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Now, let us calculate the average marks obtained by the students in each subject.

i) Mean always lie between the maximum and minimum values/observations of the data.

ii) Yes, Its always true. i.e., Mean always lies between the maximum and minimum values of the data.

(a) 295 (b) Delhi (c) Chennai (d) Patna,Jatpur,Delhi,Guwahati 

(i) Lower class limit of 15 – 18 is 15. 

(ii) Upper class limit of 24 – 30 is 30.

(iii) Upper limit of 5 – 12.5 is 12.5

(iv) If the upper and lower limits of a class interval are 16 and 10 ; the class interval is 10 – 16.

(v) If the lower and upper limits of a class interval are 7.5 and 12.5 ; the class interval is 7.5 – 12.5

Correct option is D) ₹59.29

(c) 41

The difference between the lowest and the highest observation in a given data is called its Range.

Then, Range = Highest observation – Lowest observation

= 61 – 20

= 41

Correct option is D) 8.28

First prime numbers are 2, 3, 5, 7, 11, 13 and 17.

\(\therefore\) sum of first 7 prime numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58

\(\therefore\) Mean of first 7 prime numbers = sum/7 = 58/7 = 8.28

Correct option is C) 5.5

First 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

\(\therefore\) sum of first 10 natural numbers = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

\(\therefore\) Mean Average = \(\frac{Sum}{10}=\frac{55}{10}\) = 5.5

Mean = Sum of the given numbers/Total number of given numbers

Mean = (994 + 996 + 998 + 1002 + 1000) ÷ 5

= 4990 ÷ 5

= 998.

(a) X  (b)  VIII  (c)  40  (d)  VI  (e)  160

(i) Family D has maximum members. 

(ii) Family C has minimum members. 

(iii) The difference between number of family members of families D and C = 6 – 2 = 4 

(iv) Total members in all five families = 5 + 4 + 2 + 6 + 4 = 21

(a) Class VIII has the minimum number of girls i.e. only 6 girls.

(b) No, in class V and VI, there are 10 and 16 girls respectively clearly the number of girls is more in class VI than that in V

(c) There are 12 girls in class VII.

(a) 14 bulbs were sold on friday.

(b) The maximum numbers of bulbs (i.e., 18) were sold on Sunday.

(c) Equal numbers of bulbs (i.e., 8) were sold on Wednesday and Saturday.

(d) The minimum numbers of bulbs (i.e., 8) were so Id on Wednesday and Saturday.

(e) Total bulbs sold in the week = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86

(a) 6 symbol will represent animals of village E as there were 60 animals in this village.

(b) Village B has the maximum number of animals i.e., 120.

(c) Village A and C have 80 and 90 animals in it clearly village C has more animals.

From the above pictograph, it can be observed that the number of fruit basket sold by Rahim, Lakhanpai Anwar, Martin, Ranjit sigh, and Joseph are 400, 550, 700, 950, 800 and 450 respectively.

(a) Martin sold the maximum number of baskets i.e, 950

(b) Anwar sold 700 baskets.

(c) Anwar, martin, and Ranjith singh are the 3 merchants who have sold more than 600 baskets.

Therefore they are planning to by a godown for the next season.

Given, number of students who took MDM 132, 164, 145, 182, 163, 114. 

Arithmetic Mean = \(\frac {Sum\,of\,observations}{Number\,of\,observations}\)

= \(\frac {132+164+145+182+163+114}{6} = \frac {900}{6} = 150\)

(a) Cricket has heighest value for Class VIII students in the graph.

∴ Cricket is liked the most by Class VIII students.

(b) Number of students like Hockey of Class VII = 7

Number of students like Tennis of Class VII = 10

∴ 7 + 10 = 17 students of Class VII like Hockey and Tennis.

(c) Total number of students in Class VII = 7 + 16 + 18 + 10 + 14 = 65

(d) For Cricket only, the number of students of Class VII is less than that of Class VIII.

(e) For Hockey Football, Tennis and Badminton i.e., 4 sports, students of Class VIII are less than that of Class VII.

(f) Number of students like Badminton in Class VII = 14

Number of students like Tennis in Class VIII = 7

∴ Required ratio = 14/7 = 2/1 i.e., 2:1.

i) Most of the people liked Apple.

ii) 10.

(a) The double bar graph represents the information about the number of girls and boys in different sections of Class VII

(b) Total number of boys in all sections of Class VIII = 15 + 30 + 20 + 20 + 25 =110

(c) In sections A and D, the number of girls is greater than the number of boys.

(d) In section B, the number of boys is the maximum.

(e) In section C, the number of girls is the least.

(b) 40%

From the question it is given that, in a school, only 2 out of 5 students can participate in a quiz.

So, the chance that a student picked at random = 2/5

Then percentage = 2/5 × 100

= 0.4 × 100

= 40 %

• Class intervals are shown on horizontal line.
• Length of bars shows frequency of class intervals.

1. True
2. True
3. True
4. True
5. False.

We use histogram to show the data of (ii), (iv) and (v).

