Data Handling

In a ready-made garment shop, on a particular day the following sizes of shirts were sold:

34, 38, 42, 40, 44, 32, 34, 36, 42, 40, 44, 36, 38, 42, 44, 40, 38, 40, 42, 32, 34, 38, 42, 40, 36, 42, 40, 38, 36, 40

Arrange the above data in ascending order and construct frequency distribution table. Also answer the following questions:

(i) Which shirt size had the maximum sale?

(ii) Which shirt size had the minimum sale?

(iii) The number of shirts sold of size 42 or greater than size 42.

Answer

Arranging the given data in ascending order, we get:

32, 32, 34, 34, 34, 36, 36, 36, 36, 38, 38, 38, 38, 38, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 44, 44, 44

The frequency distribution table for the given data is:

(i) The shirt size 40 had the maximum sale as it has the highest frequency (7).

Hence, shirt size 40 had the maximum sale.

(ii) The shirt size 32 had the minimum sale as it has the lowest frequency (2).

Hence, shirt size 32 had the minimum sale.

(iii) The number of shirts sold of size 42 or greater than size 42 = 6 + 3 = 9.

Hence, 9 shirts of size 42 or greater than size 42 were sold.

In a Mathematics test, the following marks were obtained by 40 students:

8, 1, 3, 7, 6, 5, 5, 4, 4, 2

4, 9, 5, 3, 7, 1, 6, 5, 2, 7

7, 3, 8, 4, 2, 8, 9, 5, 8, 6

7, 4, 5, 6, 9, 6, 4, 4, 6, 6

(i) Construct frequency distribution table for the above data.

(ii) Find how many students obtained 7 marks or more than 7 marks.

(iii) How many students obtained marks below 4?

Answer

(i) The frequency distribution table for the given data is:

(ii) The number of students who obtained 7 marks or more than 7 marks = 5 + 4 + 3 = 12.

Hence, 12 students obtained 7 marks or more than 7 marks.

(iii) The number of students who obtained marks below 4 = 2 + 3 + 3 = 8.

Hence, 8 students obtained marks below 4.

The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:

(i) Which class has the minimum number of girl students?

(ii) Is the number of girls in Class VI less than the number of girls in Class V?

(iii) How many girls are there in Class VII?

Answer

In the given pictograph, one symbol represents 4 girls.

The number of girl students in each class is:

(i) Class VIII has the minimum number of girl students (6 girls).

Hence, Class VIII has the minimum number of girl students.

(ii) Number of girls in Class VI = 16 and number of girls in Class V = 10.

Since 16 > 10, the number of girls in Class VI is not less than the number of girls in Class V.

Hence, No, the number of girls in Class VI is not less than the number of girls in Class V.

(iii) Number of girls in Class VII = 3 × 4 = 12.

Hence, there are 12 girls in Class VII.

In a village, the following pictograph shows the number of fruit baskets sold by six merchants in a particular season:

Observe the above pictograph and answer the following questions:

(i) Which merchant sold the maximum number of baskets?

(ii) How many baskets were sold by Rajinder Singh?

(iii) The merchants who have sold 700 or more number of baskets are planning to buy a cold store for the next season. Can you name them?

Answer

In the given pictograph, one symbol represents 100 baskets.

The number of fruit baskets sold by each merchant is:

(i) Anwar sold the maximum number of baskets (950 baskets).

Hence, Anwar sold the maximum number of baskets.

(ii) Rajinder Singh sold 7 $\dfrac{1}{2}$ × 100 = 750 baskets (nearly).

Hence, nearly 750 baskets were sold by Rajinder Singh.

(iii) The merchants who sold 700 or more baskets are:

• Anwar (950 baskets)
• Rajinder Singh (750 baskets)
• Joseph (700 baskets)

Hence, Anwar, Rajinder Singh and Joseph are planning to buy a cold store.

In Gurugram, the number of cars sold during a particular week was as follows:

Monday : 80, Thursday : 60

Tuesday : 70, Friday : 70

Wednesday : 90, Saturday : 40

Prepare a pictograph of the cars sold using a symbol of car representing 10 cars and answer the following questions:

(i) On which day the maximum number of cars were sold?

