Contents
GSEB Solutions for Class 10 mathematics – Statistics (English Medium)
Exercise-15.1
Question 1:
Find the mean of the following frequency distribution :
Solution :
Question 2:
Find the mean wage of 200 workers of a factory where wages are classified as follows :
Solution :
Question 3:
Marks obtained by 140 students of class X out of 50 in mathematics are given in the following distribution. Find the mean by method of assumed mean method :
Solution :
Let A = 25
Class | Frequency (f_{i}) | x_{i} | d_{i}= x_{i}-A | f_{i}d_{i} |
0-10 | 20 | 5 | -20 | -400 |
10-20 | 24 | 15 | -10 | -240 |
20-30 | 40 | 25 = A | 0 | 0 |
30-40 | 36 | 35 | 10 | 360 |
40-50 | 20 | 45 | 20 | 400 |
Total | ∑f_{i}= 140 | ∑f_{i}d_{i}= 120 |
Question 4:
Find the mean of the following frequency distribution by step-deviation method :
Solution :
Question 5:
Find the mean for the following frequency distribution :
Solution :
Question 6:
A survey conducted by a student Of B.B.A. for daily income Of 600 families is as follows, find the mean income of a family :
Solution :
Question 7:
The number of shares held by a person of various companies are as follows. Find the mean :
Solution :
Question 8:
The mean of the following frequency distribution of 100 observations is 148. Find the missing frequencies f_{1} and f_{2} :
Solution :
Question 9:
The table below gives the percentage of girls in higher secondary science stream of rural areas of various states of India. Find the mean percentage of girls by step-deviation method :
Solution :
Question 10:
The following distribution shows the number of out door patients in 64 hospitals as follows. If the mean is 18, find the missing frequencies f_{1} and f_{2} :
Solution :
Exercise-15.2
Question 1:
Find the mode for the following Frequency distribution :
Solution :
Question 2:
The data obtained for 100 shops for their daily profit per shop are as follows :
Find the modal profit per shop.
Solution :
Question 3:
Daily wages of 90 employees of a factory are as follows :
Find the modal wage of an employee.
Solution :
Question 4:
Find the mode for the following data : (4 and 5)
Solution :
Question 5:
Solution :
Question 6:
The following data gives the information of life of 200 electric bulbs (in hours) as follows :
Find the modal life of the electric bulbs.
Solution :
Exercise-15.3
Question 1:
Find the median for the following .
Solution :
We will prepare the table containing the cumulative frequency as below:
Value of variable | Frequency (f) | Cumulative frequency (cf) |
12 | 7 | 7 |
13 | 10 | 7 + 10 = 17 |
14 | 15 | 17 + 15 = 32 |
15 | 18 | 32 + 18 = 50 |
16 | 20 | 50 + 20 = 70 |
17 | 10 | 70 + 10 = 80 |
18 | 9 | 80 + 9 = 89 |
19 | 8 | 89 + 8 = 97 |
20 | 3 | 97 + 3 = 100 |
Question 2:
Find the median for the following frequency distribution :
Solution :
Let us prepare the table containing the cumulative frequencies as below :
Class | Frequency (f) | Cumulative frequency (cf) |
4 – 8 | 9 | 9 |
8 – 12 | 16 | 9 + 16 = 25 |
12 – 16 | 12 | 25 + 12 = 37 |
16 – 20 | 7 | 37 + 7 = 44 |
20 – 24 | 15 | 44 + 15 = 59 |
24 – 28 | 1 | 59 + 1 = 60 |
Question 3:
Find the median from following frequency distribution :
Solution :
We will prepare the table containing the cumulative frequencies as below :
Class | Frequency (f) | Cumulative frequency (cf) |
0 – 100 | 64 | 64 |
100 – 200 | 62 | 64 + 62 = 126 |
200 – 300 | 84 | 126 + 84 = 210 |
