ALGEBRA
Adsul Notes
5.
Statistics - I
Introduction : :
A) B)
Statistics is something that surrounds us every day - we are constantly bombarded with statistics, in the form of polls, tests, ratings etc. It is a branch of Mathematics which deals with the collection, presentation and analysis of numerical data and drawing the conclusions on the basis of the same. Statistics is used to study the problems in Biology, Psychology, Economics, Education, Sociology, Trade etc. It is widely used in industries to control the quality of products. In our Xth standard syllabus, we have only two things to learn : Measure of Central Tendency of Grouped Data. Pie-Diagram
ot es
o
MEASURE OF CENTRAL TENDENCY OF GROUPED DATA : The most basic concept in statistics is the idea of an average (arithmetic mean). An average is a single number which represents the idea of a typical value. There are three different numbers which can represent the idea of an average value. The three values are the mean, the median and the mode.
o
MEAN :
N
A)
• •
To find Mean by Direct Method
Ad s
TYPE - 1
ul
Given a set of values, the mean is sum of all the values, divided by the total number of values. Mean summarizes the properties of entire data.
Things To Remember :
Classes need not be continuous Classmark (xi) has to be found for every class of the given data. There are 3 different methods to find Mean for the given data.
o
Method 1 : Direct Method
In this method, table will have 4 columns. Step 1 : In 1st column, write the given Classes Step 2 : In 2nd column, write the Classmark/Mid-values (x i) for each of the classes. Find the class mark of the first class the remaining class marks could be found by adding class width (h) to the previous class mark. Step 3 : In 3rd column, write the Frequency (¦i)of all the classes. Find the sum of all frequencies i.e. S¦i. Step 4 : In 4th column, find the product of Classmark/Mid values (xi) and frequency (¦i) i.e. ¦ixi, for each of the classes. Find the sum of all (¦ixi) values i.e. S¦ixi f i xi Step 5 : CALCULATE, MEAN ( x ) = f i
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1
ALGEBRA
Adsul Notes
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 123) :
Sol.
Below is given distribution of money (in Rs.) collected by students for flood relief fund. Money (in Rs.)
0 - 10
10 - 20
20 - 30
No. of students
5
7
5
30 - 40 40 - 50 2
6
Find mean of money (in Rs.) collected by a student by ‘Direct Method’. (3 marks) Class width (h) = 10 Money (in Rs.)
Class Mark (x i) 5 15 25 35 45
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total Mean = Mean =
f ixi fi 595 25 Rs. 23.8
25 105 125 70 270 595
N
Mean =
f ix i
No. of students (f i) 5 7 5 2 6 25
ot es
1.
Mean of money collected is Rs. 23.8.
Following table gives age distribution of people suffering from ‘Asthama due to air pollution in certain city.
Age in years
7-11 11-15 15-19 19-23 23-27 27-31 31-35 5
9
13
Ad s
No. of people
Sol.
ul
2.
21
16
15
35-39
12
9
Find mean age of person suffering from ‘Asthama’ by ‘Direct Method’. (3 marks) Class width (h) = 4 Age in years
Class Mark (x i)
7 - 11 11 - 15 15 - 19 19 - 23 23 - 27 27 - 31 31 - 35 35 - 39
9 13 17 21 25 29 33 37
Total Mean = Mean = Mean =
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No. of people (f i)
f ix i
5 9 13 21 16 15 12 9
45 117 221 441 400 435 396 333
100
2388
f ixi fi 2388 100 23.88 years.
2
ALGEBRA
Adsul Notes
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 174) : 3.
A study related to the time (in months) taken to settle a dispute in a lower court resulted the following data. Time (in months) 0 – 2 2–4 4–6 6–8 8 – 10 10 – 12 No. of disputes
90
120
75
70
50
Mean =
f ixi fi
Mean =
2590 420
Mean = Mean =
ot es
Find mean time taken to settle a dispute in a lower court. (3 marks) Class width (h) = 2 Time in months Class Mark No. of disputes f ix i (x i) (f i) 0-2 1 15 15 2-4 3 90 270 4-6 5 120 600 6-8 7 75 525 8 - 10 9 70 630 10 - 12 11 50 550 Total 420 2590
37 16 6.17 (approximately)
N
Sol.
15
Record of no. of days of medical leave enjoyed by 30 employees within a year is given below. No. of days 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Ad s
6.
ul
Mean of time to settle a dispute is 6.17 months.
No. of employees
Sol.
5
7
11
4
3
Find mean number of days of medical leave enjoyed by an employee in a year. (3 marks) Class width (h) = 10 No. of days Class Mark No. of employees f ix i (x i) (f i) 0 - 10 5 5 25 10 - 20 15 7 105 20 - 30 25 11 275 30 - 40 35 4 140 40 - 50 45 3 135 Total 30 680 f ixi Mean = fi Mean = Mean =
680 30 22.67 (approximately)
Mean of medical leave is 22.67 days.
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3
ALGEBRA
To find Mean by Assumed Mean Method or Shift of origin method.
TYPE -2
o
Adsul Notes
Method 2 : Assumed Mean Method or shift of origin method :
ot es
In this method, table will have 5 columns. Step 1 : In 1st column, write the given classes Step 2 : In 2nd column, write the Classmark/Mid values (x i) for each of the classess. Step 3 : In 3rd column, write the Frequency (fi) of all the classes. Also find the sum of all frequencies i.e. S fi Step 4 : Select any value in the Classmark (xi) column as Assumed mean. It is denoted by A. Generally, middle value of the column is selected. Step 5 : In 4th column, find deviations using formula di = xi – A for first class and the remaining di could be found by adding class width (h) to the previous di. Step 6 : In 5th column, find the product of deviations (di) and frequency (fi) i.e. fidi, for each of the classes. Find the sum of all (fidi) values i.e. Sfidi Σ f i di Step 7 : Calculate ( d ) = Σ f i Step 8 : Calculate Mean ( x ) = A + d
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 123) :
The measurements (in mm) of the diameters of the head of screws are given below : Diameter (in mm) No. of screws
33 - 35 36 - 38 39 - 41 42 - 44 45 - 47 10
19
23
21
27
ul
Calculate mean diameter of head of a screw of ‘Assumed Mean Method’. (3 marks) Class width (h) = 3, Assumed mean (A) = 40
Ad s
Sol.
N
3.
Diameter
Class Mark
(in mm)
33 36 39 42 45
-
di = x i – A
No. of screws
(x i)
35 38 41 44 47
34 37 40 A 43 46
(f i) –6 –3 0 3 6
Total =
d
=
d
=
108 100 1.08
= = =
Ad 40 + 1.08 41.08
x
10 19 23 21 27
– 60 – 57 0 63 162
100
108
f i di fi
d
Mean
f id i
Mean of diameter of screws is 41.08 mm.
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4
ALGEBRA
Adsul Notes
4.
Below is given frequency distribution of marks (out of 100) obtained by the students. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
5.