(ii) How many pictures of cars will represent the number of cars sold on Thursday?

Answer

Since one symbol of car represents 10 cars, the number of symbols required for each day is:

The pictograph for the given information is:

(i) The maximum number of cars (90) were sold on Wednesday.

Hence, the maximum number of cars were sold on Wednesday.

(ii) The number of cars sold on Thursday = 60.

Number of pictures of cars = $\dfrac{60}{10}$ = 6.

Hence, 6 pictures of cars will represent the number of cars sold on Thursday.

Total number of animals in five villages are as follows:

Village A : 80, Village B : 120

Village C : 90, Village D : 40

Village E : 60

Prepare a pictograph of these animals using one symbol ⊗ to represent 10 animals and answer the following questions:

(i) How many symbols represent animals of village E?

(ii) Which village has the maximum number of animals?

(iii) Which village has more animals: village A or village C?

Answer

Since one symbol represents 10 animals, the number of symbols required for each village is:

The pictograph for the given information is:

(i) Number of symbols for village E = $\dfrac{60}{10}$ = 6.

Hence, 6 symbols represent animals of village E.

(ii) Village B has the maximum number of animals (120).

Hence, Village B has the maximum number of animals.

(iii) Village A has 80 animals and Village C has 90 animals.

Since 90 > 80, Village C has more animals than Village A.

Hence, Village C has more animals than Village A.

Observe the bar graph given below which is showing the number of students in a particular class of a school.

Answer the following questions:

(i) What is the scale of this graph?

(ii) How many new students are added every year?

(iii) Is the number of students in the year 2015 twice than that of in the year 2012?

Answer

From the bar graph, we observe the number of students each year:

(i) The scale of this graph is 1 unit height = 10 students.

Hence, the scale is 1 unit height = 10 students.

(ii) Number of new students added every year:

• From 2012 to 2013: 40 − 30 = 10
• From 2013 to 2014: 50 − 40 = 10
• From 2014 to 2015: 60 − 50 = 10

Hence, 10 new students are added every year.

(iii) Number of students in 2015 = 60 and number of students in 2012 = 30.

Twice the number of students in 2012 = 2 × 30 = 60 = Number of students in 2015.

Hence, yes, the number of students in 2015 is twice the number of students in 2012.

Observe the bar graph given below which is showing the sale of shirts in a ready-made garment shop from Monday to Saturday.

Answer the following questions:

(i) What information does the above bar graph give?

(ii) What is the scale chosen on the horizontal line representing number of shirts?

(iii) On which day were the maximum number of shirts sold? How many shirts were sold on that day?

(iv) On which day were the minimum number of shirts sold?

(v) How many shirts were sold on Thursday?

Answer

From the bar graph, we observe the number of shirts sold each day:

(i) The bar graph shows the number of shirts sold from Monday to Saturday.

(ii) The scale chosen on the horizontal line is 1 unit length = 5 shirts.

(iii) The maximum number of shirts was sold on Saturday, and 60 shirts were sold on that day.

Hence, the maximum number of shirts (60) was sold on Saturday.

(iv) The minimum number of shirts was sold on Tuesday (10 shirts).

Hence, the minimum number of shirts was sold on Tuesday.

(v) 35 shirts were sold on Thursday.

Make a table corresponding to the following graph:

Answer

From the bar graph, reading the height of each bar, we get the number of students for each favourite cricketer.

The table corresponding to the given bar graph is:

The following table shows the number of bicycles manufactured in a factory during the years 2011 to 2015. Illustrate this data by using a bar graph. Choose a scale of your choice.

(i) In which year were the maximum number of bicycles manufactured?

(ii) In which year were the minimum number of bicycles manufactured?

Answer

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the years 2011, 2012, 2013, 2014 and 2015 at equal distances.

3. Along the y-axis, mark the number of bicycles manufactured.

4. Choose a scale of 1 unit height = 100 bicycles manufactured.

5. The heights of the bars for various years are:

• 2011 : 800 ÷ 100 = 8 units
• 2012 : 600 ÷ 100 = 6 units
• 2013 : 900 ÷ 100 = 9 units
• 2014 : 1100 ÷ 100 = 11 units
• 2015 : 1200 ÷ 100 = 12 units

6. Draw bars of equal width keeping equal gap between adjoining bars.

The required bar graph is as given below:

(i) The maximum number of bicycles (1200) were manufactured in the year 2015.

