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Thursday, 17 October 2013
STAT
CBSE NCERT Solutions For Class 10th Mathematics
Chapter 14 : Statistics. NCERT
Exercise 14.1, Exercise
14.2, Exercise 14.3, Exercise 14.4.
Exercise 14.1
1. A survey was conducted by a
group of students as a part of their environment awareness programme, in which
they collected the following data regarding the number of plants in 20 houses
in a locality. Find the mean number of plants per house.
|
Number
of plants
|
0-2
|
2-4
|
4-6
|
6-8
|
8-10
|
10-12
|
12-14
|
|
Number
of houses
|
1
|
2
|
1
|
5
|
6
|
2
|
3
|
Which method did you use for
finding the mean, and why?
Solution:
|
Class
Interval
|
fi
|
xi
|
fixi
|
|
0-2
|
1
|
1
|
1
|
|
2-4
|
2
|
3
|
6
|
|
4-6
|
1
|
5
|
5
|
|
6-8
|
5
|
7
|
35
|
|
8-10
|
6
|
9
|
54
|
|
10-12
|
2
|
11
|
22
|
|
12-14
|
3
|
13
|
39
|
|
Sum fi = 20
|
Sum fixi = 162
|
Mean can be calculated as
follows:
In this case, the values of fi
and xi are small hence direct method has been used.
2. Consider the following
distribution of daily wages of 50 workers of a factory.
|
Daily
wages (in Rs)
|
100-120
|
120-140
|
140-160
|
160-180
|
180-200
|
|
Number
of workers
|
12
|
14
|
8
|
6
|
10
|
Find the mean daily wages of
the workers of the factory by using an appropriate method.
Solution: In this case, value of xi is quite large
and hence we should select the assumed mean method.
Let us take assumed mean a =
150
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
fidi
|
|
100-120
|
12
|
110
|
-40
|
-480
|
|
120-140
|
14
|
130
|
-20
|
-280
|
|
140-160
|
8
|
150
|
||
|
160-180
|
6
|
170
|
20
|
120
|
|
180-200
|
10
|
190
|
40
|
400
|
|
Sum fi = 50
|
Sum fidi = -240
|
Now, mean of deviations can be
calculated as follows:
Mean can be calculated as
follows:
x = d + a = -4.8 + 150 = 145.20
3. The following distribution
shows the daily pocket allowance of children of a locality. The mean pocket
allowance is Rs. 18. Find the missing frequency f.
|
Daily
pocket allowance (in Rs)
|
11-13
|
13-15
|
15-17
|
17-19
|
19-21
|
21-23
|
23-25
|
|
Number
of children
|
7
|
6
|
9
|
13
|
f
|
5
|
4
|
Solution:
|
Class
Interval
|
fi
|
xi
|
fixi
|
|
11-13
|
7
|
12
|
84
|
|
13-15
|
6
|
14
|
84
|
|
15-17
|
9
|
16
|
144
|
|
17-19
|
13
|
18
|
234
|
|
19-21
|
f
|
20
|
20f
|
|
21-23
|
5
|
22
|
110
|
|
23-25
|
4
|
24
|
96
|
|
Sum fi = 44 + f
|
Sum fixi = 752 +
20f
|
We have;
4. Thirty women were examined
in a hospital by a doctor and the number of heart beats per minute were
recorded and summarized as follows. Find the mean heart beats per minute for
these women, choosing a suitable method.
|
Number
of heart beats per min
|
65-68
|
68-71
|
71-74
|
74-77
|
77-80
|
80-83
|
83-86
|
|
Number
of women
|
2
|
4
|
3
|
8
|
7
|
4
|
2
|
Solution:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
fidi
|
|
65-68
|
2
|
66.5
|
-9
|
-18
|
|
68-71
|
4
|
69.5
|
-6
|
-24
|
|
71-74
|
3
|
72.5
|
-3
|
-9
|
|
74-77
|
8
|
75.5
|
||
|
77-80
|
7
|
78.5
|
3
|
21
|
|
80-83
|
4
|
81.5
|
6
|
24
|
|
83-86
|
2
|
84.5
|
9
|
18
|
|
Sum fi = 30
|
Sum fidi = 12
|
Now, mean can be calculated as
follows:
5. In a retail market, fruit
vendors were selling mangoes kept in packing boxes. These boxes contained
varying number of mangoes. The following was the distribution of mangoes
according to the number of boxes.