300 – 400 | 72 | 210 + 72 = 282 |
400 – 500 | 66 | 282 + 66 = 348 |
500 – 600 | 52 | 348 + 52 = 400 |
Question 4:
The following frequency distribution represents the deposits (in thousand rupees) and the number of depositors in a bank. Find the median of the data :
Solution :
We will prepare the table containing the cumulative frequencies as below:
Deposit |
Number of depositors (f) |
Cumulative frequency |
0 – 10 | 1071 | 1071 |
10 – 20 | 1245 | 1071 + 1245 = 2316 |
20 – 30 | 150 | 2316+150 = 2466 |
30 – 40 | 171 | 2466 + 171 = 2637 |
40 – 50 | 131 | 2637 + 131 = 2768 |
50 – 60 | 8 | 2768 + 8 = 2776 |
Question 5:
The median of the following frequency distribution is 38. Find the value of a and b if the sum of frequences is 400 :
Solution :
Let us prepare the table containing the cumulative frequency as below:
Class | Frequency (f) | Cumulative frequency (cf) |
10 – 20 | 42 | 42 |
20 – 30 | 38 | 42 + 38 = 80 |
30 – 40 | a | 80 + a =80+a |
40 – 50 | 54 | 80 + a + 54 = 134 + a |
50 – 60 | b | 134 + a + b=134 + a + b |
60 – 70 | 36 | 134 + a + b = 134 + a+b |
70 – 80 | 32 | 170 + a + b +32 = 202 + a + b |
Question 6:
The median of 230 observations of the following frequency distribution is 46. Find a and b :
Solution :
We will prepare the table containing the cumulative frequencies as below :
Class | Frequency (f) | Cumulative frequency (cf) |
10 – 20 | 12 | 12 |
20 – 30 | 30 | 12+30 = 42 |
30 – 40 | a | 42 + a = 42 + a |
40 – 50 | 65 | 42 + a + 65 =107 + a |
50 – 60 | b | 107 + a + b =107 + a + b |
60 – 70 | 25 | 107+a+b + 25 = 132 + a + b |
70 – 80 | 18 | 132+ a + b + 18 = 150 + a + b |
Question 7:
The following table gives the frequency distribution of marks scored by 50 students of class X in mathematics examination of 80 marks. Find the median of the data :
Solution :
Let us prepare the table containing the cumulative frequencies as below:
Class | Frequency (f) | Cumulative frequency (cf) |
0 – 10 | 2 | 2 |
10 – 20 | 5 | 2 + 5 = 7 |
20 – 30 | 8 | 7 + 8 = 15 |
30 – 40 | 16 | 15 + 16 = 31 |
40 – 50 | 9 | 31 + 9 = 40 |
50 – 60 | 5 | 40 + 5 = 45 |
60 – 70 | 3 | 45 + 3 = 48 |
70 – 80 | 2 | 48 + 2 = 50 |
Exercise-15
Question 1:
In a retail market, a fruit vendor was selling apples kept in packed boxes. These boxes contained varying number of apples. The following was the distribution of apples according to the number of boxes. Find the mean by the assumed mean number of apples kept in the box.
Solution :
Let A = 57.5
Number of apples (Class) | Number of boxes (f_{i}) | Midpoint (x_{i}) | d_{i}= x_{i }– A | f_{i}d_{i} |
50 – 53 | 20 | 51.5 | -6 | -120 |
53 – 56 | 150 | 54.5 | -3 | -450 |
56 – 59 | 115 | 57.5 = A | 0 | 0 |
59 – 62 | 95 | 60.5 | 3 | 285 |
62 – 65 | 20 | 63.5 | 6 | 120 |
Total | ∑f_{i}= 400 | ∑f_{i}d_{i}= -165 |
Question 2 :
The daily expenditure of 50 hostel students are as follows :Find the mean daily expenditure of the students of hostel using appropriate method.
Solution :
Let A = 150 and c = 20
Daily expense (in Rs.) Class | No. of students (f_{i}) | Midpoint (x_{i}) | f_{i}u_{i} | |
100 – 120 | 12 | 110 | -2 | -24 |
120 – 140 | 14 | 130 | -1 | -14 |
140 – 160 | 8 | 150 = A | 0 | 0 |
160 – 180 | 6 | 170 | 1 | 6 |
180 – 200 | 10 | 190 | 2 | 20 |
Total | ∑f_{i}= 50 | ∑f_{i}u_{i}= -12 |
Question 3 :
The mean of the following frequency distribution of 200 observations is 332. Find the value of x and y.