Hence, the maximum number of bicycles were manufactured in the year 2015.

(ii) The minimum number of bicycles (600) were manufactured in the year 2012.

Hence, the minimum number of bicycles were manufactured in the year 2012.

The number of Mathematics books sold by a shopkeeper on six consecutive days is given below:

Draw a horizontal bar graph to represent the above information choosing the scale of your choice.

Answer

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the y-axis, mark the days of the week (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday) at equal distances.

3. Along the x-axis, mark the number of books sold.

4. Choose a scale of 1 unit length = 10 books sold.

5. The lengths of the bars for various days are:

• Sunday : 65 ÷ 10 = 6.5 units
• Monday : 40 ÷ 10 = 4 units
• Tuesday : 30 ÷ 10 = 3 units
• Wednesday : 50 ÷ 10 = 5 units
• Thursday : 20 ÷ 10 = 2 units
• Friday : 70 ÷ 10 = 7 units

6. Draw bars of equal width keeping equal gap between adjoining bars.

The required horizontal bar graph is as given below:

The number of persons in various age (in years) groups in a town is given in the following table:

Draw a bar graph to represent the above information and answer the following questions (take 1 unit height = 20,000 people):

(i) Which age groups have same population?

(ii) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?

Answer

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the age groups at equal distances.

3. Along the y-axis, mark the number of persons.

4. Take scale: 1 unit height = 20,000 persons.

5. The heights of the bars for various age groups are:

• 1-14 : 2,00,000 ÷ 20,000 = 10 units
• 15-29 : 1,60,000 ÷ 20,000 = 8 units
• 30-44 : 1,20,000 ÷ 20,000 = 6 units
• 45-59 : 1,20,000 ÷ 20,000 = 6 units
• 60-74 : 80,000 ÷ 20,000 = 4 units
• 75 and above : 40,000 ÷ 20,000 = 2 units

6. Draw bars of equal width keeping equal gap between adjoining bars.

The required bar graph is as given below:

(i) The age groups 30-44 and 45-59 have the same population (1,20,000 each).

Hence, age groups 30-44 and 45-59 have the same population.

(ii) Number of senior citizens (60 and above) = 80,000 + 40,000 = 1,20,000.

Hence, there are 1,20,000 senior citizens in the town.

The number of tigers in India reduced drastically between the years 1900 and 1970, from estimated 1 lakh to less than 2000. Government of India launched Project Tiger in 1973 from the Jim Corbett National Park in Uttarakhand. Now there are 47 such tiger reserves in India. The tiger population has gone up substantially. The following table shows approximate numbers, rounded off to nearest hundreds, as per Tiger census reports:

Draw a bar chart for this data. Indicate the number of tigers on top of each bar. Do not draw vertical axis, to have a clean chart.

Answer

Steps:

1. Draw a horizontal line OX on a graph paper.

2. Along OX, mark the years 2006, 2010, 2014, 2018 and 2022 at equal distances.

3. Draw bars of equal width with heights proportional to the number of tigers for each year.

4. Write the number of tigers on top of each bar.

5. Do not draw the vertical axis to have a clean chart.

The required bar chart is as given below:

Find the mean of the following data:

(i) 40, 30, 30, 0, 26, 60

(ii) 3, 5, 7, 9, 11, 13, 15

Answer

(i) Number of observations = 6.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{40 + 30 + 30 + 0 + 26 + 60}{6} \\[1em] = \dfrac{186}{6} \\[1em] = 31$

Hence, the mean of the given data is 31.

(ii) Number of observations = 7.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{3 + 5 + 7 + 9 + 11 + 13 + 15}{7} \\[1em] = \dfrac{63}{7} \\[1em] = 9$

Hence, the mean of the given data is 9.

Find the mean of the first five even whole numbers.

Answer

The first five even whole numbers are 0, 2, 4, 6 and 8.