Ad s
ul
N
ot es
No. of 3 5 7 10 12 15 12 6 2 8 students Calculate mean marks scored by a student by ‘Assumed mean method’. (3 marks) Sol. Class width (h) = 10, Assumed Mean (A) = 55 Class Mark No. of students Marks di = x i – A f id i (x i) (f i) 0 - 10 5 – 50 3 – 150 10 - 20 15 – 40 5 – 200 20 - 30 25 – 30 7 – 210 30 - 40 35 – 20 10 – 200 40 - 50 45 – 10 12 – 120 0 15 0 50 - 60 55 A 60 - 70 65 10 12 120 70 - 80 75 20 6 120 80 - 90 85 30 2 60 90 - 100 95 40 8 320 Total 80 – 260 f id i = d fi – 260 d = 80 d = – 3.25 Mean x = A d = 55 + (– 3.25) = 55 – 3.25 = 51.75 Mean of marks obtained 51.75 marks. Following table gives frequency distribution of milk (in litres) given per week by 50 cows. (3 marks) Milk (in 24-30 30-36 36-42 42-48 48-54 54-60 60-66 66-72 72-78 78-84 84-90 litres) No. of 1 3 8 5 5 5 8 4 6 2 3 cows Find average (mean) amount of milk given by a cow by ‘shift of origin method’. Sol. Class width (h) = 6, Assumed mean (A) = 57 Milk Class Mark No. of cows di = x i – A f id i (in litres) (x i) (f i) 24 - 30 27 – 30 1 – 30 30 - 36 33 – 24 3 – 72 36 - 42 39 – 18 8 – 144 42 - 48 45 – 12 5 – 60 48 - 54 51 –6 5 – 30 0 5 0 54 - 60 57 A 60 - 66 63 6 8 48 66 - 72 69 12 4 48 72 - 78 75 18 6 108 78 - 84 81 24 2 48 84 - 90 87 30 3 90 Total 50 6
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5
ALGEBRA
d
=
f id i fi
d
=
6 50
d
=
0.12
=
Ad
= =
57 + 0.12 57.12
Mean
x
Adsul Notes
Mean of distribution of milk is 57.12 litres. Solve problem 6 by ‘Assumed Mean Method’. Following table gives frequency distribution of trees planted by different housing societies in a particular locality. (3 marks) No. of trees 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 No. of societies Sol.
2
7
9
8
6
ot es
9.
ul
N
Class width (h) = 5, Assumed mean (A) = 22.5 Class Mark No. of societies No. of trees di = x i – A (x i) (f i) 10 - 15 12.5 – 10 2 15 - 20 17.5 –5 7 20 - 25 22.5 A 0 9 25 - 30 27.5 5 8 30 - 35 32.5 10 6 35 - 40 37.5 15 4 Total 36 =
f id i fi
d
=
105 36
d
=
2.92
=
Ad
= =
22.5 + 2.92 25.42
f id i – 20 – 35 0 40 60 60 105
Ad s
d
4
Mean
x
Mean of trees planted by the societies is 25.42 trees.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 175) : 17.
For the data given in problem 10, find mean driving speed of a college student. Below is given frequency distribution of driving speed (in kms/hour) of a vehicle of 400 college students. (3 marks) Speed (in kms/hr)
20 – 30
30 – 40
40 – 50
No. of students
6
80
156
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50 – 60 60 – 70 98
60
6
ALGEBRA
Adsul Notes
Sol.
Class width (h) = 10, Assumed mean (A) = 45 Speed Class Mark (in km/hr.) (x i) 25 35 45 A 55 65
d
=
f id i fi
d
=
1260 400
d
=
3.15
=
Ad
= =
45 + 3.15 48.15
Mean
x
No. of students (f i)
– 20 – 10 0 10 20
6 80 156 98 60 400
f id i – 120 – 800 0 980 1200 1260
ot es
20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Total
di = x i – A
Mean of driving speed of vehicle is 48.15 km/hr.
Frequency distribution of daily commission received by 100 salesmen is given below.
Daily commission (in Rs.)
Sol.
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 20
45
ul
No. of salesman
N
15.
22
9
Find mean daily commission received by a salesman. Class width (h) = 20, Assumed mean (A) = 150
Ad s
Daily Class Mark commission (x i) (in Rs.) 100 120 140 160 180
-
120 140 160 180 200
110 130 150 A 170 190
Total d
=
d
=
d
=
f id i fi –1360 100 – 13.6
=
Ad
= = =
150 + (– 13.6) 150 – 13.6 136.4
Mean
x
di = x i – A
– 40 – 20 0 20 40
No. of salesman (f i)
4 (3 marks) f id i
20 45 22 9 4
– 800 – 900 0 180 160
100
– 1360
Mean of daily commission of salesman is Rs. 136.4.
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7
ALGEBRA
Adsul Notes
12.
Frequency distribution of distance travelled in kms. Per liter of a petrol by different mopeds is given below. Distance travelled 62 – 65 65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 (in km) No. of 5 8 12 28 35 10 2 mopeds Find mean distance travelled per litre of petrol by a moped. Class width (h) = 3, Assumed mean (A) = 72.5 Distance (in km) 62 65 68 71 74 77 80
-
Class Mark (x i)
65 68 71 74 77 80 83
63.5 66.5 69.5 72.5 A 75.5 78.5 81.5
Total
f i di fi
=
d
=
d
=
54 100 0.54
=
Ad
x
–9 –6 –3 0 3 6 9
f id i
5 8 12 28 35 10 2
– 45 – 48 – 36 0 105 60 18
100
54
ul
Mean
No. of mopeds (fi)
(3 marks)
N
d
di = x i – A
ot es
Sol.
Ad s
= 72.5 + 0.54 = 73.04 Mean of distance covered by moped is 73.04 km/lit.
TYPE - 3
To find Mean by Step Deviation method or Shift of origin and scale method
Things To Remember :
o Step Step Step Step
Method 1 : Step Deviation Method or shift of origin and scale method Width of class (h) = Difference between any two consecutive upper limits or lower limits. In this method, table will have 6 columns. 1 : In 1st column, write the given classes. 2 : In 2 nd column, write the Classmark/Mid-values (x i) for each of the classes. 3 : In 3rd column, write the values of the di, where di = xi – A x i – A di = 4 : In 4th column, write the value of the ui where ui = h h
Step 5 :
In 5th column, write the frequencies fi and find its sum.
Step 6 : Find mean of ui using the formula u = Step 7 : Calculate, Mean ( x ) = A + h u
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Σ fi ui Σ fi
8
ALGEBRA
Adsul Notes
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 124) : 6.
Sol.
Following table gives frequency distribution of trees planted by different housing societies in a particular locality. No. of trees 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 No. of societies 2 7 9 8 6 4 Find the number of trees planted by housing society by using ‘step deviation method’. (5 marks) Class width (h) = 5, Assumed mean (A) = 22.5 No. of trees Class Mark (x i) -
15 20 25 30 35 40
12.5 17.5 22.5 A 27.5 32.5 37.5
– 10 –5 0 5 10 15
Total
–2 –1 0 1 2 3
2 7 9 8 6 4
–4 –7 0 8 12 12
36
=
u
=
u
=
f iui fi 21 36 0.583
= = = =
A hu 22.5 + 5 (0.583) 22.5 + 2.92 25.42
ul
x
21
N
u
Mean
di No. of societies f iu i h (f i)
ot es
10 15 20 25 30 35
di = xi – A ui =
7.
Ad s
Mean of trees planted by societies 25.42 trees. Solve problem 2 by ‘Step Deviation Method’. Following table gives age distribution of people suffering from ‘Asthama due to air pollution in certain city. (5 marks)
Age in years
7-11 11-15 15-19 19-23 23-27 27-31 31-35
No. of people Sol.
