students in the group 80 – 90.
2. Show the following data by a frequency polygon.
Solution:
3. The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.
Solution:
Practice set 6.6
1. Draw a histogram of the following data.
Solution:
2. The table below shows the yield of jowar per acre. Show the data by histogram.
Solution:
3. In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.
Solution:
4. Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
Solution:
Problem Set 6
1. Find the correct answer from the alternatives given.
(1) The persons of O- blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O- blood group?
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Ans: Option (D) : 144⁰
(2) Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure ₹ 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction ?
(A) 2,16,000
(B) 3,60,000
(C) 4,50,000
(D) 7,50,000
Ans: Option (A) : 2,16,000
(3) Cumulative frequencies in a grouped frequency table are useful to find _____.
(A) Mean
(B) Median
(C) Mode
(D) All of these
Ans: Option (B) : Median
(4) The formula to find mean from a grouped frequency table is X̅ = A + \(\large \frac {Σfiui}{Σfi}\) × h g. In the formula ui = ____.
(A) \(\large \frac {xi\,+\,A}{g}\)
(B) \((xi – a)\)
(C) \(\large \frac {xi\,–\,A}{g}\)
(D) \(\large \frac {A\,–\,xi}{g}\)
Ans: Option (C) : \(\large \frac {xi\,–\,A}{g}\)
(5)
The median of the distances covered per litre shown in the above data is in the group _____
(A) 12-14
(B) 14-16
(C) 16-18
(D) 18-20
Ans: Option (C) : 16-18
(6)
The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4-6 are _____
(A) (4, 8)
(B) (3, 5)
(C) (5, 8)
(D) (8, 4)
Ans: Option (C) : (5, 8)
2. The following table shows the income of farmers in a grape season. Find the mean of their income.
Solution:
Mean (\(\overline {x}\))
= \(\large \frac {Σfixi}{Σfi}\)
= \(\large \frac {4410}{84}\)
= 52.5
Mean of production
= 52.5 × 1000
= 52,500
Ans: Mean of production in rupees is ₹ 52,500.
3. The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.
Solution:
Assumed Mean (A) = 65
\(\overline {d}\)
= \(\large \frac {Σfidi}{Σfi}\)
= \(\large \frac {40}{100}\)
= 0.4
Mean (\(\overline {x}\))
= A + \(\overline {d}\)
= 65 + 0.4
= 65.4
Total amount
= 65.4 × 1000
= 65,400
Ans: Mean of loan given by bank is ₹ 65,400.
4. The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
Solution:
Assumed Mean (A) = 75
\(\overline {d}\)
= \(\large \frac {Σfidi}{Σfi}\)
= \(\large \frac {–\,130}{50}\)
= 2.6
Mean (\(\overline {x}\))
= A + \(\overline {d}\)
= 75 + (– 2.6)
= 72.4
Total amount
= 72.4 × 1000
= ₹ 72,400
Ans: Mean of money given is ₹ 72,400.
5. The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.
Solution:
g = 2000, Assumed Mean (A) = 3000
\(\overline {u}\)
= \(\large \frac {Σfiui}{Σfi}\)
= \(\large \frac {75}{120}\)
= 0.625
Mean (\(\overline {x}\))
= A + \(\overline {u}\) g
= 3000 + 2000 × 0.625
= 3000 + 1250
= 4250
Ans: Mean of weekly salary is ₹ 4250.
6. The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.
Solution:
Here, total frequency = Σfi = N = 250 and h = 10
∴ \(\large \frac {N}{2}\) = \(\large \frac {250}{2}\) = 125
Cumulative frequency (less than type) which is just greater than 125 is 180.
∴ Corresponding class 220 – 230 is the median class.
∴ f = 80, c. f. = 100, L = 220
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 220 + \(\large [\frac {125\,–\,100}{80}]\) × 10
= 220 + \(\large [\frac {25}{80}]\) × 10
= 220 + 3.125
= 223.13
Ans: Median distance covered is 223.13 km.
7. The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.
Solution:
Here, total frequency = Σfi = N = 400 and h = 20
∴ \(\large \frac {N}{2}\) = \(\large \frac {400}{2}\) = 200
Cumulative frequency (less than type) which is just greater than 200 is 240.
∴ Corresponding class 20 – 40 is the median class.
∴ f = 100, c. f. = 140, L = 20
Median
= L + \(\large [\frac {\frac {N}{2}\,–\,c.f.}{f}]\) × h
= 20 + \(\large [\frac {200\,–\,140}{100}]\) × 20
= 20 + \(\large [\frac {60}{100}]\) × 20
= 20 + 12
= 32
Ans: Median of Amount is ₹ 32.
8. The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.
Solution:
Here, the maximum frequency f₁ = 60.
∴ The corresponding class 250 – 500 is the Modal class.
f₁ = 60, f₀ = 10, f₂ = 25, L = 250 and h = 250
Mode
= L + \(\large [\frac {f₁\,–\,f₀}{2f₁\,–\,f₀\,–\,f₂}]\) × h
= 250 + \(\large [\frac {60\,–\,10}{2(60)\,–\,10\,–\,25}]\) × 250
= 250 + \(\large [\frac {50}{120\,–\,35}]\) × 250
= 250 + \(\large [\frac {50}{85}]\) × 250
= 250 + 147.06
= 397.06
Ans: Mode of weight of sweets is 397.06 gm.
9. Draw a histogram for the following frequency distribution.
Solution:
10. In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.
Solution:
11. The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.
Solution:
12. Draw a frequency polygon for the following grouped frequency distribution table.
Solution:
13. The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.
Solution:
14. Observe the adjacent pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am
(1) Find the central angle for each type of vehicle.
Ans:
(2) If the number of two-wheelers is 1200, find the number of all vehicles.
Ans:
Total no. of two wheelers = 1200
Central angle of two wheeler = \(\large \frac {Number\,of\,two\,wheelers}{Total\,number\,of\,vehicles}\) × 360
∴ 144 = \(\large \frac {1200}{Total\,number\,of\,vehicles}\) × 360
∴ Total number of vehicles = \(\large \frac {1200\,×\,360}{144}\)
∴ Total number of vehicles = 3000
15. The following table shows causes of noise pollution. Show it by a pie diagram.
Solution:
16. A survey of students was made to know which game they like. The data obtained in the survey is presented in the adjacent pie diagram. If the total number of students are 1000,
(1) How many students like cricket?
Ans:
(i) Total no. of students = 1000
Central angle for cricket = 81°
∴ No. of students interested in cricket
= \(\large \frac {Central\,angle}{360}\) × Total no. of students
= \(\large \frac {81}{360}\) × 1000
= 225
∴ No. of students interested in cricket = 225
(2) How many students like football?
Ans:
Central angle for Football = 63°
∴ No. of students interested in football
= \(\large \frac {Central\,angle}{360}\) × Total no. of students
= \(\large \frac {63}{360}\) × 1000
= 175
∴ No. of students interested in football = 175
(3) How many students prefer other games?
Ans:
Central angle for other sports = 72°
∴ No. of students interested in other sports
= \(\large \frac {Central\,angle}{360}\) × Total no. of students
= \(\large \frac {72}{360}\) × 1000
= 200
∴ No. of students interested in other sports = 200.
17. Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of haemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Solution:
18. On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.
Solution:
