14 : Statistics
Statistics
14. STATISTICS
SECTION = A
Q1. In a frequency distribution , the class mark of a class is 10 and its width is 5 .The lower limit of class is :
(a) 5 (b) 7.5 (c) 10 (d) 12.5
Solution: 9. (b) 7.5
[ let 
and 
be the lower term and upper term respectively .
So, Class mark 
and Class width 
From 
, we get 
The class interval is 
]
Q2. The median class of the following data is :
|
Classes |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
50 – 60 |
60 – 70 |
|
Frequency |
4 |
4 |
8 |
10 |
12 |
8 |
4 |
(a) 30 – 40 (b) 40 – 50 (c) 20 – 30 (d) 50 – 60
Solution: (b) 40 – 50
[ Here , 
This observation lies in the class 30 – 40 .
The median is 12 , which lies in the class 40 – 50 . ]
Q3. The median class of the following data is :
|
Class interval |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
50 – 60 |
|
Frequency |
14 |
11 |
13 |
15 |
17 |
(a) 30 – 40 (b) 10 – 20 (c) 50 – 60 (d) 20 – 30
Solution: (a) 30 – 40
[ we construct the table :
|
Class interval |
Frequency |
C.f |
|
10 – 20 20 – 30 30 – 40 40 – 50 50 - 60 |
14 11 13 15 17 |
14 25 38 53 70 |
This observation lies in the class 30 – 40 . ]
Q4. The model class of the following data is :
|
Class interval |
15 – 25 |
25 – 35 |
35 – 45 |
45 – 55 |
55 – 65 |
65 – 75 |
|
Frequency |
6 |
11 |
7 |
4 |
4 |
2 |
(a) 6 (b) 7 (c) 4 (d) 11
Solution: (d) 11
[ The mode is the most frequency occurring observation . ]
Q5. The mean of 
observations is 
. If the first item is increased by 1 , second by 2 and so on ,
then the new mean is :
(A) 
(B) 
(C) 
(D) 
Solution : 748 kb
Q6. The mean of first 10 odd numbers is :
(A) 10 (B) 100 (C) 20 (D) 30
Solution :
Q7. The abscissa of the point of intersection of the less than type and of the more than type
cumulative frequency curves of a grouped data gives its :
(a) mean (b) median (c) mode (d) all the three above
Solution: (b) median .
Q8. The median of the observations 11 , 12 , 14 , 18 , 
, 
, 30 , 32 , 35 , 41 arranged in
ascending order is 24 , then the value of 
:
(a) 31 (b) 11 (c) 18 (d) 21
Solution: (d) 21
[ Here, 
.
A/Q, Median 
]
Q9. The following table gives the literacy rate ( in percentage ) of 35 cities .
|
Literacy rate(in |
45 – 55 |
55 – 65 |
65 – 75 |
75 – 85 |
85 – 95 |
|
Number of cities |
3 |
10 |
11 |
8 |
3 |
)The upper limit of the median class in the given data is :
(a) 55 (b) 75 (c) 65 (d) 85
Solution: (c) 65
[ Here , 
; 
=
17.5 .
We observe that the cumulative frequency just greater than 17.5 is 24
and the corresponding class is 60 – 70 . The upper limit is 65 . ]
Q10. If the mean of 15, 12 , , 10, 16 , 19 is 12 ,then the value of .
(a) 1 (b) 7 (c) 2 (d) 0
Solution: (d) 0 .
[ We have , 
; The value of is 0 . ]
Fill in the blanks
Q1. The cumulative frequency table is useful in determining the 
.
Answer : Median .
Answers following the questions
Q1. If upper class limit and lower class limit are 55 and 40 , then find the class mark .
OR If the class interval of the data is 40 – 55 , then find the class mark .
Solution : We know that , Class mark 
Q2. If the class mark and lower class limit are 70 and 60 respectively , then find upper class limit .
Solution : We know that, Class mark 
Q3. If the class mark and the class size of the data are 40 and 20 respectively , then find the
class interval of the data .
Solution : let lower class limit and upper class limit of the data are 
and
respectively.
A/Q, 
[ 
Class mark 
]
[ Class size U.C.L – L.C.L ]
From 
we get , 
Therefore, the class interval of the data is 30 – 50 .