5
9
13
21
16
15
35-39
12
9
Class width (h) = 4, Assumed mean (A) = 25 No. of trees 7 - 11 11 - 15 15 - 19 19 - 23 23 - 27 27 - 31 31 - 35 35 - 39 Total
" Learn from the Best "
Class Mark (x i) 9 13 17 21 25 A 29 33 37
di = xi – A ui = – 16 – 12 –8 –4 0 4 8 12
– – – –
di h 4 3 2 1 0 1 2 3
No. of people (f i) 5 9 13 21 16 15 12 9 100
f iu i – – – –
20 27 26 21 0 15 24 27
– 28
9
ALGEBRA
Adsul Notes
f iui fi
u
=
u
=
u
=
– 28 100 – 0.28
= = = =
A hu 25 + 4 (– 0.28) 25 – 1.12 23.88 years
Mean
x
Mean of age is 23.88 years. Solve problem 3 by ‘Step Deviation Method’. The measurements (in mm) of the diameters of the head of screws are given below : Diameter (in mm) 33 - 35 36 - 38 39 - 41 42 - 44 45 - 47 No. of screws Sol.
10
19
23
21
ot es
8.
27 (5 marks)
Class width (h) = 3, Assumed mean (A) = 40 Class Mark
(in mm)
(x i)
34 37 40 A 43 46
–6 –3 0 3 6
di No. of screws h (f i)
–2 –1 0 1 2
10 19 23 21 27 100
f iu i – 20 – 19 0 21 54 36
Ad s
ul
33 - 35 36 - 38 39 - 41 42 - 44 45 - 47 Total
di = xi – A ui =
N
Diameter
f iui fi
u
=
u
=
u
=
36 100 0.36
= = = =
A hu 40 + 3 (0.36) 40 + 1.08 41.08
Mean
x
Mean of diameter of the screws is 41.08 mm. 10.
Solve problem 4 by ‘Step Deviation Method’. Below is given frequency distribution of marks (out of 100) obtained by the students. (5 marks)
Marks No. of students Sol.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 3
5
7
10
12
15
12
6
2
8
Class width (h) = 10, Assumed mean (A) = 45
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10
ALGEBRA
Adsul Notes Class Mark
Class mark
(x i)
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100
di No. of students h (f i)
di = xi – A ui =
5 15 25 35 45 A 55 65 75 85 95
– – – –
40 30 20 10 0 10 20 30 40 50
– – – –
4 3 2 1 0 1 2 3 4 5
3 5 7 10 12 15 12 6 2 8
Total
– – – –
80
f iui fi
=
u
=
u
=
54 80 0.675
= = = =
A hu 45 + 10 (6.25) 45 + 6.75 51.75
54
N
x
12 15 14 10 0 15 24 18 8 40
ot es
u
Mean
f iu i
ul
Mean of marks obtained by students is 51.75 marks.
Ad s
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 175) : 11. Following table shows frequency distribution of duration (in seconds) of advertisements on T.V. Duration (in sec.) 25 – 30 30 – 35 35 – 40 40 – 45 occupied No. of advertisements
Sol.
10
32
15
9
45 – 50
50 – 65
7
2
Obtain mean duration of advertisement on T.V. by shift of origin and scale method. (5 marks) Class width (h) = 5, Assumed mean (A) = 37.5 Classes (Duration in sec) 25 30 35 40 45 50
-
30 35 40 45 50 55
Total
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Class Mark 27.5 32.5 37.5 A 42.5 47.5 52.5
di = xi – A ui = – 10 –5 0 5 10 15
di h
–2 –1 0 1 2 3
No. of advertisements
f iu i
10 32 15 9 7 2
– 20 – 32 0 9 14 6
75
– 23
11
ALGEBRA
Adsul Notes
f iui fi
u
=
u
=
u
=
– 23 75 – 0.306
=
A hu
= = =
37.5 + 5 (– 0.306) 37.5 – 1.533 35.97
Mean
x
Mean of duration of advertisement is 35.97 seconds. 9.
Frequency distribution of duration of 500 telephone calls received at a telephone exchange on a certain day is given below :
No. of calls
12
61
153
190
57
105-119 120-134
10
9
Find mean duration of telephone call received at the telephone exchange. (3 marks) Class width (h) = 15, Assumed mean (A) = 52
N
Sol.
8
75-89 90-104
ot es
Duration of call 15-29 30-44 45-59 60-74 (in sec.)
call in sec.
(x i)
22 37 52 A 67 82 97 112 127
Ad s
15 - 29 30 - 44 45 - 59 60 - 74 75 - 89 90 - 104 105 - 119 120 - 134
di = xi – A ui =
ul
Duration of Class Mark
– 30 – 15 0 15 30 45 60 75
Total u
=
u
=
u
=
Mean
x
= = = =
di h
–2 –1 0 1 2 3 4 5
No. of calls
f iu i
(f i) 8 12 61 153 190 57 10 9
– 16 – 12 0 153 380 171 40 45
500
761
f i ui fi 761 500 1.522 A hu 52 + 15 (1.522) 52 + 22.83 74.83
Mean of call duration is 74.83 second.
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12
ALGEBRA 8.
Adsul Notes Below is given frequency distribution of dividend in percentage declared by 120 companies.
Dividened 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 (in %) No. of companies
15
28
42
15
12
3
Obtain mean dividend declared by a company by step deviation method. (3 marks) Class width (h) = 10, Assumed mean (A) = 44.5 Dividend
Class Mark
10 20 30 40 50 60 70
-
(xi)
19 29 39 49 59 69 79
14.5 24.5 34.5 44.5 A 54.5 64.5 74.5
Total
u
=
f i ui fi
–3 –2 –1 0 1 2 3
f iu i
5 15 28 42 15 12 3
– 15 – 30 – 28 0 15 24 9
120
– 25
ul
=
No. of companies (f i)
– 25 120 – 0.208
Ad s
u
– 30 – 20 – 10 0 10 20 30
di h
ot es
(in %)
di = xi – A ui =
N
Sol.
5
70 – 79
u
Mean
x
= =
A hu 44.5 + 10 (– 0.208)
=
44.5 – 2.08
=
42.42
=
Mean of dividend is 42.42%. 13.
Number of calories (in’ 00) consumed daily by a sample of 15 years old boys are given below.
Calories No. of boys
1000 – 1500
1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 4500
5
13
16
18
27
10
4
Obtain mean calories consumed daily by a boy by step deviation method. (5 marks) Sol.
Class width (h) = 500, Assumed mean (A) = 2750
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13
ALGEBRA
Calories
Class Mark
di = xi – A ui =
(xi) 1000 1500 2000 2500 3000 3500 4000
-
1500 2000 2500 3000 3500 4000 4500
1250 1750 2250 2750 A 3250 3750 4250
– 1500 – 1000 – 500 0 500 1000 1500
di No. of boys h (f i)
–3 –2 –1 0 1 2 3
Total
5 13 16 18 27 10 4
– 15 – 26 – 16 0 27 20 12
93
2
u
=
u
=
2 93 0.0215
=
A hu
= = =
2750 + 500 (0.0215) 2750 + 10.75 2760.75
N
ot es
=
x
f iu i
f i ui fi
u
Mean
Adsul Notes
TYPE -4 Median :
To find Median for the given data.