|
Number
of mangoes
|
50-52
|
53-55
|
56-58
|
89-61
|
62-64
|
|
Number
of boxes
|
15
|
110
|
135
|
115
|
25
|
Find the mean number of mangoes
kept in a packing box. Which method of finding the mean did you choose?
Solution:
|
Class
Interval
|
fi
|
xi
|
di
= x – a
|
fidi
|
|
50-52
|
15
|
51
|
-6
|
90
|
|
53-55
|
110
|
54
|
-3
|
-330
|
|
56-58
|
135
|
57
|
||
|
59-61
|
115
|
60
|
3
|
345
|
|
62-64
|
25
|
63
|
6
|
150
|
|
Sum fi = 400
|
Sum fidi = 75
|
Mean can be calculated as
follows:
In this case, there are wide
variations in fi and hence assumed mean method is used.
6. The table below shows the
daily expenditure on food of 25 households in a locality.
|
Daily
expenditure (in Rs)
|
100-150
|
150-200
|
200-250
|
250-300
|
300-350
|
|
Number
of households
|
4
|
5
|
12
|
2
|
2
|
Find the mean daily expenditure
on food by a suitable method.
Solution:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
ui
= di/h
|
fiui
|
|
100-150
|
4
|
125
|
-100
|
-2
|
-8
|
|
150-200
|
5
|
175
|
-50
|
-1
|
-5
|
|
200-250
|
12
|
225
|
|||
|
250-300
|
2
|
275
|
50
|
1
|
2
|
|
300-350
|
2
|
325
|
100
|
2
|
4
|
|
Sum fi = 25
|
Sum fiui = -7
|
Mean can be calculated as
follows:
7. To find out the concentration of SO2 in the
air (in parts per million, i.e. ppm), the data was collected for 30 localities
in a certain city and is presented below:
|
Concentration
of SO2 (in ppm)
|
Frequency
|
|
0.00-0.04
|
4
|
|
0.04-0.08
|
9
|
|
0.08-0.12
|
9
|
|
0.12-0.16
|
2
|
|
0.16-0.20
|
4
|
|
0.20-0.24
|
2
|
Find the mean concentration of SO2 in the
air.
Solution:
|
Class
Interval
|
fi
|
xi
|
fixi
|
|
0.00-0.04
|
4
|
0.02
|
0.08
|
|
0.04-0.08
|
9
|
0.06
|
0.54
|
|
0.08-0.12
|
9
|
0.10
|
0.90
|
|
0.12-0.16
|
2
|
0.14
|
0.28
|
|
0.16-0.20
|
4
|
0.18
|
0.72
|
|
0.20-0.24
|
2
|
0.22
|
0.44
|
|
Sum fi = 30
|
Sum fixi = 2.96
|
Mean can be calculated as
follows:
8. A class teacher has the
following absentee record of 40 students of a class for the whole term. Find
the mean number of days a student was absent.
|
Number
of days
|
0-6
|
6-10
|
10-14
|
14-20
|
20-28
|
28-38
|
38-40
|
|
Number
of students
|
11
|
10
|
7
|
4
|
4
|
3
|
1
|
Solution:
|
Class
Interval
|
fi
|
xi
|
fixi
|
|
0-6
|
11
|
3
|
33
|
|
6-10
|
10
|
8
|
80
|
|
10-14
|
7
|
12
|
84
|
|
14-20
|
4
|
17
|
68
|
|
20-28
|
4
|
24
|
96
|
|
28-38
|
3
|
33
|
99
|
|
38-40
|
1
|
39
|
39
|
|
Sum fi = 40
|
Sum fixi = 499
|
Mean can be calculated as
follows:
9. The following table gives
the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
|
Literacy
rate (in %)
|
45-55
|
55-65
|
65-75
|
75-85
|
85-98
|
|
Number
of cities
|
3
|
10
|
11
|
8
|
3
|
Solution:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
ui
= di/h
|
fiui
|
|
45-55
|
3
|
50
|
-20
|
-2
|
-6
|
|
55-65
|
10
|
60
|
-10
|
-1
|
-10
|
|
65-75
|
11
|
70
|
|||
|
75-85
|
8
|
80
|
10
|
1
|
8
|
|
85-95
|
3
|
90
|
20
|
2
|
6
|
|
Sum fi = 35
|
Sum fiui = -2
|
Mean can be calculated as
follows:
Exercise 14.2
1. The following table shows
the ages of the patients admitted in a hospital during a year.
