Question 1: Use a graph paper for this question. The daily pocket expenses of students in a school are given below;

Pocket expenses (in Rs.) | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |

Number of students (frequency) | 10 | 14 | 28 | 42 | 50 | 30 | 14 | 12 |

Draw a histogram representing the above distribution and estimate the mode from the graph. **[2014]**

Answer:

Question 2: The marks obtained by 100 students in a Mathematics test are given below;

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

No. of Students | 3 | 7 | 12 | 17 | 23 | 14 | 9 | 6 | 5 | 4 |

Draw an ogive for the distribution on a graph sheet. (Use a scale of 2 cm = 10 units on both axis)

Use the ogive to estimate the:

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the if the pass percentage was 35 **[2014]**

Answer:

Marks | No. of Students | Cumulative Frequency (c.f) |

0-10
10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 |
8
7 12 17 23 14 9 6 5 4 |
3
10 22 39 62 76 85 91 96 100 |

On the graph paper, we plot the following points:

(i) Medium

From the graph

(ii) Lower quartile

From the graph

(iii) The number of students who obtained more than marks in test students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 students.

Question 3: The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks |
0 | 1 | 2 | 3 | 4 | 5 |

No. of Students |
1 | 3 | 6 | 10 | 5 | 5 |

Calculate the mean, median and mode of the above distribution. **[2015]**

Answer:

Below table

0 | 1 | 0 | 1 |

1 | 3 | 3 | 4 |

2 | 6 | 12 | 10 |

3 | 10 | 30 | 20 |

4 | 5 | 20 | 25 |

5 | 5 | 25 | 30 |

Mean

Median

Mode = 3 marks (as highest frequency is 10)

Question 4: Calculate the mean of the following distribution. ** [2015]**

Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 8 | 5 | 12 | 35 | 24 | 16 |

Answer:

Class | Mid Value | ||

0-10 | 5 | 8 | 40 |

10-20 | 15 | 5 | 75 |

20-30 | 25 | 12 | 300 |

30-40 | 35 | 35 | 1225 |

40-50 | 45 | 24 | 1080 |

50-60 | 55 | 16 | 880 |

Total |

Mean

Question 5: The weight of 50 workers is given below:

Weight in kg | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 | 100-110 | 110-120 |

No. of workers | 4 | 7 | 11 | 14 | 6 | 5 | 3 |

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight. ** [2015]**

Answer:

Below the table

Weight | ||

50-60 | 4 | 4 |

60-70 | 7 | 11 |

70-80 | 11 | 22 |

80-90 | 14 | 36 |

90-100 | 6 | 42 |

100-110 | 5 | 47 |

110-120 | 3 | 50 |

Total | 50 |

(i) Lower quartile=

Upper quartile =

(ii) No. of over weight workers

Question 6: The number and are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of ** [2014]**

Answer:

Arrange numbers in ascending order are

Mean

No. of terms

Median

Median

According to given condition

or

Question 8: Calculate the mean of the distribution given below using the short cut method; **[2014]**

Marks | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |

No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |

Answer:

Table as follows

Mean Value | ||||

11-20
21-30 31-40 41-50 51-60 61-70 71-80 |
2
6 10 12 9 7 4 |
15-5
25-5 35-5 45-5 55-5 65-5 75-5 |
-30
-20 -10 0 10 20 30 |
-60
-120 -100 0 90 140 120 |

Question 8: The median of the following observation arranged in ascending order is . Find the value of and hence find the mean. ** [2013]**

Answer:

Given observation are and mediam

Since which is odd, therefore

Therefore,

Now, Mean

Question 9: Draw a histogram from the following frequency distribution and find the made from the graph: **[2013]**

Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |

Frequency | 2 | 5 | 18 | 14 | 8 | 5 |

Answer:

Question 10: Find the mean of the following distribution by step deviation method: ** [2013]**

Class Internal | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency | 10 | 6 | 8 | 12 | 5 | 9 |

Answer:

C.I | Frequency | Mid-Value | ||

20-30
30-40 40-50 50-60 60-70 70-80 |
10
6 8 12 5 9 |
25
35 45 55 65 75 |
-2
-1 0 1 2 3 |
-20
-6 0 12 10 27 |

Here, and

Mean

Question 11: The mark obtained by 120 students in a test are given below;

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

Students | 5 | 9 | 16 | 22 | 26 | 18 | 11 | 6 | 4 | 3 |

Draw an ogive for the given distribution on a graph sheet; Using suitable scale for ogive to estimate the following:

(i) The mediam

(ii) The number of students who obtained more than 75% marks in the test.

(iii) The number of students who did not pass the test if minimum marks required to pass is 40. **[2013]**

Answer:

Marks | No. of students | Cumulative Frequency |

0-10
10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 |
5
9 16 22 26 18 11 6 4 3 |
5
14 30 52 78 96 107 113 117 120 |

N=120 |

On the graph paper we plot the following points:

(i) Mediam

From the graph 60^{th} term

(ii) The number of students who obtained more than marks in test

(iii) The number of students who did not pass the test if the minimum pass marks

Question 12: Marks obtained by students in a short assessment is given below, where are two missing data:

Marks | 5 | 6 | 7 | 8 | 9 |

No. of Students | 6 | A | 16 | 13 | B |

If the mean of the distribution is find . **[2012]**

Answer:

Given, the total number of students

Therefore

… … … (i)

Given mean

Therefore

… … … (ii)