Ad s
o
ul
Mean of calories consumed by boy is 2760.75 calories.
Median is a measure of the ’Middle’ value of the given data. Median is often a better representative of a typical member of a group. If we take all values in a list and arrange them in increasing order the number at the centre will be the median. Median gives you the central member of the group. Median is a number which divides data into two equal parts.
o Step Step Step Step
Steps to calculate to Median 1 2 3 4
: : : :
Make classes continuous if they are not In next column, write the frequencies In next column, calculate the ‘Cumulative frequency less than type’. Determine the median class, frequency of median class (f) and cumulative frequency of the class preceeding the median class (c.f.)
N h Step 5 : Calculate Median = L + 2 – c. f . f L = Lower limit of median class. N = Total frequency. c.f. = Cumulative frequency of class preceeding the median class. f = Frequency of median class. h = Width of the class.
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14
ALGEBRA
o
Adsul Notes
Steps to determine Median class
N , where N = Total Frequency. 2 Step 2 : Find the cumulative freuqency less than type which is just greater Step 1 : Find the value of
than or equal to
N the corresponding class is a median class. 2
Things To Remember : For Median,
•
Classes has to be continuous
•
Cumulative frequency less than type has to be found for every sum
EXERCISE - 5.2 (TEXT BOOK PAGE NO. 130) : Following is the distribution of the size of certain farms from a taluka (tehasil) :
ot es
1.
Size of farm 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 (in acres) No. of farms
7
12
Size of form (in acres) 5 - 15 15 - 25 25 - 35 45 - 55 55 - 65 65 - 75 Total
31
7 19 36
25 f
61
31 5 3
5
3 (5 marks)
Cumulative frequency less than type
7 12 17
Ad s
35 - 45
Frequency (fi) (No. of farms)
ul
Sol.
25
N
Find median size of a farm.
17
c.f.
92 97 100
100 N
Here total frequency = fi = N = 100
N 100 = = 50 2 2 Cumulative frequency (less than type) which is just greater than 50 is 61. Therefore corresponding class 35 - 45 is median class. L = 35, N = 100, c.f. = 36, f = 25, h = 10 Median
N h = L – c.f. 2 f
100 10 – 36 = 35 25 2 = 35 + (50 – 36)
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10 25
15
ALGEBRA
= 35 + (14)
Adsul Notes
10 25
140 25 = 35 + 5.6 = 40.6 = 35 +
Median of size of farm is 40.6 acres. 2.
Below is given distribution of profit in Rs. per day of a shop in certain town : Profit (in Rs.) 500 - 1000 - 1500 - 2000 - 2500 - 3000 - 3500 No. of shops
900
1400
1900
2400
2900
3400
3900
8
18
27
21
20
18
8 (5 marks)
Calculate median profit of a shop.
500 - 900 1000 - 1400 1500 - 1900 2000 - 2400 2500 - 2900 3000 - 3400 3500 - 3900
450 - 950 950 - 1450 1450 - 1950 1950 - 2450 2450 - 2950 2950 - 3450 3450 - 3950
8 18 27 21 20 18 8
f
120
8 26 53 74 94 112 120
c.f.
N
ul
Total
ot es
Class Continuous Frequency (fi) Cumulative frequency (Profit in Rs.) Classes (No. of shops) less than type
N
Sol.
Here total frequency = fi = N = 120
N 120 = = 60 2 2 Cumulative frequency (less than type) which is just greater than 60 is 74. Therefore corresponding class 1950 - 2450 is median class. L = 1950, N = 120, c.f. = 53, f = 21, h = 500
Ad s
Median
N h = L – c.f. 2 f
120 500 – 53 = 1950 2 21 = 1950 + (60 – 53) = 1950 + (7) ×
500 21
500 21
500 3 = 1950 + 166.67 = 2116.67 = 1950 +
Median of profit is Rs. 2116.67.
" Learn from the Best "
16
ALGEBRA 3.
Adsul Notes Following table shows distribution of monthly expenditure (in Rs.) done by households in a certain village on electricity : Monthly expenditure
150 - 225 - 300 - 375 - 450 225 300 375 450 525
525 600
600 and above
No. of 65 171 196 75 53 26 14 households Find median expenditure done by a household on electricity per month. (5 marks) Classes
F requen cy (fi)
Cumulative frequency
(Monthly exp.)
(No. of households)
less than type
150 225 300 375 450 525 600
- 225 - 300 - 375 - 450 - 525 - 600 and above
65 171 196 f 75 53 26 14
65 236 432 507 560 586 600
c.f.
ot es
Sol.
600 N
Total
N
Here total frequency = fi = N = 600 N 600 = = 300 2 2 Cumulative frequency (less than type) which is just greater than 300 is 432. Therefore corresponding class 300 - 375 is median class. L = 300, N = 600, c.f. = 236, f = 196, h = 75
N h = L – c.f. 2 f
ul
Median
Ad s
600 75 – 236 = 300 2 196 = 300 + (300 – 236) = 300 + (64)
75 196
75 196
75 = 300 + 16 49 1200 49 = 300 + 24.49 = 324.49 = 300 +
Median of monthly expenditure is Rs. 324.49. 4.
The following table shows ages of 300 patients getting medical treatment in a hospital on a particular day. (5 marks) Age (in years) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 No. of patients
" Learn from the Best "
60
42
55
70
53
20
17
ALGEBRA Sol.
Classes (Age in years) 10 20 30 40 50 60
Frequency (fi) (No. of patients)
- 20 - 30 - 40 - 50 - 60 -70
Adsul Notes
Cumulative frequency less than type
60 42 55 f 70 53 20
60 102 157 227 280 300
c.f.
300 N
Total
ot es
Here total frequency = fi = N = 300 N 300 = = 150 2 2 Cumulative frequency (less than type) which is just greater than 150 is 157. Therefore corresponding class 30 - 40 is median class. L = 30, N = 300, c.f. = 102, f = 55, h = 10 N h Median = L – c.f. 2 f
300 10 – 102 = 30 2 55 = 30 + (150 – 102)
N
10 55 2 30 + (48) 11 96 30 + 11 30 + 8.73 38.73
10 55
= =
Ad s
= =
ul
= 30 + (48)
Median of age is 38.73 years.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 174) : 1.
Below is given frequency distribution of I.Q. (Intelligent Quotient) of 80 candidates. I.Q, 70–80 80–90 90–100 100–110 110 – 120 120–130 130 – 140
No. of candidates
Sol.
7
16
20
17
Find median I.Q. of a candidate. Classes Frequency (fi) (I.Q.) (No. of candidates) 70 - 80 7 80 - 90 16 90 - 100 20 f 100 - 110 17 110 - 120 11 120 - 130 7 130 - 140 2 Total 80 N
" Learn from the Best "
11
7
2
(5 marks) Cumulative frequency less than type 7 23 c.f. 43 60 71 78 80
18
ALGEBRA
Adsul Notes Here total frequency = fi = N = 80
N 80 = = 40 2 2 Cumulative frequency (less than type) which is just greater than 40 is 43. Therefore corresponding class 90 - 100 is median class. L = 90, N = 80, c.f. = 23, f = 20, h = 10 Median
N h = L – c.f. 2 f
80 10 – 23 = 90 2 20 = 90 + (40 – 23)
1 2
= 90 + 8.5 = 98.5 Median of I.Q.is 98.5.
No. of rooms occupied No. of days
0 – 10
N
Following table shows frequency distribution of no. of rooms occupied in a hotel per day. 10 – 20 20 – 30
5
27
17
ul
7.
ot es
= 90 + (17)
1 2
30 – 40 40 – 50 11
9
Find median number of rooms occupied per day in hotel. No. of rooms occupied
Frequency (fi) (No. of days)
Ad s
Sol.
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Total Here total frequency
5 27 17 f 11 9 1 70 N = fi = N = 70
50 – 60 1 (5 marks)
Cumulative frequency less than type 5 32 49 60 69 70
c.f.
N 70 = = 35 2 2 Cumulative frequency (less than type) which is just greater than 35 is 49. Therefore corresponding class 20 - 30 is median class. L = 20, N = 70, c.f. = 32, f = 17, h = 10 Median
N h = L – c.f. 2 f
70 10 – 32 = 20 2 17
" Learn from the Best "
19
ALGEBRA
Adsul Notes
10 17
= 20 + (35 – 32)
30 17 = 20 + 1.76 = 21.76 = 20 +
Median of rooms occipied is 21.76 rooms. For the data of problem 13, find median calories consumed daily by a boy. Number of calories (in’ 00) consumed daily by a sample of 15 years old boys are given below. (5 marks)
Calories No. of boys
Sol.
1000 – 1500
1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 4500
5
13
16
18
27
10
4
ot es
19.
Frequency (fi) (No. of boys) 1000 - 1500 5 1500 - 2000 13 2000 - 2500 16 2500 - 3000 18 f 3000 - 3500 27 3500 - 4000 10 4000 - 4500 4 Total 93 N Here total frequency = fi = N = 93
Cumulative frequency less than type 5 18 34 c.f. 52 79 89 93
ul
N
Calories
N 93 = = 46.5 2 2 Cumulative frequency (less than type) which is just greater than 46.5 is 52. Therefore corresponding class 2500 - 3000 is median class. L = 2500, N = 93, c.f. = 34, f = 18, h = 500
Ad s
Median
N h = L – c.f. 2 f
93 500 – 34 = 2500 2 18 = 2500 + (46.5 – 34) = 2500 + (12.5)
500 18
500 18
6250 18 = 2500 + 347.22 = 2847.22 = 2500 +
Median of calories consumed by boys is 2847.22 calories.
" Learn from the Best "
20
ALGEBRA 20.
Adsul Notes For the data of problem 14, find median number of packages received per day by a post office. Below is given frequency distribution of no. of packages received at a post office per day. (5 marks)
No. of packages No. of days Sol.
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 2
8
16
Classes Frequency (fi) (No. of packages) (No. of days) 10 20 30 40 50 60
-
20 30 40 50 60 70
2 8 16 24 30 f 20
24
30
60 – 70 20
Cumulative frequency less than type 2 10 26 50 80 100
c.f.
100 N
ot es
Total
Here total frequency = fi = N = 100
N 100 = = 50 2 2 Cumulative frequency (less than type) which is just greater than 50 is 80. Therefore corresponding class 50 - 60 is median class. L = 50, N = 100, c.f. = 50, f = 30, h = 10 N h = L – c.f. 2 f
ul
Median
N
Ad s
100 10 – 50 = 50 2 30 = 50 + (50 – 50) = 50 + (0)
10 30
10 30
= 50 + 0 = 50
Median of package received by post office is 50. 4.
Following table gives frequency distribution of amount of bonus paid to the workers in a certain factory.
Bonus paid (in Rs.)
Below 500
Below 600
Below 700
Below 800
Below 900
Below 1000
Below 1100
No. of workers
4
12
24
41
51
58
60
Find median amount of bonus paid to the worker.
" Learn from the Best "
(5 marks)
21
ALGEBRA Sol.
Bonus paid (in Rs.)
Frequency (fi) (No. of students)
Below 500 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000 1000 - 1100 Total
Adsul Notes
Cumulative frequency less than type
4 8 12 17 f 10 7 2 60 N
4 12 24 41 51 58 60
c.f.
ot es
Here total frequency = fi = N = 60 N 60 = = 30 2 2 Cumulative frequency (less than type) which is just greater than 30 is 41. Therefore corresponding class 700 - 800 is median class. L = 700, N = 60, c.f. = 24, f = 17, h = 100 N h Median = L – c.f. 2 f
ul
N
60 100 – 24 = 700 2 17 100 = 700 + (6) 17 600 = 700 + 17 = 700 + 35.29 = 735.29
Median of bonus paid is Rs.735.29.
Mode :
Ad s
o
Mode is observation which occurs most frequently. The observation having maximum frequency is called mode. A data set has no mode when all the observations appearing in the data have the frequency 1. A data set has multiple modes when two or more values appear with the same highest frequency.
TYPE - 5
To Find mode for the give data.
Things To Remember : For Mode
• •
Classes has to be continuous
o
Steps to calculate Mode :
If maximum frequency occurs twice in the data then data will have two Modal class. Hence, Mode has to be found for both modal class.
Step 1 : Make the classes continuous if they are not Step 2 : Determine Modal class, Maximum frequency (f m), frequency of class preceding modal class (f 1), frequency of class succeeding the modal class (f 2)
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22
ALGEBRA
Adsul Notes
fm f1 × h Step 3 : Calculate, Mode = L + 2 fm f1 f2 L = Lower limit of modal class. N = Total frequency. fm = Maximum frequency. f1 = Frequency of class preceding modal class. f2 = Frequency of class succeeding the modal class. h = Width of the class.
o
Steps to determine Modal class :
Step 1 : The class having maximum frequency is termed as Modal class.
EXERCISE - 5.3 (TEXT BOOK PAGE NO. 133) : The weight of coffee (in gms) in 70 packets is given below : Weight (in gms)
200 201
201 202
202 203
203 204
204 205
205 206
No. of packets
12
26
20
9
2
1
ot es
1.
Determine the modal weight of coffee in a packet. Sol.
Weight (in gms)
(3 marks)
No. of packets
ul
N
200 - 201 12 f1 201 - 202 26 fm 202 - 203 20 f2 203 - 204 9 204 - 205 2 205 - 206 1 Here the maximum frequency fm = 26. The corresponding class 201 - 202 is the modal class. L = 201, fm = 26, f1 = 12, f2 = 20, h = 1 =
f m – f1 L h 2 f m – f 1 – f 2
=
26 – 12 201 1 2 (26) – 12 – 20
Ad s Mode
= = = = =
14 201 52 – 32 14 201 20
7 201 10 201 + 0.7 201.7
Mode of weight of coffee is 201.7 gms. 2.
Forty persons were examined for their Hemoglobin % in blood (in mg per 100 ml) and the results were grouped as below : Hemoglobin % 13.1 - 14.1- 15.1- 16.1- 17.1(mg/100 ml) 14 15 16 17 18 No. of persons
" Learn from the Best "
8
12
10
6
4
23
ALGEBRA Sol.