Q4. Find the mode of the following marks (out of 10) obtained by 20 students :
4 , 6 , 5 , 9 , 3 , 2 , 7 , 7 , 6 , 5 , 4 , 9 , 10 , 10 , 3 , 4 , 7 , 6 , 9 , 9
Solution : We arrange the given data in the following form :
2 , 3 , 3 , 4 , 4 ,4 , 5 , 5 , 6 , 6 , 6 , 7 , 7 , 7 , 9 , 9 , 9 , 9 , 10 , 10
Here, 9 occurs most frequently [ i.e., four time] . So, the mode is 9 .
Q5. Find the mode of the following data :
51 , 34 , 17 , 58 , 41 , 51 , 17 , 29 , 39 , 29 , 41 , 29
Solution : We arrange the given data in the following form :
17 , 17 , 29 , 29 , 29 , 34 , 39 , 41 , 41 , 51 , 51 , 58
Here, 29 occurs most frequently [ i.e., three time] . So, the mode is 29 .
Q6. Find the median of the following data : 41 , 39 , 48 , 52 , 46 , 62 , 54 , 40 , 96 , 98 ,60
Solution : First of all we arrange the data in ascending order , as follows :
39 , 40 , 41 , 46 , 48 , 52 , 54 , 60 , 62 , 96 , 98 ; Here 
is odd
Therefore, the median 
obs.
obs.
obs.
obs.
Q7. Find the median of the following data : 48 , 29 , 84 , 32 , 78 , 95 , 72 , 53
Solution : First of all we arrange the data in ascending order , as follows :
29 , 32 , 48 , 53 , 72 , 78 , 84 , 95 ; Here 
is even
The median 
Q8. If the mode of a data is 45 and mean is 27 , the find the median of the data.
Solution : We know that , 3 Median = Mode + 2 Mean
3 Median = 45 + 2 × 27 = 45 + 54 = 99
3 Median = 99
Median 
Therefore, the median of the data is 33 .
SECTION = B
Q1. Find the mean of the following data :
|
Class interval |
10 – 25 |
25 – 40 |
40 – 55 |
55 – 70 |
70 – 85 |
85 – 100 |
|
Number of students |
2 |
3 |
7 |
6 |
6 |
6 |
Solution : We construct the table :
|
Class interval |
Class marks |
Number of students |
|
|
10 – 25 25 – 40 40 – 55 55 – 70 70 – 85 85 – 100 |
17.5 32.5 47.5 62.5 77.5 92.5 |
2 3 7 6 6 6 |
35 97.5 332.5 375.5 465.5 555.0 |
|
Total |
|
|
|
Mean 
SECTION = C
Q1. The table below shows the daily expenditure on food of 25 household 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 the assumed mean method .
Solution : We constructed the table :
|
Daily exp. (in Rs.) |
No. of households |
Class mark |
|
|
|
100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 |
4 5 12 2 2 |
125 175 275 325 |
– 100 – 50 0 50 100 |
– 400 – 250 0 100 200 |
|
Total |
|
|
|
|
225 
The assumed mean method 
Required the daily expenditure on the food is Rs. 211 .
Q2. The distribution below shows the number of wickets taken by bowlers in one-day cricket matches .
Find the mean number of wickets by choosing a suitable method . What does the mean signify ?
[Using the assumed mean method .]
|
Number of wickets |
20 – 60 |
60 – 100 |
100 – 150 |
150 – 250 |
250 – 350 |
350 – 450 |
|
Number of bowlers |
7 |
5 |
16 |
12 |
2 |
3 |
Solution : We constructed the table :
|
No. of wickets |
No. of bowlers |
|
|
|
|
20 – 60 60 – 100 100 – 150 150 – 250 250 – 350 350 – 450 |
7 5 16 12 2 3 |
40 80 125 200 300 400 |
– 160 – 120 – 75 0 100 200 |
– 1120 – 600 – 1200 0 200 600 |
|
Total |
|
|
|
|
Using the assumed mean method :
Mean
Therefore, the number of wickets taken by these 45 bowlers in one day cricket is 152.89 .
