Data Handling

(i) On arranging the given data in ascending order of their magnitudes, we get :

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

Hence, the data in ascending order is 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6.

(ii) From the above arrangement,

Maximum number of rooms = 6

Minimum number of rooms = 1

By formula,

Range of the data = Maximum value − Minimum value

Substituting the values, we get :

Range = 6 − 1 = 5

Hence, the range of the data is 5.

(iii) Writing the number of rooms from the lowest to the highest, marking the tally marks against each number and then counting the tally marks, we get the required frequency distribution table as shown below :

Number of rooms (variate)	Tally marks	Number of houses (Frequency)
1	III	3
2	IIII	4
3	IIII III	8
4	IIII	5
5	III	3
6	II	2
Total		25

Hence, the above table is the required frequency distribution table.

(iv) Houses having 4 or more than 4 rooms are the houses with 4, 5 and 6 rooms, whose frequencies are 5, 3 and 2 respectively.

∴ Number of such houses = 5 + 3 + 2 = 10

Hence, 10 houses have 4 or more than 4 rooms.

(v) Take the number of rooms along X-axis and the number of houses along Y-axis.

Scale : Take 1 unit height along Y-axis = 1 house.

The heights of bars for various number of rooms are :

1 room : 3 ÷ 1 = 3 units

2 rooms : 4 ÷ 1 = 4 units

3 rooms : 8 ÷ 1 = 8 units

4 rooms : 5 ÷ 1 = 5 units

5 rooms : 3 ÷ 1 = 3 units

6 rooms : 2 ÷ 1 = 2 units

The bar graph representing the given data is shown below :

Scale : 1 cm = 100 books.

(i) From the bar graph,

Number of books sold in 2008 = about 140

Number of books sold in 2009 = about 360

Number of books sold in 2011 = about 180

Hence, about 140, 360 and 180 books were sold in the years 2008, 2009 and 2011 respectively.

(ii) From the bar graph,

The bar of height about 475 books corresponds to the year 2012.

The bar of height about 225 books corresponds to the year 2010.

Hence, about 475 books were sold in 2012 and about 225 books were sold in 2010.

Take the favourite colours along X-axis and the number of students along Y-axis.

Scale : Take 1 unit height along Y-axis = 10 students.

The heights of bars for various colours are :

Red : 43 ÷ 10 = 4.3 units

Green : 19 ÷ 10 = 1.9 units

Blue : 55 ÷ 10 = 5.5 units

Yellow : 49 ÷ 10 = 4.9 units

Orange : 34 ÷ 10 = 3.4 units

The bar graph representing the given data is shown below :

(i) From the bar graph, the bar for blue colour is the highest, i.e. 55 students.

Hence, the most preferred colour is blue.

(ii) From the bar graph, the bar for green colour is the lowest, i.e. 19 students.

Hence, the least preferred colour is green.

(iii) From the bar graph, there are 5 bars representing the colours red, green, blue, yellow and orange.

Hence, there are 5 colours in all and they are red, green, blue, yellow and orange.

Take the years along X-axis and the number of books sold along Y-axis.

Scale : Take 1 unit height along Y-axis = 100 books.

The double bar graph representing the given data is shown below :

(i) Difference in the sale of two language books :

In 2008 = 500 − 350 = 150

In 2009 = 525 − 400 = 125

In 2010 = 600 − 450 = 150

In 2011 = 650 − 620 = 30

The difference is least in the year 2011.

Hence, the difference in the sale of two language books was least in the year 2011.

(ii) Increase in the sale of English books from 2008 to 2011 = 620 − 350

= 270

Increase in the sale of Hindi books from 2008 to 2011 = 650 − 500

= 150

Since, 270 > 150, the sale of English books increased more than the sale of Hindi books.

Hence, yes, the demand for English books rose faster.

Take the favourite sports along X-axis and the number of persons along Y-axis.

Scale : Take 1 unit height along Y-axis = 200 persons.

The double bar graph representing the given data is shown below :

(i) From the bar graph, the bars for cricket are the highest in both watching (1240) and participating (620).

Hence, cricket is the most popular sport.

(ii) From the bar graph, the bar for watching is higher than the bar for participating in every sport.

Total number of persons watching sports = 1240 + 470 + 510 + 430 + 250 = 2900

Total number of persons participating in sports = 620 + 320 + 320 + 250 + 110 = 1620

Since, 2900 > 1620, the total number of persons watching sports is greater than the total number of persons participating in sports.

Hence, watching sports is more preferred than participating in sports.

Sum of all observations = 40 + 30 + 30 + 0 + 26 + 60

= 186

and, number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{186}{6}$ = 31

Hence, the mean of the given data is 31.

Sum of all observations = 3 + 5 + 7 + 9 + 11 + 13 + 15

= 63

and, number of observations = 7

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{63}{7}$ = 9

Hence, the mean of the given data is 9.

First five whole numbers are 0, 1, 2, 3 and 4.

Sum of all observations = 0 + 1 + 2 + 3 + 4

= 10

and, number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{10}{5}$ = 2

Hence, the mean of the first five whole numbers is 2.

Sum of all observations = 36 + 35 + 50 + 46 + 60 + 55

= 282

and, number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean runs = $\frac{282}{6}$ = 47

Hence, the mean runs scored by him in an inning = 47.

Sum of all observations = 1555 + 1670 + 1750 + 2013 + 2540 + 2825

= 12353

and, number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean enrolment = $\frac{12353}{6}$

= 2058.83

= 2059 (approx.)