Number of observations = 5.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{0 + 2 + 4 + 6 + 8}{5} \\[1em] = \dfrac{20}{5} \\[1em] = 4$

Hence, the mean of the first five even whole numbers is 4.

A batsman scored the following number of runs in six innings:

36, 35, 50, 46, 60, 55

Calculate the mean runs scored by him in an inning.

Answer

Number of innings = 6.

$\text{Mean runs} = \dfrac{\text{Sum of all runs scored}}{\text{Number of innings}} \\[1em] = \dfrac{36 + 35 + 50 + 46 + 60 + 55}{6} \\[1em] = \dfrac{282}{6} \\[1em] = 47$

Hence, the mean runs scored by the batsman in an inning is 47.

The enrolment in a school during six consecutive years was as follows:

1555, 1670, 1750, 2013, 2540, 2825

Find the mean enrolment of the school for this period.

Answer

Number of years = 6.

$\text{Mean enrolment} = \dfrac{\text{Sum of enrolments}}{\text{Number of years}} \\[1em] = \dfrac{1555 + 1670 + 1750 + 2013 + 2540 + 2825}{6} \\[1em] = \dfrac{12353}{6} \\[1em] \approx 2059$

Hence, the mean enrolment of the school for this period is 2059 (nearly).

The marks (out of 100) obtained by a group of students in a science test are:

85, 76, 90, 85, 39, 48, 56, 95, 81, 75

Find the:

(i) highest and lowest marks obtained by the students.

(ii) mean marks obtained by the students.

Answer

Arranging the given marks in ascending order, we get:

39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) Highest marks = 95 and lowest marks = 39.

Hence, the highest marks obtained by the students is 95 and the lowest marks is 39.

(ii) Number of students = 10.

$\text{Mean marks} = \dfrac{\text{Sum of all marks}}{\text{Number of students}} \\[1em] = \dfrac{85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75}{10} \\[1em] = \dfrac{730}{10} \\[1em] = 73$

Hence, the mean marks obtained by the students is 73.

Find the median of the following data:

(i) 3, 1, 5, 6, 3, 4, 5

(ii) 3, 1, 5, 6, 3, 4, 5, 6

Answer

(i) Arranging the given data in ascending order, we get:

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

Total number of observations = 7 (odd).

The middle observation is the 4^th observation, which is 4.

Hence, the median of the given data is 4.

(ii) Arranging the given data in ascending order, we get:

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

Total number of observations = 8 (even).

There are two middle items: 4^th term and 5^th term, which are 4 and 5.

$\text{Median} = \dfrac{\text{Sum of middle terms}}{2} \\[1em] = \dfrac{4 + 5}{2} \\[1em] = \dfrac{9}{2} \\[1em] = 4.5$

Hence, the median of the given data is 4.5.

Calculate the mean and the median of the following numbers:

1, 3, 2, 6, 2, 3, 1, 3

Answer

Number of observations = 8.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{1 + 3 + 2 + 6 + 2 + 3 + 1 + 3}{8} \\[1em] = \dfrac{21}{8} \\[1em] = 2.625$

Arranging the given data in ascending order, we get:

1, 1, 2, 2, 3, 3, 3, 6

Total number of observations = 8 (even).

There are two middle items: 4^th term and 5^th term, which are 2 and 3.

$\text{Median} = \dfrac{\text{Sum of middle terms}}{2} \\[1em] = \dfrac{2 + 3}{2} \\[1em] = \dfrac{5}{2} \\[1em] = 2.5$

Hence, the mean of the given data is 2.625 and the median is 2.5.

Calculate the mean and the median of the following numbers:

3, 7, 2, 5, 3, 4, 1, 5, 3, 6

Answer

Number of observations = 10.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{3 + 7 + 2 + 5 + 3 + 4 + 1 + 5 + 3 + 6}{10} \\[1em] = \dfrac{39}{10} \\[1em] = 3.9$

Arranging the given data in ascending order, we get:

1, 2, 3, 3, 3, 4, 5, 5, 6, 7

Total number of observations = 10 (even).

There are two middle items: 5^th term and 6^th term, which are 3 and 4.