Adsul Notes
Determine modal value of Hemoglobin % in blood of a person. (3 marks) Here the class given are discontinuous they have to be made continuous. The difference between lower limit of a class and upper limit of previous class is 0.1 0.1 = 0.05 2 Hence we subtract 0.05 from lower limit of every class and add 0.05 to upper limit of every class. Hemoglobin % (mg / 100ml) 13.1 14.1 15.1 16.1 17.1
-
14 15 16 17 18
Continuous class 13.05 14.05 15.05 16.05 17.05
-
No. of persons 8 f1 12 fm 10 f2 6 4
14.05 15.05 16.05 17.05 18.05
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
12 – 8 14.05 + ×1 2 (12) – 8 – 10
=
4 14.05 + 24 – 18
=
ul
N
=
=
14.05 +
4 14.05 + 6
Ad s
Mode
ot es
Here the maximum frequency fm = 12 The corresponding class 14.05 - 15.05 is the modal class. L = 14.05, fm = 12, f1 = 8, f2 = 10, h = 1
= =
2 3 14.05 + 0.67 14.72
Mode of hemoglobin is 14.72 mg/100 ml. 3.
The maximum bowling speed (kms/hour) of 33 players at a cricket coaching centre is given below : Bowling speed (kms / hr) 85 - 100 100 -115 115 - 130 130 - 145 No. of players
9
11
Find the modal bowling speed of a player. Sol.
Bowling speed (kms / hr.) 85 - 100 100 - 115 115 - 130 130 - 145
8
5 (3 marks)
No. of players 9 f1 11 fm 8 f2 5
Here the maximum frequency fm = 11.
" Learn from the Best "
24
ALGEBRA
Adsul Notes The corresponding class 100 - 115 is the modal class. L = 100, fm = 11, f1 = 9, f2 = 8, h = 15. Mode
=
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
11 – 9 100 + × 15 2 (11) – 9 – 8
=
2 100 + × 15 22 – 17
=
30 100 + 5
= =
100 + 6 106
4.
The following table shows frequency distribution of body weight (in gms) of fish in a pond. Body weight (in gms)
15-15.9 16-16.9 17-17.9 18-18.9 19-19.9 20-20.9 2
4
8
6
6
4
N
No. of fish
Find modal body weight of a fish in a pond. (3 marks) Here the class given are discontinuous they have to be made continuous. The difference between lower limit of a class and upper limit of previous class is 0.1 0.1 = 0.05 2 Hence we subtract 0.05 from lower limit of every class and odd 0.05 to upper limit of every class.
Ad s
ul
Sol.
ot es
Mode of bowling speed is 106 km/hr.
Body weight (in gms) 15 16 17 18 19 20
-
15.9 16.9 17.9 18.9 19.9 20.9
Continuous class 14.95 15.95 16.95 17.95 18.95 19.95
-
15.95 16.95 17.95 18.95 19.95 20.95
No. of fish 2 4 f1 8 fm 6 f2 6 4
Here the maximum frequency fm = 8 The corresponding class 16.95 - 17.95 is the modal class. L = 16.95, fm = 8, f1 = 4, f2 = 6, h = 1 Mode
" Learn from the Best "
=
f m – f1 L+ ×h f f f 2 – – m 1 2
=
8–4 16.95 + ×1 2 (8) – 4 – 6
25
ALGEBRA
=
4 16.95 + 6
= =
16.95 + 0.67 17.62
Adsul Notes
Mode of body weight is 17.62 gm.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 174) : 2.
Following table gives frequency distribution of time (in minutes) taken by a person in watching T.V. on a day
Time (in min) 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 No. of persons
4
6
19
14
8
7
2
Obtain modal time taken for watching a T.V. by a person on a day. (3 marks) Time (in min)
4 6 19 14 8 7 2
f1 fm f2
N
30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100
No. of persons
ot es
Sol.
=
f m – f1 L+ ×h 2 f – f – f m 1 2
Ad s
Mode
ul
Here the maximum frequency fm = 19 The corresponding class 50 - 60 is the modal class. L = 50, fm = 19, f1 = 6, f2 = 14, h = 10
=
19 – 6 50 + × 10 2 (19) – 6 – 14
=
13 50 + × 10 38 – 20
=
13 50 + × 10 18
=
65 9 50 + 7.22
=
57.22
=
50 +
Mode of time is 57.22 minutes. 5.
Following table gives frequency distribution of electricity consumption of a household in certain area in amonth.
No. of units of electricity No. of households
" Learn from the Best "
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 4
16
41
65
8
26
ALGEBRA
Sol.
Adsul Notes Find modal no. of units of electricity consumed by a household in amonth. (3 marks) No. of units of electricity No. of household 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
4 16 47 f1 65 fm 8 f2
Here the maximum frequency fm = 65. The corresponding class 60 - 80 is the modal class. L = 60, fm = 65, f1 = 47, f2 = 8, h = 20
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
65 – 47 60 + × 20 2 (65) – 47 – 8
=
18 60 + × 20 130 – 55
=
18 60 + × 20 75
=
18 60 + × 4 15
= =
N
=
ot es
=
6 60 + × 4 5 60 + 4.8 64.8
ul
Mode
10.
Sol.
Ad s
Mode of electricity units is 64.8 units. Below is given frequency distribution of driving speed (in kms/hour) of a vehicle of 400 college students. Speed (in kms/hr)
20 – 30
30 – 40
40 – 50
No. of students
6
80
156
50 – 60 60 – 70 98
Find modal driving speed of a college student. Class width (h) = 10 Speed in (km / hr) 20 30 40 50 60
-
30 40 50 60 70
60 (3 marks)
No. of students 6 80 f1 156 fm 98 f2 60
Here the maximum frequency fm = 156 The corresponding class 40 - 50 is the modal class. L = 40, fm = 156, f1 = 80, f2 = 98, h = 10
" Learn from the Best "
27
ALGEBRA
Mode
=
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
156 – 80 40 + × 10 2 (156) – 80 – 98
=
76 40 + × 10 312 – 178
=
76 40 + × 10 134
= =
40 + 5.67 45.67
Adsul Notes
Mode of driving speed is 45.67 km/hr. Below is given frequency distribution of no. of packages received at a post office per day.
No. of packages
ot es
14.
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of days
2
8
16
24
30
60 – 70 20
Find modal number of packages received by the post office per day. (3 marks) -
20 30 40 50 60 70
2 8 16 24 f1 30 fm 20 f2
Ad s
10 20 30 40 50 60
No. of days
N
No. of packages
ul
Sol.
Here the maximum frequency fm = 30 The corresponding class 50 - 60 is the modal class. L = 50, fm = 30, f1 = 24, f2 = 20, h = 10 Mode
=
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
30 – 24 50 + × 10 2 (30) – 24 – 20
=
6 50 + × 10 60 – 44
=
6 50 + × 10 16
=
50 +
= =
15 4 50 + 3.75 53.75
Mode of packages is 53.75.
" Learn from the Best "
28
ALGEBRA
Sol.
For the data given in problem 6, find modal number of days of medical leave enjoyed by on employee in a year. (3 marks) No. of days No. of employees 0 - 10 5 10 - 20 7 f1 20 - 30 11 fm 30 - 40 4 f2 40 - 50 3 Here the maximum frequency fm = 11 The corresponding class 20 - 30 is the modal class. L = 20, fm = 11, f1 = 7, f2 = 4, h = 10 Mode
=
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
11 – 7 20 + × 10 2 (11) – 7 – 4
=
4 20 + × 10 22 – 11
= = =
40 20 + 11 20 + 3.64 23.64
ot es
16.
Adsul Notes
Sol.