Q3. Month pocket money of 50 students of a class are given in the following distribution :
|
Monthly pocket money (Rs.) |
0 – 50 |
50 – 100 |
100 – 150 |
150 – 200 |
200 – 250 |
250 – 300 |
|
Number of student |
2 |
7 |
8 |
30 |
12 |
1 |
Find modal class and also give class mark of the modal class .
Solution: We construct the table :
|
Monthly pocket money (Rs.) |
Class marks |
No. of student |
|
0 – 50 50 – 100 100 – 150 150 – 200 200 – 250 250 – 300 |
25 75 125 175 225 275 |
2 7 8 30 12 1 |
Here the maximum class frequency is 30 . So, the modal class is 150 – 200 .
, 
, 
, 
, 
Mode 
Therefore, the mode is 177.5
Q4. 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: We construct the table :
|
Weight (in kg) |
Frequency |
Cumulative frequency |
|
40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 |
2 3 8 6 6 3 2 |
2 5 13 19 25 28 30 |
Now , 
, 
. This observation lies in the class 55 – 60 .
Here , 
, 
, 
and 
Median 
Therefore, the weight of the students is 56.67 .
SECTION = D
Q1. If the mode of the following distribution is 57.5 , then find the value of :
|
Class interval |
30 – 40 |
40 – 50 |
50 – 60 |
60 – 70 |
70 – 80 |
80 – 90 |
90 – 100 |
|
Frequency |
6 |
10 |
16 |
|
10 |
5 |
2 |
Solution: We construct the table :
|
Class interval |
30 – 40 |
40 – 50 |
50 – 60 |
60 – 70 |
70 – 80 |
80 – 90 |
90 – 100 |
|
Frequency |
6 |
10 |
16 |
|
10 |
5 |
2 |
Here , the maximum class frequency is 16 and the modal class is 50 – 60 .
, 
, 
, 
Mode 
Q4. Find the missing frequencies in following frequency distribution table , if median is 32 .
|
Class interval |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
50 – 60 |
Total |
|
Number of students |
10 |
|
25 |
30 |
|
10 |
100 |
Solution: We construct the frequency distribution table :
|
Class interval |
No . of students |
C.F. |
|
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 |
10 25 30 10 |
10 |
|
Total |
|
|
The median is 32 , which lies in the class 30 – 40 .
Here , 
, 
, 
, 
Using the formula :
Median 
Putting the value of 
in , we get 
Q6. Determine the missing frequency , from the following data , when mode is 67 .
|
Class interval |
40 – 50 |
50 – 60 |
60 – 70 |
70 – 80 |
80 – 90 |
|
Frequency |
5 |
|
15 |
12 |
7 |
Solution: We construct the distribution frequency table of the given data as follows :
|
Class interval |
Frequency |
|
40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 |
5 15 12 7 |
Here , 
, 
, 
, 
and
Using mode formula :
Mode
Therefore, the value of x is 13.7 .
SECTION = E
Q12. If the mean of the following frequency distribution is 145 , find the missing frequencies and 
.
|
Class interval |
0 – 50 |
50 – 100 |
100 – 150 |
150 – 200 |
200 – 250 |
250 – 300 |
Total |
|
Frequency |
8 |
12 |
|
25 |
|
5 |
80 |
Solution: We construct the table :
|
Class interval |
Frequency ( |
Class mark ( |
|
|
0 – 50 50 – 100 100 – 150 150 – 200 200 – 250 250 – 300 |
8 12 25 5 |
25 75 125 175 225 275 |
125 900 125 4375 225 1375 |
|
Total |
|
|
|
)
Here , 
Mean 
From and 
, we get 
From , we have 
Therefore, the value of 
and 
.
Q16. 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 |
Find the mean , mode and median of the given distribution .
Solution : 40. We construct the table :
|
Daily income (in Rs.) |
|
|
|
C.f. |
|
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 |
12 14 8 6 10 |
110 130 150 170 190 |
1320 1820 1200 1020 1900 |
12 26 34 40 50 |
|
Total |
|
|
|
|
Using the formula of mean :
Mean (
) 
Using the formula of mode :
Here , 
; 
; 
; 
; 
Mode 
× 
× 
+ 
Using the formula of median :
Here , 
. So, 
will be in 120 – 140 .
This observation lies in the class 120 – 140 . So ,The frequency of the median class is 14 .
; 
; 
; 
Median 
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