Q11. During the medical checkup of 35 students of a class, their weights were recorded as follows:
Weight (in kg)  Number of students 
3840  3 
4042  2 
4244  4 
4446  5 
4648  14 
4850  4 
5052  3 
Draw a ‘less than type’ and ‘more than type’ ogive from the given data. Hence, obtain the median weight from the graph.
Sol.
Weight (in kg)  No. of students  Less than cumulative frequency  More than cumulative frequency 
3840  3  Less than 40 = 3  38 and more than 38 = 35 
4042  2  Less than 42 = 5  40 and more than 40 = 32 
4244  4  Less than 44 = 9  42 and more than 42 = 30 
4446  5  Less than 46 = 14  44 and more than 44 = 26 
4648  14  Less than 48 = 28  46 and more than 46 = 21 
4850  4  Less than 50 = 32  48 and more than 48 = 7 
5052  3  Less than 52 = 35  50 and more than 50 = 3 
Total  ∑ f_{i} = 35  ⇒ n = 35 

∴ For less than type ogive, we plot the points (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35).
∴ For more than type ogive, we plot the points (38, 35), (40, 32), (42, 30), (44, 26), (46, 21), (48, 7) and (50, 3) on the same graph.
From the graph, we obtain the median = 46.5 kg.
Q12. The following table gives daily income of 50 workers of a factory:
Daily income (in Rs)  100120  120140  140160  160180  180200 
Number of workers  12  14  8  6  10 
Find the mean, mode and median of the above data.
Sol. Let assumed mean a = 150. Here, h = 20
Daily income (in Rs)  No. of workers (f_{i})  ^{x}i  ^{f}i ^{u}i  ^{cf}  
100120  12  110  2  24  12 + 0 = 12 
120140  14  130  1  14  12 + 14 = 26 
140160  8  150  0  0  26 + 8 = 34 
160180  6  170  1  6  34 + 6 = 40 
180200  10  190  2  20  40 + 10 = 50 
Total  ∑ f_{i} = 50 

 ∑ f_{i}u_{i} =12 

(i) Mean
Thus, mean income is Rs 145.2
(ii) For finding the mode,
We have the greatest frequency = 14 which lies in the class 120−140
∴ Modal class = 120−140
Therefore, l = 120
f_{1} = 14
f_{0} = 12
f_{2} = 8
and h = 20
(iii) For finding median,
And 25 lies in the class 120−140
Median class is 120−140
Since n/2 = 25, cf = 12, f = 14 and h = 20
Median income = Rs138.57
Q13. Find the mode, median and mean for the following data:
Marks obtained  2535  3545  4555  5565  6575  7585 
Number of students  7  31  33  17  11  1 
Sol. Let the assumed mean a = 60. Here h = 10, we have:
Marks obtained  Class marks x_{i}  ^{f}i  ^{cf}  ^{f}i ^{u}i  
2535  30  7  7  3  21 
3545  40  31  38  2  62 
4555  50  33  71  1  33 
5565  60  17  88  0  0 
6575  70  11  99  1  11 
7585  80  1  100  2  2 
Total 
 ∑ f = 100 ⇒ n = 100 

 ∑ f_{i}u_{i} =103 
(i)
⇒
(ii) Median
Here,
∴ Median class is 45−55.
l = 45
cf = 38
f = 33 and h = 10
(iii) Mode: Greatest frequency is 33 which corresponds to the class 45−55.
l = 45, h = 10
f_{1} = 33, f2 = 17
f_{0} = 31
Q14. A survey regarding the heights (in cm) of 50 girls of Class X of a school was conducted and the following data was obtained:
Height (in cm)  120130  130140  140150  150160  160170  Total 
Number of girls  2  8  12  20  8  50 
Find the mean, median and mode of the above data.
Sol. We have:
Height (in cm)  f  ^{cf}  x_{i}  ^{f}i^{x}i 
120130  2  2 + 0 = 2  125  250 
130140  8  2 + 8 = 10  135  1080 
140150  12  10 + 12 = 22  145  1740 
150160  20  22 + 20 = 42  155  3100 
160170  8  42 + 8 = 50  165  1320 
Total  50 

