III. LONG ANSWER TYPE QUESTIONS
Q1. The median of the following frequency distribution is 35. Find the value of x.
Class Interval  Frequency  Cumulative frequency 
020  7  7 
2040  8  15 
4060  12  27 
6080  10  37 
80100  8  45 
100120  5  50 
Total  50 

Also find the modal class.
Sol. Let us prepare the cumulative frequency table:
Class intervals  f  ^{cf} 
010  2  2 + 0 = 2 
1020  3  2 + 3 = 5 
2030  x  5 + x = (5 + x) 
3040  6  (5 + x) + 6 = 11 + x 
4050  5  (11 + x) + 5 = 16 + x 
5060  3  (16 + x) + 3 = 19 + x 
6070  2  (19 + x) + 2 = 21 + x 
Total  (21 + x) 

Here,
Obviously, lies in the class interval 30−40.
∴
l = 30, Cf = x + 5, f = 6 and h = 10
⇒ 5 × 12 = 110 − 10x
⇒ 10x = 110 − 60
⇒ 10x =50 ⇒ x = 5
(i) ∴ The required value of x is 5.
(ii) ∵ The maximum frequency is 6
∴ The modal class is 30−40.
Q2. The mean of the following data is 53, find the missing frequencies.
Age (in years)  020  2040  4060  6080  80100  Total 
Number of people  15  f  21  f2  17  100 
Sol.
Age (in years)  Number of People (f_{i})  Mid value (xi)  f_{i} × x_{i} 
020  15  10  15 x 10 = 150 
2040  fa  30  f_{1} x 30 = 30 f_{1} 
4060  21  50  21 x 50 = 1050 
6080  f2  70  h x 70 = 70 f2 
80100  17  90  17 x 90 = 1530 
Total  100 
 2730 + 30 f_{1} + 70 f_{2} 
Since, 15 + f_{1} + 21 + f_{2} + 17 = 100
∴ 53 + (f_{1} + f_{2}) = 100
⇒ f_{1} + f_{2} = 100 − 53 = 47 ...(1)
⇒ 2730 + 30 f_{1} + 70 f_{2} = 5300
⇒ 30 f_{1} + 70 f_{2} = 5300 − 2730 = 2570
⇒ 3 f_{1} + 7 f_{2} = 257
⇒ 3 (47 − f_{2}) + 7 f_{2} = 257 ∵ f_{1} = 47 − f_{2}
⇒ 141 − 3 f_{2} + 7 f_{2} = 257
⇒ 4 f_{2} = 116 or f_{2} = 29
∴ f_{1} = 41 − 29 = 18
Thus, f_{1} = 18 and f_{2} = 29.
Q3. The mean of the following distribution is 18.
Class interval  1113  1315  1517  1719  1921  2123  2325 
Frequencies  3  6  9  13  f  5  4 
Find f
Sol. Let the assumed mean (a) = 18
∴ We have the following table:
Class interval  x_{i}  f_{i}  f_{i} u_{i}  
1113  12  3  3  3 x (3) = 9 
1315  14  6  2  6 x (2) = 12 
1517  16  9  1  9 x (1) = 9 
1719  18  13  0  13 x 0 = 0 
1921  20  ^{f}  1  f x 1 = f 
2123  22  5  2  5 x 2 = 10 
2325  24  4  3  3 x 3 = 9 