|
Age
(in years)
|
5-15
|
15-25
|
25-35
|
35-45
|
45-55
|
55-65
|
|
Number
of patients
|
6
|
11
|
21
|
23
|
14
|
5
|
Find the mode and the mean of
the data given above. Compare and interpret the two measures of central
tendency.
Solution: Solution: Modal class = 35 – 45, l = 35, h
= 10, f1 = 23, f0 = 21 and f2 = 14
Calculations for Mean:
|
Class
Interval
|
fi
|
xi
|
fixi
|
|
5-15
|
6
|
10
|
60
|
|
15-25
|
11
|
20
|
220
|
|
25-35
|
21
|
30
|
630
|
|
35-45
|
23
|
40
|
920
|
|
45-55
|
14
|
50
|
700
|
|
55-65
|
5
|
60
|
300
|
|
Sum fi = 80
|
Sum fixi = 2830
|
The mode of the data shows that
maximum number of patients is in the age group of 26.8, while average age of
all the patients is 35.37.
2. The following data gives the
information on the observed lifetime (in hours) of 225 electrical components:
|
Lifetime
(in hours)
|
0-20
|
20-40
|
40-60
|
60-80
|
80-100
|
100-120
|
|
Frequency
|
10
|
35
|
52
|
61
|
38
|
29
|
Determine the modal lifetimes
of the components.
Solution: Modal class = 60-80, l = 60, f1 = 61, f0 =
52, f2 = 38 and h = 20
3. The following data gives the
distribution of total monthly household expenditure of 200 families of a
village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure.
|
Expenditure
|
Number
of families
|
|
1000-1500
|
24
|
|
1500-2000
|
40
|
|
2000-2500
|
33
|
|
2500-3000
|
28
|
|
3000-3500
|
30
|
|
3500-4000
|
22
|
|
4000-4500
|
16
|
|
4500-5000
|
7
|
Solution: Modal class = 1500-2000, l = 1500, f1 = 40,
f0 = 24, f2 = 33 and h = 500
Calculations for mean:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
ui
= di/h
|
fiui
|
|
1000-1500
|
24
|
1250
|
-1500
|
-3
|
-72
|
|
1500-2000
|
40
|
1750
|
-1000
|
-2
|
-80
|
|
2000-2500
|
33
|
2250
|
-500
|
-1
|
-33
|
|
2500-3000
|
28
|
2750
|
|||
|
3000-3500
|
30
|
3250
|
500
|
1
|
30
|
|
3500-4000
|
22
|
3750
|
1000
|
2
|
44
|
|
4000-4500
|
16
|
4250
|
1500
|
3
|
48
|
|
4500-5000
|
7
|
4750
|
2000
|
4
|
28
|
|
fi = 200
|
fiui = -35
|
4. The following distribution gives the
state-wise teacher-student ratio in higher secondary schools of India. Find the
mode and mean of this data. Interpret the two measures.