(c) Since, Range = Highest observation – Lowest observation

(i) 10 families.

(ii) 80 families.

(iii) 130 families (= 10 + 30 + 50 + 40).

(a) Number of students scored marks below 40 i.e., 30-39 ate 4 . 4 students have failed.

(b) Number of students got marks from 50 to 59 is 7 and number of students got marks from 60 to 69 is 11.

∴ 7 + 11 = 18 students got marks from 50 to 69.

(c) 4 students scored marks from 90 and above.

(d) Number of students got marks from 80 to 89 is 6. Number of students got marks from 90 to 92 is 4.

∴ 6 + 4 = 10 students are merit holders.

(e) Total number of students in the class = 4 + 2 +7 + 11 + 8 + 6 + 4 = 42

(b) Mode

Mode is the observation that occurs most frequently in the data.

(i) The maximum number of students watched T.V. for 4 – 5 hours.

(ii) 34 students watched T.V. for less than 4 hours.

(iii) 14 students spent more than 5 hours in watching T.V.

(i) 10% represents 100 people.

Therefore 20% represents = 100 x 20/10 = 200 pepole

Hence, 200 people were surveyed.

(ii) Light music is liked by the maximum number of people.

(iii) CD’s of classical music = 10 x 1000/100 = 100

CD’s of semi-classical music = 20 x 1000/100 = 100

CD’s of light music = 40 x 1000/100 = 400

CD’s of folk music = 30 x 1000/100 = 300

Frequency distribution table:

The required frequency table will be as shown below :

(i) Workers getting less than Rs 850 per week

No. of workers getting Rs 750 = 3 workers

No. of workers getting Rs 800 = 4 works

∴ Workers getting less than Rs 850 = 4 + 3 = 7 workers

(ii) Workers are getting more than Rs 800 per week

No. of workers getting Rs 850 = 3

No. of workers getting Rs 900 = 3

No. of workers getting Rs 950 = 2

∴ Workers getting more than Rs 800 = 3 + 3 + 2 = 8 worker

Required frequency table for given data:

Given that,

6, 3, 5, 6, 4, 3, 2, 5, 4, 2, 4, 2, 1, 2, 2,

0, 5, 4, 6, 1, 6, 0, 5, 3, 6, 1, 5, 5, 2, 6

From the above set we can observe that 0 occurs for 2 times,1 occurs for 3times, 2 occurs for 6times, 3 occurs for 3times, 4 occurs for 4times, 5 occurs for 6times, 6 occurs for 6 times.

The frequency table for the number of accidents per day for a period of 30 days is shown below:

False.

In a bar graph, bars of uniform width are drawn horizontally or vertically with equal spacing between them.

In a bar graph, bars of Uniform width can be drawn horizontally or vertically with Equal spacing between them.

False.

In a bar graph, bars of uniform width are drawn horizontally or vertically with equal spacing between them.

In a bar graph, bars of uniform width are drawn vertically only.False

False.

In a bar graph, bars of uniform width are drawn horizontally or vertically with equal spacing between them.

In a bar graph, bars of uniform width can be drawn horizontally or vertically with equal spacing between them.

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34, 35, 35, 38, 42

Here, n = 7

Therefore median = ((n+1)/2)^th term

Median = value of 4^th term

Median = 35

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

True.

The difference between the upper and lower limit of a class interval is called the

Size of the class interval. i.e. 68 – 60 = 8

First we have to arrange given data into ascending order,

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (45 + 54)/2

= 49.5

Hence median for given data = 49.5

First we have to arrange given data into ascending order

25, 27, 28, 31, 35, 36, 38, 40

Given number of observations, n = 8 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 4^th term + value of 5^th term)/2

= (31 + 35)/2

= 33

Hence median for given data = 33

Correct answer is (c) 68.

First we have to arrange given data into ascending order,

73, 85, 89, 94, 100, 104, 108, 120, 133

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (94 + 100)/2

= 97

Hence median for given data = 97

(i) 830 – 840 group has the maximum number of workers. 

(ii) 10 workers can earn more than ₹ 850. 

(iii) 20 workers earn less than ₹ 850.

It is certain that sun setting every evening

∴ Probability of sun setting tomorrow is 1.

The data is 5, 7, 7, 8, 1,5, 4, 3, 1, 2

Mean = Sum of all observations/ Number of observations

4.5 = (5+7+7+8+x+5+4+3+1+2)/10

45 = x +42

x = 45-42 = 3

Now, arranging the data in ascending order, we get
1, 2, 3, 3, 4, 5, 5, 7, 7, 8

Since, observations are even in number.

∴ Median = Mean of two middle most observations

= (4+5)/2

= 9/2

= 4.5

False

Median of the data: 4, 5, 9, 2, 6, 8, 7 is 2.

False.

The mode of the observations 23, 26, 15, 12, 28, 38, 19, 23, 26, 23 is 28.

Mean = Sum of runs/ No. of innings

= (68 + 03 + 17 + 78 + 12 + 104)/6

= 282/6

= 47