For the data given in problem 12, obtain modal distance travelled per litre by a moped. (3 marks) Disitance in km.
No. of mopeds
ul
18.
N
Mode of medical leave enjoyed is 23.64 days.
Ad s
62 - 65 5 65 - 68 8 68 - 71 12 71 - 74 28 f1 74 - 77 35 fm 77 - 80 10 f2 80 - 83 2 Here the maximum frequency fm = 35 The corresponding class 74 - 77 is the modal class. L = 74, fm = 35, f1 = 28, f2 = 10, h = 3 Mode
=
f m – f1 L+ ×h 2 f m – f 1 – f 2
=
35 – 28 74 + ×3 2 (35) – 28 – 10
= = = =
7 74 + ×3 70 – 38 21 74 + 32 74 + 0.66 74.66
Mode of distance covered by moped is 74.66 kms/lit.
" Learn from the Best "
29
ALGEBRA
o
Adsul Notes
Relation between measures of central tendency : We have studied three mesures of central tendency, mean, median and mode. There exists an empirical relationship among these three measures of central tendency. This relationship was stated by prof. Karl Pearson on the basis on the his vast experienece. The relationship is Mean – Mode = 3 (Mean – Median) It can not be proved theoritically but is observed to be valid for number of data sets after actualc alculations. We know that in some cases mode cannot be determined. In these cases, we can use the above KARL PEARSON relationship to determine value of mode approximately. (1847 - 1936) If values of two measures of central tendency are known, we can use above relationship to find the value of remaining measures of central tendency, approximately.
EXERCISE - 5.4 (TEXT BOOK PAGE NO. 133) :
Median
Sol.
3. Sol.
= 54.4
For a certain frequency distribution the values of Median and Mode is 95.75 and 95.5 respectively, find the mean. (3 marks) Median = 95.75, Mode = 95.5 [Given] We know, Mean – Mode = 3 (Mean – Median) Mean – 95.5 = 3 (Mean – 95.75) Mean – 95.5 = 3 Mean – 287.25 – 95.5 – 287.25= 3 Mean – Mean 191.75 = 2 Mean 191.75 = Mean 2 Mean = 95.875
Ad s
2.
ul
N
Sol.
For a certain frequency distribution the values of Mean and Mode are (3 marks) 54.6 and 54 respectively. Find the value of median. Mean = 54.6, Mode = 54 [Given] We know, Mean – Mode = 3 (Mean – Median) 54.6 – 54 = 3 (54.6 – Median) 0.6 = 3 (54.6 – Median) 0.6 = 54.6 – Median 3 0.2 = 54.6 – Median Median = 54.6 – 0.2
ot es
1.
For a certain frequency distribution the value of Mean is 101 and Median is 100. Find the value of Mode. (3 marks) Mean = 101, Median = 100 [Given] We know, Mean – Mode = 3 (Mean – Median) 101 – Mode = 3 (101 – 100) 101 – Mode = 3 (1) 101 – 3 = Mode Mode
" Learn from the Best "
= 98
30
ALGEBRA
Adsul Notes
MCQ’s 1.
2.
For finding mean by direct method the f i x i (a) f (b) i f iui (c) (d) fi
formula is f idi fi Ad
For finding mean by assumed mean method, the formula is f idi (a) f (b) A d i (c) A nu
(d)
x1 – A
For finding mean by step deviation method, the formula is f iui (b) A d (a) f i di (c) A hu (d) h
4.
The formula for median is
N h (a) L – c.f. . 2 f
(d)
The formula for mode is
N h L c.f. . 2 f
N f L c.f. . 2 h
N
(b)
N f (c) L – c.f. . 2 h f1 – f2 (a) L 2f – f . f . h 1 2 m
(b)
fm – f2 L .h 2fm – f1 – f2
fm – f1 (c) L f – f – f . h 1 2 m
(d)
fm – f1 L .h 2fm – f1 – f2
Ad s
ul
5.
ot es
3.
6.
For certain frequency distribution the value of mean and median is 101 and 100 respectively the mode is (a) 99 (b) 98 (c) 1 (d) 201
7.
The common relation (a) Mode – mean = 3 (b) Mean – mode = 3 (c) Mean – median = (d) Median – mean =
8.
The mode of the data 2, 4, 8, 6, 6, 4 is (a) 4 and 6 (b) 8 (c) 6 (d) 2
9.
If 12, 26, 20, 9, 2, 1 is frequency distribution in the data the frequency of class preceding modal class is (a) 12 (b) 20 (c) 26 (d) none of above
" Learn from the Best "
between (median (mean – 3 (mean 3 (mode
mean, median and mode is – mean) median) – mode) – mean)
31
ALGEBRA
Adsul Notes
10.
For certain frequency distribution the value of median and mode is 95.75 and 95.5 then mean is (a) 98.575 (b) 95.875 (c) 97.585 (d) none of above
11.
If class interval is given in the range 0 - 8, 8 - 16, m – 24, 24 - 32 then value of m is (a) 8 (b) 24 (c) 32 (d) none of these
12.
If class interval is given in the range 31 - 33, 34 - 36, 37 - 39, 40 - 42 then class width of above data is (a) 2 (b) 3 (c) 32 (d) not defined
13.
The class mark of 0 to 9 is (a) 4.5 (c) 9
0 not defined
ot es
(b) (d)
For finding out mean by step deviation and assumed mean method, A should be from the class mark coloumn (a) middle one (b) starting one (c) last one (d) any one
15.
The number of student for particular range is 15, 90, 120, 75, 70, 50 then total number of student is (a) 370 (b) 470 (c) 420 (d) 405
16.
In case of mean, the class mark is given by Lower limit upper limit (a) (b) Lower limit + upper limit 2 Upper lim it – lower limit (c) (d) 2 (lower limit + upper limit) 2
17.
In step deviation method ui is given by
Ad s
ul
N
14.
(a) A d (c)
xi – A h
(b)
xi – A
(d)
xi A h
fi di fi used in which of following method (a) Assume mean method (b) median (c) step deviation method (d) mode
18.
The formula d =
19.
fi in the mode formula defines (a) frequency of pre modal class (b) (c) modal class (d)
20.
frequency of post modal class none of these
In statistics there are .............. measure of central tendency. (a) 4 (b) 2 (c) 3 (d) only 1
" Learn from the Best "
32
ALGEBRA
Adsul Notes
: ANSWERS : f i x i fi
2.
(b) A d
(c) A hu
4.
N h (a) L – c.f. . 2 f
5.
fm – f1 (d) L 2f – f – f . h 1 2 m
6.
(b) 98
7.
(b) Mean – mode = 3 (mean – median)
8.
(a) 4 and 6
10.
(b) 95.875
11.
(d) none of these
12.
(b) 3
13.
(a) 4.5
14.
(d) any one
15.
(c) 420
16.
(a)
17.
(c)
(a)
3.
9.
20.
(c) 3
xi – A h (d) none of these
19.
Ad s
ul
N
18.
Lower lim it upper lim it 2 (a) Assume mean method
(b) 20
ot es
1.
" Learn from the Best "
33
ALGEBRA
Adsul Notes
S.S.C.
Marks : 30
CHAPTER 5 : Statistics - I SET - A
ALGEBRA
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
2
For a certain frequency distribution the values of Mean and Mode are 54.6 and 54 respectively. Find the value of median.