 7490 
(i)
(ii) ∵
∴ Median class is 150−160.
∴ We have,
l = 150
f = 20
cf = 2
2h = 10
∴
⇒
(iii) ∵ Greatest frequency = 20
∴ Modal class = 150−160
So, we have
l = 150, f_{0} = 12, f_{1} = 20
f_{2} = 8 and h = 10
Q15. Find the mean, mode and median of the following data:
Classes  Frequency 
010  5 
1020  10 
2030  18 
3040  30 
4050  20 
5060  12 
6070  5 
Sol.
(i) Mean:
Let the assumed mean ‘a’ = 35
Now we have the following data:
Class  Class mark x_{i}  fi  Cf  f_{i}u_{i}  
010  5  5  5 + 0 = 5  3  (3) x 5 = 15 
1020  15  10  5 + 10 = 15  2  (2) x 10 = 20 
2030  25  18  15 + 18 = 33  1  (1) x 18 = 18 
3040  35  30  33 + 30 = 63  0  0 x 30 = 0 
4050  45  20  63 + 20 = 93  1  1 x 20 = 20 
5060  55  12  83 + 12 = 95  2  2 x 12 = 24 
6070  65  5  95 + 5 = 100  3  3 x 5 = 15 
Here ∑f_{i} = 100 and ∑f_{i}u_{i} = 6
(ii) Mode:
Here, the maximum frequency is 30.
∴ Modal class is 30−40.
So, we have
l = 30, h = 10
f_{1} = 30,
f_{0} = 18
f_{2} = 20
(iii) Median:
∴ Median class = 30−40
So, we have:
l = 30, h = 10, cf = 33, f = 30
Q16. Find the mean, mode and median for the following data:
Classes  Frequency 
010  3 
1020  8 
2030  10 
3040  15 
4050  7 
5060  4 
6070  3 
Sol. Let the assumed mean = 35; h = 10
Classes  x_{i}  f_{i}  cf  f_{i}u_{i}  
010  5  3  3 + 0 = 3  3  (3) x 3 = 9 
1020  15  8  3 + 8 = 11  2  (2) x 8 = 16 
2030  25  10  11 + 10 = 21  1  (1) x 10 = 10 
3040  35  15  21 + 15 = 36  0  0 x 15 = 0 
4050  45  7  36 + 7 = 43  1  1 x 7= 7 
5060  55  4  43 + 4 = 47  2  2 x 4= 8 
6070  65  3  47 + 3 = 50  3  3 x 3= 9 
Total 
 ∑f_{i} = 50 

 ∑f_{i}u_{i}= 11 
Now,
(i)
(ii) To find mode
Here, highest frequency is 15.
∴ Modal class is 30−40.
Here,
l = 30, f_{1} = 15, f_{0} = 10
f_{2} = 7 and h = 10
(iii) ∵
So, the median class is 30−40.
∴ We have l = 30, cf = 21, f = 15 and h = 10
We have
Q18. Find the mean, mode and median of the following data:
Class  Frequency 
010  3 
1020  4 
2030  7 
3040  15 
4050  10 
5060  7 
6070  4 
Sol. Let the assumed mean ‘a’ = 35
Here, h =10
Classes  x_{i}  f_{i}  cf  f_{i}u_{i}  
010  5  3  3 + 0 = 3  3  (3) x 3 = 9 
1020  15  4  4 + 3 = 7  2  (2) x 4 = 8 
2030  25  7  7 + 7 = 14  1  (1) x 7 = 7 
3040  35  15  15 + 14 = 29  0  0 x 15 = 0 
4050  45  10  10 + 29 = 39  1  1 x 10 = 10 
5060  55  7  7 + 39 = 46  2  2 x 7 = 14 
6070  65  4  4 + 46 = 50  3  3 x 4 = 12 
Total 
 ∑f_{i} = 50 

 ∑u_{i}f_{i} = 12 
(i)
(ii) To find mode
Here, greatest frequency = 15
∴ Modal class = 30 − 40
l = 30, f_{1} = 15, f_{0} = 7, f_{2} = 10 and h = 10
So,
⇒
(iii) To find median
Here,
∴ Median class is 30−40.
Such that l = 30, cf = 14, f = 15 and h = 10