$\text{Median} = \dfrac{\text{Sum of middle terms}}{2} \\[1em] = \dfrac{3 + 4}{2} \\[1em] = \dfrac{7}{2} \\[1em] = 3.5$

Hence, the mean of the given data is 3.9 and the median is 3.5.

Fill in the blanks:

(i) A ..... is a collection of numerical figures to give some information.

(ii) Each numerical figure in a data is called an .....

(iii) The number of times a particular observation occurs in a data is called ...... of the observation.

(iv) A data arranged in ascending or descending order is called an ..... data.

(v) A pictorial representation of a data is called a .....

(vi) If ☺ represents 5 students then ☺ ☺ ☺ represent ..... students.

(vii) In a pictograph, if one bicycle represents 20 bicycles then 140 bicycles can be represented by ..... bicycles.

(viii) In a pictograph, if one electric bulb represents 10 bulbs then the picture of half a bulb will represent ..... bulbs.

(ix) In a pictograph, if the symbol □□□□□ represents 50 fruit baskets, then □□□□□ □□□□□ □□ represent ..... fruit baskets.

(x) The frequency of 8 is written symbolically as ..... using tally marks.

(xi) In a bar graph, the bars are of uniform .....

(xii) The mean of the first 8 natural numbers is .....

Answer

(i) A data is a collection of numerical figures to give some information.

(ii) Each numerical figure in a data is called an observation.

(iii) The number of times a particular observation occurs in a data is called frequency of the observation.

(iv) A data arranged in ascending or descending order is called an arrayed data.

(v) A pictorial representation of a data is called a pictograph.

(vi) If ☺ represents 5 students then ☺ ☺ ☺ represent 3 × 5 = 15 students.

(vii) In a pictograph, if one bicycle represents 20 bicycles then 140 bicycles can be represented by $\dfrac{140}{20}$ = 7 bicycles.

(viii) In a pictograph, if one electric bulb represents 10 bulbs then the picture of half a bulb will represent $\dfrac{10}{2}$ = 5 bulbs.

(ix) Since □□□□□ represents 50 fruit baskets, each □ represents 10 fruit baskets.

So, □□□□□ □□□□□ □□ represents 12 × 10 = 120 fruit baskets.

(x) The frequency of 8 is written symbolically as |||| ||| using tally marks.

(xi) In a bar graph, the bars are of uniform width.

(xii) The mean of the first 8 natural numbers is:

$= \dfrac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \dfrac{36}{8} \\[1em] = \mathbf{4.5}$

State whether the following statements are true (T) or false (F):

(i) If the data is large, it is difficult to get information from raw data.

(ii) Pictographs and bar graphs help us in understanding and analyzing a data.

(iii) In a bar graph, bars can be drawn either vertically or horizontally.

(iv) In a bar graph, width of a bar has no significance. It is only for eye attraction.

(v) Usually, the tally marks are recorded in bunches of 5.

(vi) The frequency 9 is represented as |||| |||| | using tally marks.

(vii) Mentioning of scale is necessary in pictographs.

(viii) Mean is always one of the number in a given data.

(ix) Median is always one of the numbers in a given data.

Answer

(i) True. When the data is large, it is difficult to get useful information from raw data without organising it.

(ii) True. Pictographs and bar graphs are visual representations that help in understanding and analyzing the data.

(iii) True. Bar graphs can be drawn either vertically or horizontally.

(iv) True. In a bar graph, only the height (or length) of a bar has significance. The width of a bar is for eye attraction.

(v) True. Tally marks are usually recorded in bunches of 5, with the fifth tally mark crossing the previous four.

(vi) False. The frequency 9 should be represented as ||||‖|||| (5 + 4 = 9), but in tally marks the correct representation is one group of 5 (with the slash) and 4 more single marks.

(vii) True. Mentioning of scale is necessary in pictographs as it indicates what each symbol represents.

(viii) False. Mean may or may not be one of the numbers in a given data.

(ix) False. Median may or may not be one of the numbers in a given data, especially when the number of observations is even.

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Choose the correct answer from the given four options for questions 3 to 7:

The minimum number of ice cream cones were sold on:

1. Monday

2. Saturday

3. Tuesday

4. Thursday

Answer

From the given pictograph (1 symbol = 2 cones), the number of ice cream cones sold on each day is:

The minimum number of cones (7) was sold on Thursday.