(ii)
Following table gives frequency distribution of amount of bonus paid to the workers in a certain factory. Below
Below
Below
500
600
700
No. of workers
4
12
24
Below
Below
Below
Below
800
900
1000
1500
41
51
58
60
N
Bonus paid (in Rs.)
ot es
(i)
The maximum bowling speed (kms / hour) of 33 players at a cricket coaching centre is given below :
Ad s
(iii)
ul
Locate median class and hence find L, N, c.f., f, h.
Bowling speed (kms / hr)
No. of players
85 - 100 100 -115 115 - 130 130 - 145 9
11
8
5
Locate the modal class and hence find f m, f 1, f 2, L and h. Q.2. Attempt any THREE of the following : (i)
12
Below is given distribution of money (in Rs.) collected by students for flood relief fund. Money (in Rs.)
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
No. of students
5
7
5
2
6
Find mean of money (in Rs.) collected by a student by ‘Direct Method’. (ii)
The weight of coffee (in gms) in 70 packets is given below :
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ALGEBRA
Adsul Notes
Weight (in gms)
200 201
201 202
202 203
203 204
204 205
205 206
No. of packets
12
26
20
9
2
1
Determine the modal weight of coffee in a packet. (iii)
Record of no. of days of medical leave enjoyed by 30 employees within a year is given below. No. of days
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
No. of employees
5
7
11
4
3
(iv)
ot es
Find mean number of days of medical leave enjoyed by an employee in a year. Following table gives frequency distribution of electricity consumption of a household in certain area in amonth.
No. of units of electricity
0 – 20
No. of households
4
20 – 40
40 – 60
60 – 80
80 – 100
16
41
65
8
N
Find modal no. of units of electricity consumed by a household in a month.
Below is given distribution of profit in Rs. per day of a shop in certain town :
Ad s
(i)
20
ul
Q.3. Attempt any FIVE of the following :
Profit (in Rs.)
No. of shosp
500 - 1000 - 1500 - 2000 - 2500 - 3000 900 1400 1900 2400 2900 3400 8
18
27
21
20
3500 3900
18
8
Calculate median profit of a shop. (ii)
Frequency distribution of duration of 500 telephone calls received at a telephone exchange on a certain day is given below :
Duration of call 15-29 30-44 45-59 60-74 (in sec.) No. of calls
8
12
61
153
75-89
90-104
190
57
105-119 120-134 10
9
Find mean duration of telephone call received at the telephone exchange. Using step deviation method.
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ALGEBRA
Adsul Notes
(iii)
Below is given frequency distribution of marks (out of 100) obtained by the students. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of studnets 3 5 7 10 12 15 12 6 2 8 Calculate mean marks scored by a student by ‘Assumed mean method’. (iv) Following table shows frequency distribution of no. of rooms occupied in a hotel per day. No. of rooms occupied
0 – 10
No. of days
10 – 20 20 – 30
5
27
30 – 40
40 – 50
50 – 60
11
9
1
17
Below is given frequency distribution of dividend in percentage declared by 120 companies.
Dividen (in %) No. of companies
10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 5
15
N
(v)
ot es
Find median number of rooms occupied per day in hotel.
28
42
60 – 69
70 – 79
12
3
15
Forty persons were examined for their Hemoglobin % in blood (in mg per 100 ml) and the results were grouped as below :
Ad s
(vi)
ul
Obtain mean dividend declared by a company by step deviation method.
Hemoglobin % (mg/100 ml)
13.1 - 14.114 15
15.116
16.117
17.118
No. of persons
8
10
6
4
12
Determine modal value of Hemoglobin % in blood of a person.
Best Of Luck
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36
ALGEBRA
Adsul Notes
S.S.C.
Marks : 30
CHAPTER 5 : Statistics - I SET - B
ALGEBRA
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
2
For a certain frequency distribution the values of Median and Mode is 95.75 and 95.5 respectively, find the mean.
(ii)
Below is given frequency distribution of no. of packages received at a post office per day.
No. of packages
ot es
(i)
10 – 20 20 – 30 30 – 40
40 – 50
50 – 60
20
N
No. of days 2 8 16 24 30 Find the median class and also find values of L, N, c.f., f, h.
60 – 70
Below is given frequency distribution of driving speed (in kms/hour) of a vehicle of 400 college students. Speed (in kms/hr) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
ul
(iii)
Ad s
No. of students 6 80 156 Find modal class and also find L, fm, f1, f2, h.
98
60
Q.2. Attempt any THREE of the following : (i)
12
A study related to the time (in months) taken to settle a dispute in a lower court resulted the following data.
Time (in months)
0–2
2–4
4–6
6–8
8 – 10
10 – 12
No. of disputes
15
90
120
75
70
50
Find mean time taken to settle a dispute in a lower court. (ii)
Forty persons were examined for their Hemoglobin % in blood (in mg per 100 ml) and the results were grouped as below : Hemoglobin % (mg/100 ml)
13.1 - 14.114 15
15.116
16.117
17.118
No. of persons
8
10
6
4
12
Determine modal value of Hemoglobin % in blood of a person.
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37
ALGEBRA
Adsul Notes
(iii)
Following table gives age distribution of people suffering from ‘Asthama due to air pollution in certain city. Age in years 7-11 11-15 15-19 19-23 23-27 27-31 31-35 35-39
No. of people
5
9
13
21
16
15
12
9
Find mean age of person suffering from ‘Asthama’ by ‘Direct Method’. (iv)
Below is given frequency distribution of driving speed (in kms/hour) of a vehicle of 400 college students. Speed (in kms/hr)
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
No. of students
6
80
156
98
60
ot es
Find modal driving speed of a college student. Q.3. Attempt any FIVE of the following :
Below is given distribution of profit in Rs. per day of a shop in certain town : Profit (in Rs.) 500 - 1000 - 1500 - 2000 - 2500 - 3000 - 3500 900
N
(i)
1400
1900
2400
2900
3400
3900
21
20
18
8
ul
No. of shosp 8 18 27 Calculate median profit of a shop.
Following table shows frequency distribution of duration (in seconds) of advertisements on T.V. Duration (in sec.) 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 65 occupied No. of 10 32 15 9 7 2 advertisements
Ad s
(ii)
20
Obtain mean duration of advertisement on T.V. by shift of origin and scale method. (iii)
Find mean by using ‘Assumed Mean Method’ : Following table gives frequency distribution of trees planted by different housing societies in a particular locality. No. of trees 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 No. of students
(iv)
2
7
9
8
6
4
Find median calories consumed daily by a boy. Number of calories (in’
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38
ALGEBRA
Adsul Notes
00) consumed daily by a sample of 15 years old boys are given below. Calories No. of boys
(v)
1000 – 1500
1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 4500
5
13
16
18
27
10
4
Number of calories (in’ 00) consumed daily by a sample of 15 years old boys are given below :
Calories No. of boys
1000 – 1500
1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 45003
5
13
16
18
27
10
4
The following table shows frequency distribution of body weight (in gms) of fish in a pond. Body weight (in gms) No. of fish
15-15.9 16-16.9 17-17.9 18-18.9 19-19.9 20-20.9 2
N
(vi)
ot es
Obtain mean calories consumed daily by a boy by step deviation method.
4
8
6
6
4
Ad s
ul
Find modal body weight of a fish in a pond.
Best Of Luck
" Learn from the Best "
39