Solution :
Let A = 375 and c = 50
Class | Frequency (f_{i}) | Midpoint (x_{i}) | f_{i}u_{i} | |
100-150 | 4 | 125 | -5 | -20 |
150-200 | 8 | 175 | -4 | -32 |
200-250 | X | 225 | -3 | -3x |
250-300 | 42 | 275 | -2 | -84 |
300-350 | 50 | 325 | -1 | -50 |
350-400 | Y | 375 = A | 0 | 0 |
400-450 | 32 | 425 | 1 | 32 |
450-500 | 6 | 475 | 2 | 12 |
500-550 | 4 | 525 | 3 | 12 |
Total | ∑f_{i}= 146 + x + y | ∑fiui = -130 – 3x |
Question 4 :
Find the mode of the following frequency distribution :
Solution :
Question 5 :
Find the mode of the following data :
Solution :
Question 6 :
The mode of the following frequency distribution of 165 observations is 34.5. Find the value of a and b.
Solution :
Question 7 :
Find the mode of the following frequency distribution :
Solution :
Question 8 :
Find the median of the following frequency distribution :
Solution :
Class | Frequency (f) | Cumulative Frequency (cf) |
10-20 | 9 | 9 |
20-30 | 11 | 9 + 11 = 20 |
30-40 | 15 | 20 + 15 = 35 |
40-50 | 24 | 35 + 24 = 59 |
50-60 | 19 | 59 + 19 = 78 |
60-70 | 9 | 78 + 9 = 87 |
70-80 | 8 | 87 + 8 = 95 |
80-90 | 5 | 95+ 5 = 100 |
Question 9 :
The median of the following data is 525. Find the value of x and y, if the sum of frequency is 100 :
Solution :
We will prepare the table containing cumulative frequencies as below:
Class | Frequency (f) | Cumulative frequency (cf) |
0 – 100 | 3 | 3 |
100 – 200 | 4 | 3+4=7 |
200 – 300 | x | 7+x=7+x |
300 – 400 | 12 | 7+x+12=19+x |
400 – 500 | 17 | 19+x+17=36+x |
500 – 600 | 20 | 36+x+20=56+x |
600 – 700 | 9 | 56+x+9=65+x |
700 – 800 | y | 65+x+y=65+x+y |
800 – 900 | 8 | 65+x+y+8=73+x+y |
900 – 1000 | 3 | 73+x+y+3=76+x+y |
Question 10:
Select a proper option (a), (b), (c) or (d) from given options :
Question 10(1) :
For some data, if Z = 25 and = 25, then M = ………
Solution :
Question 10(2) :
For some data Z – M = 2.5. If the mean of the data is 20, then Z = …….
Solution :
Question 10(3) :
If – Z = 3 and + Z = 45, then M = …….
Solution :
Question 10(4) :
If Z = 24, = 18, then M = …….
Solution :
Question 10(5) :
If M = 15, = 10, then Z = …….
Solution :
Question 10(6) :
If M = 22, Z = 16, then = …….
Solution :
Question 10(7) :
If = 21.44 and Z = 19.13, then M = ………
Solution :
Question 10(8) :
If M = 26, = 36, then Z = ……..
Solution :
Question 10(9) :
The modal class of the frequency distribution given below is ……
Solution :
c. 30 – 40
Here, the maximum frequency is 17 and the corresponding class of this frequency is 30 – 40. Hence, the modal class is 30 – 40.
Question 10(10) :
The cumulative frequency of class 20-30 of the frequency distribution given in (9) is ………
Solution :
b. 35
Cumulative frequency of class 20 – 30 = frequency of class 20 – 30 and the frequencies of all the classes preceding class 20 – 30 = 7 + 15 + 13 = 35
Question 10(11) :
The median class of the frequency distribution given in (9) is ………
Solution :
Kishan patel says
Nice solutions love it
sherine says
you have solved by stepdeviation method can you help me by assumed method plz
Bhavesh says
Nice
Deep says
Nice