Hence, option 4 is the correct option.

The maximum number of ice cream cones were sold on:

1. Tuesday

2. Friday

3. Wednesday

4. Thursday

Answer

From the pictograph, the maximum number of cones (16) was sold on Tuesday.

Hence, option 1 is the correct option.

Ratio of the number of ice cream cones sold on Saturday to the number of ice cream cones sold on Wednesday is:

1. 3 : 2

2. 2 : 3

3. 4 : 5

4. 4 : 7

Answer

Number of cones sold on Saturday = 8.

Number of cones sold on Wednesday = 12.

$\text{Required ratio} = 8 : 12 \\[1em] = \dfrac{8}{12} \\[1em] = \dfrac{2}{3} \\[1em] = 2 : 3$

Hence, option 2 is the correct option.

Total number of ice cream cones sold during the whole week was:

1. 33

2. 67

3. 65

4. 57

Answer

Total number of ice cream cones sold during the week

= 10 + 16 + 12 + 7 + 14 + 8

= 67

Hence, option 2 is the correct option.

If the cost of one ice cream cone is ₹ 40, then the sale value on Thursday was:

1. ₹ 140

2. ₹ 200

3. ₹ 280

4. ₹ 2680

Answer

Number of cones sold on Thursday = 7.

Cost of one ice cream cone = ₹ 40.

Sale value on Thursday = 7 × ₹ 40 = ₹ 280.

Hence, option 3 is the correct option.

Observe the following bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options for questions 8 to 11:

Which country played maximum number of matches?

1. India

2. England

3. Pakistan

4. Australia

Answer

From the bar graph, the number of matches played by each country (scale: 1 unit = 4 matches) is:

Australia played the maximum number of matches (32).

Hence, option 4 is the correct option.

How many matches did South Africa play?

1. 16

2. 18

3. 20

4. 24

Answer

From the bar graph, South Africa played 18 matches.

Hence, option 2 is the correct option.

How many more matches were played by India than Pakistan?

1. 6

2. 12

3. 24

4. 30

Answer

Number of matches played by India = 30.

Number of matches played by Pakistan = 24.

Difference = 30 − 24 = 6 matches.

Hence, option 1 is the correct option.

Ratio of the number of matches played by India to the number of matches played by Sri Lanka is

1. 4 : 5

2. 5 : 4

3. 4 : 3

4. 7 : 6

Answer

Number of matches played by India = 30.

Number of matches played by Sri Lanka = 24.

$\text{Required ratio} = 30 : 24 \\[1em] = \dfrac{30}{24} \\[1em] = \dfrac{5}{4} \\[1em] = 5 : 4$

Hence, option 2 is the correct option.

The mean of the first 6 odd natural numbers is

1. 5

2. 5.5

3. 6

4. 6.5

Answer

The first 6 odd natural numbers are 1, 3, 5, 7, 9 and 11.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{1 + 3 + 5 + 7 + 9 + 11}{6} \\[1em] = \dfrac{36}{6} \\[1em] = 6$

Hence, option 3 is the correct option.

The median of the numbers 4, 4, 7, 5, 7, 6, 7, 3, 11 is

1. 7

2. 6

3. 5

4. 4

Answer

Arranging the given data in ascending order, we get:

3, 4, 4, 5, 6, 7, 7, 7, 11

Total number of observations = 9 (odd).

The middle observation is the 5^th observation, which is 6.

Hence, option 2 is the correct option.

Statement I: In a bar graph, the breadth of the bar graph has no significance.

Statement II: The number of times a particular entry occurs in a given data is called its frequency.

1. Statement I is true but statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

In a bar graph, the breadth of the bars has no significance. It is only for visual representation. So, Statement I is true.

The number of times a particular entry occurs in a given data is called its frequency. So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Statement I: The mean of 10, 20, 30, 40 is 20.