 ∑ f_{i} = 40 + f 
 ∑f_{i} u_{i} = f− 8 
⇒
Thus, the required frequency = 8
Q4. The percentage of marks obtained by 100 students in an examination are given below:
Marks  30  35  35  40  40  45  45  50  50 55  55  60  6065 
Frequency  14  16  18  23  18  8  3 
Find the median of the above data.
Sol.
Here n = 100 ⇒ n/2 = 50, which lies in the class 45 – 50, where
l_{1} (lower limit of the median – class) = 45
c (The cumulative frequency of the class preceding the median class) = 48
f (The frequency of the median class) = 23
h (The class size) = 5
∴ The median percentage of marks is 45.4.
Q5. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.
Class interval  0  6  6 12  12 18  18  24  24  30 
Frequency  4  x  5  ^{y}  1 
Sol.
Class interval  Frequency  Cumulative Frequency 
06  4  4 + 0 = 4 
612  x  4 + x = (4 + x) 
1218  5  5 + (4 + x) = 9 + x 
1824  ^{y}  y + (9 + x) = 9 + x + y 
2430  1  1 + (9 + x + y) = 10 + x + y 
Since, n = 20
∴ 10 + x + y = 20
⇒ x + y = 20 − 10
⇒ x + y = 10. ...(1)
Also, we have
Median = 14.4,
which lies in the class interval 12−18.
∴ The median class is 12−18, such that
l = 12, f = 5, Cf = 4 + x and h = 6
⇒ 24 + 6x = (9.6) × 5
⇒ 24 + 6x =48
⇒ 6 x = 48 − 24 = 24
⇒ x = 24/6 = 4
Now, from (1), we have:
x + y = 10
⇒ 4 + y = 10
⇒ y = 10 − 4 = 6
Thus, x =4 and y = 6
Q6. The distribution below gives the weights of 30 students of a class. Find the mean and the median weight of the students:
Weight (in kg)  Number of students 
4045  2 
4550  3 
5055  8 
5560  6 
6065  6 
6570  3 
7075  2 
Sol. Let the assumed mean, a = 57.5
∴ h = 5
∴
We have the following table:
Weight of students  Class mark  Frequency ^{f}i  Cumulative frequency  ^{f}i ^{u}i  






4045  42.5  2  2 + 0 = 2  3  6 
4550  47.5  3  2 + 3 = 5  2  6 
5055  52.5  8  5 + 8 = 13  1  8 
5560  57.5  6  13 + 6 = 19  0  0 
6065  62.5  6  19 + 6 = 25  1  6 
6570  67.5  3  25 + 3 = 28  2  6 
7075  72.5  2  28 + 2 = 30  3  6 
Total 
 ∑ f_{i} = 30 

 ^{∑f}i^{u}i = ^{ 2} 
∴ The mean of the given data
For finding the median:
And it lies in the class 55–60.
Q7. The lengths of 40 leaves of a plant are measured correct upto the nearest millimetre and the data is as under:
Length (in mm)  Numbers of Leaves 
118126  4 
126134  5 
134142  10 
142150  12 
150158  4 
158166  5 
Find the mean and median length of the leaves.
Sol. For finding the mean:
Let the assumed mean a = 146
h = 8
Now, we have the following table:
Length (in mm)  Class mark (x_{i})  Frequency (f_{i})  Cumulative frequency Cf  fi ui  
118126  122  4  4 + 0 = 4
 3  12  
126134  130  5  4 + 5 = 9
 2  10  
134142  138  10  9 + 10 = 19
 1  10  
142150  146  12  19 + 12 = 31
 0  0  
150158  154  4  31 + 4 = 35
 1  4  
158166  162  5  35 + 5 = 40
 2  10  
Total 
 ∑ f_{i} = ^{40} 

 ^{∑ f}i ^{u}i =^{18} 
∴ Mean
= 142.4 mm
Median: Since [Median class is 142150]
l = 142
cf = 19
f = 12 and h = 8
Q8. The table below shows the daily expenditure on food of 30 households in a locality:
Daily expenditure (in Rs)  Numbers of households 
100150  6 
150200  7 
200250  12 
250300  3 
300350  2 
Find the mean and median daily expenditure on food.
Sol. For finding the mean
Let the assumed mean a = 225.
h = 50
We have the following table:
Daily expenditure  x_{i}  f_{i}  c.f.  f_{i} u_{i}  
100150  125  6  6 + 0 = 6  2  (2) x 6 = 12 
150200  175  7  6 + 7 = 13  1  (1) x 7 = 7 
200250  225  12  13 + 12 = 25  0  (0) x 12 = 0 
250300  275  3  25 + 3 = 28  1  (1) x 3 = 3 
300350  325  2  28 + 2 = 30  2  (2) x 2 = 4 
Total 
 ∑ f_{i} = 30 

 ∑ f_{i}u_{i} =12 
∴ Mean
⇒
To find median:
And 15 lies in the class 200−250.
∴ Median class is 200−250.
∴ l = 200
cf = 13
f = 12 and h = 50
Thus,
124 videos457 docs77 tests

1. What is statistics and why is it important? 
2. What are the different types of data in statistics? 
3. How is mean, median, and mode used in statistics? 
4. What is the difference between probability and statistics? 
5. How can statistics be misused or misrepresented? 

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