|
Number
of students per teacher
|
Number
of states/UT
|
|
15-20
|
3
|
|
20-25
|
8
|
|
25-30
|
9
|
|
30-35
|
10
|
|
35-40
|
3
|
|
40-45
|
|
|
45-50
|
|
|
50-55
|
2
|
Solution: Modal class = 30-35, l = 30, f1 = 10, f0 =
9, f2 = 3 and h = 5
Calculation for mean:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
ui
= di/h
|
fiui
|
|
15-20
|
3
|
17.5
|
-15
|
-3
|
-9
|
|
20-25
|
8
|
22.5
|
-10
|
-2
|
-16
|
|
25-30
|
9
|
27.5
|
-5
|
-1
|
-9
|
|
30-35
|
10
|
32.5
|
|||
|
35-40
|
3
|
37.5
|
5
|
1
|
3
|
|
40-45
|
42.5
|
10
|
2
|
||
|
45-50
|
47.5
|
15
|
3
|
||
|
50-55
|
2
|
52.5
|
20
|
4
|
8
|
|
Sum fi = 35
|
Sum fiui = -23
|
The mode shows that maximum
number of states has 30-35 students per teacher. The mean shows that average
ratio of students per teacher is 29.22
5. The given distribution shows
the number of runs scored by some top batsmen of the world in one-day
international cricket matches.
|
Runs
scored
|
Number
of batsmen
|
|
3000-4000
|
4
|
|
4000-5000
|
18
|
|
5000-6000
|
9
|
|
6000-7000
|
7
|
|
7000-8000
|
6
|
|
8000-9000
|
3
|
|
9000-1000
|
1
|
|
10000-11000
|
1
|
Find the mode of the data.
Solution: Modal class = 4000-5000, l = 4000, f1 = 18,
f0 = 4, f2 = 9 and h = 1000
6. A student noted the number
of cars passing through a spot on a road for 100 periods each of 3 minutes and
summarized it in the table given below. Find the mode of the data.
|
Number
of cars
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
50-60
|
60-70
|
70-80
|
|
Frequency
|
7
|
14
|
13
|
12
|
20
|
11
|
15
|
8
|
Solution: Modal class = 40 – 50, l = 40, f1 = 20, f0
= 12, f2 = 11 and h = 10
Exercise 14.3
1. The following frequency
distribution gives the monthly consumption of electricity of 68 consumers of a
locality. Find the median, mean and mode of the data and compare them.
|
Monthly
consumption (in units)
|
Number
of customers
|
|
65-85
|
4
|
|
85-105
|
5
|
|
105-125
|
13
|
|
125-145
|
20
|
|
145-165
|
14
|
|
165-185
|
8
|
|
185-205
|
4
|
Solution:
|
Class
Interval
|
Frequency
|
Cumulative
frequency
|
|
65-85
|
4
|
4
|
|
85-105
|
5
|
9
|
|
105-125
|
13
|
22
|
|
125-145
|
20
|
42
|
|
145-165
|
14
|
56
|
|
165-185
|
8
|
64
|
|
185-205
|
4
|
68
|
|
N = 68
|
Here; n = 68 and hence n/2 = 34
So, median class is 125-145
with cumulative frequency = 42
now, l = 125, n = 68, cf = 22,
f = 20, h = 20
Median can be calculated as
follows:
Calculations for Mode:
Modal class = 125-145, f1 = 20,
f0 = 13, f2 = 14 and h = 20
Calculations for Mean:
|
Class
Interval
|
fi
|
xi
|
di
= xi – a
|
ui
= di/h
|
fiui
|
|
65-85
|
4
|
75
|
-60
|
-3
|
-12
|
|
85-105
|
5
|
95
|
-40
|
-2
|
-10
|
|
105-125
|
13
|
115
|
-20
|
-1
|
-13
|
|
125-145
|
20
|
135
|
|||
|
145-165
|
14
|
155
|
20
|
1
|
14
|
|
165-185
|
8
|
175
|
40
|
2
|
16
|
|
185-205
|
4
|
195
|
60
|
3
|
12
|
|
Sum fi = 68
|
Sum fiui = 7
|
Mean, median and mode are more
or less equal in this distribution.