Statement II: Mean = $\dfrac{\text{Sum of items}}{\text{Number of items}}$

1. Statement I is true but statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

$\text{Mean of 10, 20, 30, 40} = \dfrac{10 + 20 + 30 + 40}{4} \\[1em] = \dfrac{100}{4} \\[1em] = 25$

So, the mean is 25, not 20. Statement I is false.

Mean = $\dfrac{\text{Sum of items}}{\text{Number of items}}$ — this is the correct formula for mean. So, Statement II is true.

∴ Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Statement I: 5 army aspirants have heights of 160 cm, 163 cm, 167 cm, 150 cm and 170 cm. The aspirant with the maximum height is 20 cm taller than the shortest aspirant.

Statement II: The median of the heights is 167 cm.

1. Statement I is true but statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Maximum height = 170 cm and minimum height = 150 cm.

Difference = 170 − 150 = 20 cm.

So, the aspirant with the maximum height is 20 cm taller than the shortest. Statement I is true.

Arranging the heights in ascending order: 150, 160, 163, 167, 170.

Total number of observations = 5 (odd).

The middle observation is the 3^rd observation, which is 163 cm.

So, the median is 163 cm, not 167 cm. Statement II is false.

∴ Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Statement I: Writing entries in ascending or descending order does not change the value of median.

Statement II: When the number of observations of given numerical data is even, then the median is the average of the two middle terms.

1. Statement I is true but statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Writing entries in ascending or descending order does not change the value of median. Whichever order we use, the median value remains the same. So, Statement I is true.

When the number of observations is even, the median is the average (mean) of the two middle terms. So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

A die is thrown 25 times and the numbers appearing were as given below:

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

(a) Construct data array.

(b) Construct tally chart and frequency distribution table.

Answer

(a) Arranging the given data in ascending order, we get the data array:

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

(b) The tally chart and frequency distribution table for the given data is:

In a primary school, the number of students in different classes are as follows:

Represent this data by a pictograph, using 1 symbol = 20 students

Answer

Since 1 symbol represents 20 students, the number of symbols required for each class is:

The pictograph for the given information is:

Observe the following bar graph, showing the marks scored by Avneet in the annual examination in different subjects:

Answer the following questions:

(i) What is the scale of this bar graph?

(ii) In which subject Avneet obtained maximum marks?

(iii) In which subject she obtained minimum marks?

(iv) Name the subject(s) in which she got 80 or more marks.

Answer

From the given bar graph, the marks obtained by Avneet in each subject are:

(i) The scale of this bar graph is 1 unit length = 10 marks obtained.

(ii) Avneet obtained maximum marks (100) in Mathematics.

Hence, Avneet obtained maximum marks in Mathematics.

(iii) Avneet obtained minimum marks (50) in Hindi.

Hence, Avneet obtained minimum marks in Hindi.

(iv) Avneet got 80 or more marks in Science (80) and Mathematics (100).

Hence, Avneet got 80 or more marks in Science and Mathematics.

The following table shows the monthly expenditure of a family on various items:

Represent the data by a bar graph.

Answer

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the items (Rent, Food, Education, Transport, Miscellaneous) at equal distances.

3. Along the y-axis, mark the expenditure in ₹.

4. Choose a scale of 1 unit height = ₹ 1000.

5. The heights of the bars for various items are:

• Rent : 4000 ÷ 1000 = 4 units
• Food : 6500 ÷ 1000 = 6.5 units
• Education : 3000 ÷ 1000 = 3 units
• Transport : 1500 ÷ 1000 = 1.5 units
• Miscellaneous : 5000 ÷ 1000 = 5 units

6. Draw bars of equal width keeping equal gap between adjoining bars.

The required bar graph is as given below:

Find the mean and the median of the following data:

5, 3, 12, 0, 7, 11, 4, 3, 9

Answer

Number of observations = 9.

$\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \dfrac{5 + 3 + 12 + 0 + 7 + 11 + 4 + 3 + 9}{9} \\[1em] = \dfrac{54}{9} \\[1em] = 6$

Arranging the given data in ascending order, we get:

0, 3, 3, 4, 5, 7, 9, 11, 12

Total number of observations = 9 (odd).

The middle observation is the 5^th observation, which is 5.

So, Median = 5.

Hence, the mean of the given data is 6 and the median is 5.