class interval 25-35,35-45,45-55,55-65,65-75,75-85,85,95.frequency 1,5, 5,4,0,8,2, read the frequency distribution table given belowand answer the questions that follow.questions::
1)class interval which has the lowest frequency?
2)class interval which has the highest frequency?
3)what is the class size of the intervals ?
4)what is the upper limit of the 5th class?
5)what is the lower limit of the last class?
Answers
Answer:
1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.
Solution:
Arranging the given data in ascending order, we have
694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745
As the number of terms is an old number i.e., N = 15
We use the following procedure to find the median.
Median = (N + 1)/2 th term
= (15 + 1)/2 th term
= 8th term
So, the 8th term in the arranged order of the given data should be the median.
Therefore, 716 is the median of the data.
2. The following is the distribution of height of students of a certain class in a certain city:
Height (in cm):
160 – 162
163 – 165
166 – 168
169 – 171
172 – 174
No of students:
15
118
142
127
18
Find the median height.
Solution:
Class interval (exclusive)
Class interval (inclusive)
Class interval frequency
Cumulative frequency
160 – 162
159.5 – 162.5
15
15
163 – 165
162.5 – 165.5
118
133(F)
166 – 168
165.5 – 168.5
142(f)
275
169 – 171
168.5 – 171.5
127
402
172 – 174
171.5 – 174.5
18
420
N = 420
Here, we have N = 420,
So, N/2 = 420/ 2 = 210
The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such, that
L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3
R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 1
= 165.5 + 1.63
= 167.13
3. Following is the distribution of I.Q of 100 students. Find the median I.Q.
I.Q:
55 – 64
65 – 74
75 – 84
85 – 94
95 – 104
105 – 114
115 – 124
125 – 134
135 – 144
No of students:
1
2
9
22
33
22
8
2
1
Solution:
Class interval (exclusive)
Class interval (inclusive)
Class interval frequency
Cumulative frequency
55 – 64
54.5 – 64-5
1
1
65 – 74
64.5 – 74.5
2
3
75 – 84
74.5 – 84.5
9
12
85 – 94
84.5 – 94.5
22
34(F)
95 – 104
94.5 – 104.5
33(f)
67
105 – 114
104.5 – 114.5
22
89
115 – 124
114.5 – 124.5
8
97
125 – 134
124.5 – 134.5
2
98
135 – 144
134.5 – 144.5
1
100
N = 100
Here, we have N = 100,
So, N/2 = 100/ 2 = 50
The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) such that L = 94.5, F = 33, h = (104.5 – 94.5) = 10
R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 2
= 94.5 + 4.85
= 99.35
4. Calculate the median from the following data:
Rent (in Rs):
15 – 25
25 – 35
35 – 45
45 – 55
55 – 65
65 – 75
75 – 85
85 – 95
No of houses:
8
10
15
25
40
20
15
7
Solution:
Class interval
Frequency
Cumulative frequency
15 – 25
8
8
25 – 35
10
18
35 – 45
15
33
45 – 55
25
58(F)
55 – 65
40(f)
98
65 – 75
20
118
75 – 85
15
133
85 – 95
7
140
N = 140
Here, we have N = 140,
So, N/2 = 140/ 2 = 70
The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10
R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 3
= 55 + 3 = 58
5. Calculate the median from the following data:
Marks below:
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
70 – 80
85 – 95
No of students:
15
35
60
84
96
127
198
250
Swipe left
Solution:
Marks below
No. of students
Class interval
Frequency
Cumulative frequency
10
15
0 – 10
15
15
20
35
10 – 20
20
35
30
60
20 – 30
25
60
40
84
30 – 40
24
84
50
96
40 – 50
12
96(F)
60
127
50 – 60
31(f)
127
70
198
60 – 70
71
198
80
250
70 – 80
52
250
N = 250
Here, we have N = 250,
So, N/2 = 250/ 2 = 125
The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10
R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 4
= 50 + 9.35
= 59.35
Answer:
A=1 B=5 try 3questions on