2. If the median of the distribution
given below is 28.5, find the value of x and y.
|
Class
Interval
|
Frequency
|
|
0-10
|
5
|
|
10-20
|
x
|
|
20-30
|
20
|
|
30-40
|
15
|
|
40-50
|
y
|
|
50-60
|
5
|
|
Total
|
60
|
Solution: n = 60 and hence n/2 = 30
Median class is 20 – 30 with
cumulative frequency = 25 + x
lower limit of median class =
20, cf = 5 + x , f = 20 and h = 10
Now, from cumulative frequency,
we can find the value of x + y as follows:
Hence, x = 8 and y = 7
3. A life insurance agent found
the following data for distribution of ages of 100 policy holders. Calculate
the median age, if policies are given only to persons having age 18 years
onwards but less than 60 years.
|
Age
(in years)
|
Number
of policyt hodlers
|
|
Below 20
|
2
|
|
Below 25
|
6
|
|
Below 30
|
24
|
|
Below 35
|
45
|
|
Below 40
|
78
|
|
Below 45
|
89
|
|
Below 50
|
92
|
|
Below 55
|
98
|
|
Below 60
|
100
|
Solution:
|
Class
interval
|
Frequency
|
Cumulative
frequency
|
|
15-20
|
2
|
2
|
|
20-25
|
4
|
6
|
|
25-30
|
18
|
24
|
|
30-35
|
21
|
45
|
|
35-40
|
33
|
78
|
|
40-45
|
11
|
89
|
|
45-50
|
3
|
92
|
|
50-55
|
6
|
98
|
|
55-60
|
2
|
100
|
Here; n = 100 and n/2 = 50,
hence median class = 35-45
In this case; l = 35, cf = 45,
f = 33 and h = 5
4. The lengths of 40 leaves of
a plant are measured correct to the nearest millimeter, and the data obtained
is represented in the following table:
|
Length
(in mm)
|
Number
of leaves
|
|
118-126
|
3
|
|
127-135
|
5
|
|
136-144
|
9
|
|
145-153
|
12
|
|
154-162
|
5
|
|
163-171
|
4
|
|
172-180
|
2
|
Find the median length of
leaves.
Solution:
|
Class
Interval
|
Frequency
|
Cumulative
frequency
|
|
117.5-126.5
|
3
|
3
|
|
126.5-135.5
|
5
|
8
|
|
135.5-144.5
|
9
|
17
|
|
144.5-153.5
|
12
|
29
|
|
153.5-162.5
|
5
|
34
|
|
162.5-171.5
|
4
|
38
|
|
171.5-180.5
|
2
|
40
|
We have; n = 40 and n/2 = 20
hence median class = 144.5-153.5
Thus, l = 144.5, cf = 17, f =
12 and h = 9
5. The following table gives
distribution of the life time of 400 neon lamps.
|
Lifetime
(in hours)
|
Number
of lamps
|
|
1500-2000
|
14
|
|
2000-2500
|
56
|
|
2500-3000
|
60
|
|
3000-3500
|
86
|
|
3500-4000
|
74
|
|
4000-4500
|
62
|
|
4500-5000
|
48
|
Find the median life time of a
lamp.
Solution:
|
Class
Interval
|
Frequency
|
Cumulative
Frequency
|
|
1500-2000
|
14
|
14
|
|
2000-2500
|
56
|
70
|
|
2500-3000
|
60
|
130
|
|
3000-3500
|
86
|
216
|
|
3500-4000
|
74
|
290
|
|
4000-4500
|
62
|
352
|
|
4500-5000
|
48
|
400
|
We have; n = 400 and n/2 = 200
hence median class = 3000 – 3500
So, l = 3000, cf = 130, f = 86
and h = 500
6. 100 surnames were randomly
picked up from a local telephone directory and the frequency distribution of
the number of letters in the English alphabets in the surnames was obtained as
follows:
|
Number
of letters
|
1-4
|
4-7
|
7-10
|
10-13
|
13-16
|
16-19
|
|
Number
of surnames
|
6
|
30
|
40
|
16
|
4
|
4
|
Determine the median number of
letters in the surnames. Find the mean number of letters in the surnames. Also,
find the modal size of the surnames.