Solving (i) and (ii) we get and

Question 13: The following distribution represents the height of 160 students of a school

Height | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 | 175-180 |

Students | 12 | 20 | 30 | 38 | 24 | 16 | 12 | 8 |

Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm=20 students on the other axis. Using the graph, determine:

(i) The medium height

(ii) The interquartile range

(iii) The number of students whose height is above 172 cm. ** [2012]**

Answer:

Following table:

Height | F | c.f. |

140-145
145-150 150-155 155-160 160-165 165-170 170-175 175-180 |
12
20 30 38 24 16 12 8 |
12
32 62 100 124 140 152 160 |

(i) Mean

(ii) Interquartile range

(iii) No. of students above

Question 14: Find the mode and medium of the following frequency distribution: ** [2012]**

X | 10 | 11 | 12 | 13 | 14 | 15 |

f | 1 | 4 | 7 | 5 | 9 | 3 |

Answer:

Mode is the value of the highest frequency.

Therefore Mode

For Median, first write the data in ascending order as follows:

.

Since Median is the middle most value.

Median

Question 15: A mathematics aptitude test of 50 students was recorded as follows:

Marks | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

No. of students | 4 | 8 | 14 | 19 | 5 |

Draw a histogram for the above data using a graph paper and locate the mode. **[2011]**

Answer:

Question 16: Using step division method, calculate the mean marks of the following distribution: State the modal class: **[2011]**

Class Interval | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 |

Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |

Answer:

C.I | |||||

50-55 | 5 | 52.5 | -15 | -3 | -15 |

55-60 | 20 | 57.5 | -10 | -2 | -40 |

60-65 | 10 | 62.5 | -5 | -1 | -10 |

65-70 | 10 | 67.5 | 0 | 0 | 0 |

70-75 | 9 | 72.5 | 5 | 1 | 9 |

75-80 | 6 | 77.5 | 10 | 2 | 12 |

80-85 | 12 | 82.5 | 15 | 3 | 36 |

85-90 | 8 | 87.5 | 20 | 4 | 32 |

(i)

(ii) Modal class is (class with highest freq.)

Question 17: Marks obtained by 200 students in an examination are given below:

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

No. of students | 5 | 11 | 10 | 20 | 28 | 37 | 40 | 29 | 14 | 6 |

Draw an ogive for the given distribution taking marks on one axis and students on the other axis. Using the graph, determine.

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is .

(iii) If scoring and more marks is considered as grade one, find the number of students who secured grade on in the examination; ** [2011]**

Answer:

Class Interval | Frequency | Cumulative Frequency |

0-10 | 5 | 5 |

10-20 | 11 | 16 |

20-30 | 10 | 20 |

30-40 | 20 | 46 |

40-50 | 28 | 74 |

50-60 | 37 | 111 |

60-70 | 40 | 151 |

70-80 | 29 | 180 |

80-90 | 14 | 194 |

90-100 | 6 | 200 |

(i) (even)

Median observation observation observation

(ii) Number of student who failed

(iii) Number of students who secured grade one

Question 18: The distribution given below shows the marks obtained by students in an aptitude test. Find the mean, median and mode of the distribution.** [2010]**

Marks obtained | 5 | 6 | 7 | 8 | 9 | 10 |

No. of students | 3 | 9 | 6 | 4 | 2 | 1 |

Answer:

5 | 3 | 15 | 3 |

6 | 9 | 54 | 12 |

7 | 6 | 42 | 18 |

8 | 4 | 32 | 22 |

9 | 2 | 18 | 24 |

10 | 1 | 10 | 25 |

Mean =

Since is odd,

Median =

Mode (maximum frequency)

Question 19: The mean of the following distribution is 52 and the frequency of class interval 30-40 is . Find . ** [2010]**

Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency | 5 | 3 | 7 | 2 | 6 | 13 |

Answer:

Interval | Frequency | |||

10-20 | 5 | 15 | -30 | -150 |

20-30 | 3 | 25 | -20 | -60 |

30-40 | F | 35 | -10 | -10f |

40-50 | 7 | 45 (A) | 0 | 0 |

50-60 | 2 | 55 | 10 | 20 |

60-70 | 6 | 65 | 20 | 120 |

70-80 | 13 | 75 | 30 | 390 |

Question 20: The monthly income of a group of 320 employees in a company is given below:

Monthly Income | No. of Employees |

6000-7000 | 20 |

7000-8000 | 45 |

8000-9000 | 65 |

9000-10000 | 95 |

10000-11000 | 60 |

11000-12000 | 30 |

12000-13000 | 5 |

Draw an ogive of the given. distribution on a graph sheet taking on one axis and on the other axis. From the graph determine:

(i) the median wage

(ii) the number of employees whose income is below

(iii) If the salary of a senior employee is above find the number of senior employees in the company.

(iv) The upper quartile** [2010]**

Answer:

Monthly Income | No. of Employees | Cumulative Frequency |

6000-7000 | 20 | 20 |

7000-8000 | 45 | 65 |

8000-9000 | 65 | 130 |

9000-10000 | 95 | 225 |

10000-11000 | 60 | 285 |

11000-12000 | 30 | 315 |

12000-13000 | 5 | 320 |

Here n (no. of employees) (even)

(i) Median

Required median (from graph)

(ii) Number of employees whose income is below approx

(iii) Number of senior employees in the company

(iv) Upper Quartile term

Upper Quartile