Solution: Calculations for median:
|
Class
Interval
|
Frequency
|
Cumulative
Frequency
|
|
1-4
|
6
|
6
|
|
4-7
|
30
|
36
|
|
7-10
|
40
|
76
|
|
10-13
|
16
|
92
|
|
13-16
|
4
|
96
|
|
16-19
|
4
|
100
|
Here; n = 100 and n/2 = 50
hence median class = 7-10
So, l = 7, cf = 36, f = 40 and
h = 3
Calculations for Mode:
Modal class = 7-10,
Here; l = 7, f1 = 40, f0 = 30,
f2 = 16 and h = 3
Calculations for Mean:
|
Class
interval
|
fi
|
xi
|
fixi
|
|
1-4
|
6
|
2.5
|
15
|
|
4-7
|
30
|
5.5
|
165
|
|
7-10
|
40
|
8.5
|
340
|
|
10-13
|
16
|
11.5
|
184
|
|
13-16
|
4
|
14.5
|
51
|
|
16-19
|
4
|
17.5
|
70
|
|
Sum fi = 100
|
Sum fixi = 825
|
7. The distribution below gives
the weights of 30 students of a class. Find the median weight of the students.
|
Weight
(in kg)
|
40-45
|
45-50
|
50-55
|
55-60
|
60-65
|
65-70
|
70-75
|
|
Number
of students
|
2
|
3
|
8
|
6
|
6
|
3
|
2
|
Solution:
|
Class
Interval
|
Frequency
|
Cumulative
frequency
|
|
40-45
|
2
|
2
|
|
45-50
|
3
|
5
|
|
50-55
|
8
|
13
|
|
55-60
|
6
|
19
|
|
60-65
|
6
|
25
|
|
65-70
|
3
|
28
|
|
70-75
|
2
|
30
|
We have; n = 30 and n/2 = 15
hence median class = 55-60
So, l = 55, cf = 13, f = 6 and
h = 5
Exercise 14.4(NCERT)
1. The following distribution
gives the daily income of 50 workers of a factory.
|
Daily
income (in Rs)
|
100-120
|
120-140
|
140-160
|
160-180
|
180-200
|
|
Number
of workers
|
12
|
14
|
8
|
6
|
10
|
Convert the distribution above
to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
|
Daily
income
|
Cumulative
frequency
|
|
Less than 120
|
12
|
|
Less than 140
|
26
|
|
Less than 160
|
34
|
|
Less than 180
|
40
|
|
Less than 200
|
50
|
2. During the medical checkup of 35
students of a class, their weights were recorded as follows:|
Weight
(in kg)
|
Number
of students
|
|
Less than 38
|
|
|
Less than 40
|
3
|
|
Less than 42
|
5
|
|
Less than 44
|
9
|
|
Less than 46
|
14
|
|
Less than 48
|
28
|
|
Less than 50
|
32
|
|
Less than 52
|
35
|
Draw a less than type ogive for
the given data. Hence obtain the median weight from the graph and verify the
result by using the formula.
Solution:
|
Weight
(in kg)
|
Frequency
|
Cumulative
Frequency
|
|
36-38
|
||
|
38-40
|
3
|
3
|
|
40-42
|
2
|
5
|
|
42-44
|
4
|
9
|
|
44-46
|
5
|
14
|
|
46-48
|
14
|
28
|
|
48-50
|
4
|
32
|
|
50-52
|
3
|
35
|
Since N = 35 and n/2 = 17.5
hence median class = Less than 46-48
Here; l = 46, cf = 14, f = 14
and h = 2
Median can be calculated as
follows:
This value of median verifies
the median shown in ogive.
3. The following table gives
production yield per hectare of wheat of 100 farms of a village.
|
Production
yield (in kg)
|
50-55
|
55-60
|
60-65
|
65-70
|
70-75
|
75-80
|
|
Number
of farms
|
2
|
8
|
12
|
24
|
38
|
16
|
Change this distribution to a
more than type distribution, and draw its ogive.
Solution:
|
Production
yield
|
Cumulative
frequency
|
|
More than 50
|
100
|
|
More than 55
|
98
|
|
More than 60
|
90
|
|
More than 65
|
78
|
|
More than 70
|
54
|
|
More than 75
|
16
|
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