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

Arranging the marks in ascending order, we get :

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

(i) From the above arrangement,

Highest marks = 95

Lowest marks = 39

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

(ii) By formula,

Range = Highest marks − Lowest marks

Substituting the values, we get :

Range = 95 − 39

= 56

Hence, the range of the marks obtained = 56.

(iii) Sum of all observations = 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95

= 730

and, number of observations = 10

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean marks = $\frac{730}{10}$ = 73

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

Arranging the heights in ascending order, we get :

128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) From the above arrangement, the greatest height is 151 cm.

Hence, the height of the tallest girl = 151 cm.

(ii) From the above arrangement, the least height is 128 cm.

Hence, the height of the shortest girl = 128 cm.

(iii) Sum of all observations = (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151) cm

= 1414 cm

and, number of observations = 10

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean height = $\frac{1414}{10}$ cm = 141.4 cm

Hence, the mean height of the girls = 141.4 cm.

(iv) The heights more than the mean height (141.4 cm) are 143 cm, 146 cm, 149 cm, 150 cm and 151 cm.

∴ Number of such girls = 5

Hence, 5 girls have heights more than the mean height.

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow 5 = \frac{8 + 4 + 6 + x + 2 + 7}{6} \\[1em] \Rightarrow 5 = \frac{27 + x}{6} \\[1em] \Rightarrow 27 + x = 5 \times 6 \\[1em] \Rightarrow 27 + x = 30 \\[1em] \Rightarrow x = 30 - 27 \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, the value of $\bm{x = 3}$.

To calculate the mean, we construct the following table :

Marks obtained (xi)	Number of students (fi)	fixi
2	3	6
3	2	6
4	6	24
7	7	49
10	2	20
Total	Σfi = 20	Σfixi = 105

By formula,

Mean = $\frac{\Sigma f_i x_i}{\Sigma f_i}$

Substituting the values, we get :

Mean = $\frac{105}{20}$ = 5.25

Hence, the mean of the given data is 5.25.

On arranging the given data in ascending order of their magnitudes, we get :

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

Since, the number of terms $(n)$ is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 4\end{aligned}$$

Hence, the median of the given data is 4.

On arranging the given data in ascending order of their magnitudes, we get :

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

As the number of terms $(n)$ is 8 and 8 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(4 + 5) \\[1em] = \frac{1}{2} \times 9 \\[1em] = 4.5\end{aligned}$$

Hence, the median of the given data is 4.5.

Arranging the given data in ascending order, we get :

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

In the given data, the number 3 appears maximum times, i.e. 3 times.

∴ Mode = 3

Hence, the mode of the given data is 3.

In the given frequency distribution, the observation 22 has maximum frequency, i.e. 18.

∴ Mode = 22

Hence, the mode of the given data is 22.

On arranging the given data in ascending order of their magnitudes, we get :

12, 12, 13, 13, 14, 14, 14, 16, 19

Since, the number of terms $(n)$ is 9 and 9 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{9 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 5^{\text{th}}\text{ term} \\[1em] = 14\end{aligned}$$

In the given data, the number 14 appears maximum times, i.e. 3 times.

∴ Mode = 14

Hence, the median of the given data is 14 and the mode is 14.

On arranging the given data in ascending order of their magnitudes, we get :

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

In the given data, the number 20 appears maximum times, i.e. 4 times.

∴ Mode = 20

Since, the number of terms $(n)$ is 15 and 15 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{15 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 8^{\text{th}}\text{ term} \\[1em] = 20\end{aligned}$$

Hence, the mode of the given data is 20 and the median is 20. Yes, they are same.

(i) On arranging the given data in ascending order of their magnitudes, we get :

32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50

Since, the number of terms $(n)$ is 15 and 15 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{15 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 8^{\text{th}}\text{ term} \\[1em] = 40 \text{ kg}\end{aligned}$$

In the given data, the numbers 38 and 43 both appear maximum times and equal number of times, i.e. 3 times each.

∴ Mode = 38 kg and 43 kg

Hence, the median of the given data is 40 kg and the modes are 38 kg and 43 kg.

(ii) Since, both 38 and 43 occur maximum number of times (3 times each), both are modes of the given data.

Hence, yes, there is more than one mode (there are two modes : 38 kg and 43 kg).

Sum of all observations = 6 + 15 + 120 + 50 + 100 + 80 + 10 + 15 + 8 + 10 + 15

= 429

and, number of observations = 11

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{429}{11}$

= 39

On arranging the given data in ascending order of their magnitudes, we get :

6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120

In the given data, the number 15 appears maximum times, i.e. 3 times.

∴ Mode = 15

Since, the number of terms $(n)$ is 11 and 11 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{11 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 6^{\text{th}}\text{ term} \\[1em] = 15\end{aligned}$$

Hence, the mean is 39, the mode is 15 and the median is 15. No, the three are not same.

As the given data has large number of observations, we put it in tabular form :

Observations	Tally marks	Frequency
12	III	3
13	IIII	4
14	IIII	5
15	IIII IIII	10
16	IIII I	6
17	II	2
18	I	1
19	I	1
Total		32

Looking at the table, we find that the observation 15 has maximum frequency, i.e. 10.

∴ Mode = 15

Hence, the mode of the given data is 15.

(i) Every day we grow older, so we are always older today than yesterday.

Hence, it is certain to happen.

(ii) A car can seat only a few people, so two hundred people can never sit in a car.

Hence, it is impossible to happen.

(iii) A tossed coin may land either heads up or tails up.

Hence, it can happen but not certain.

(iv) A die has only the numbers 1, 2, 3, 4, 5 and 6 on its faces, so 8 can never come on top.

Hence, it is impossible to happen.

(v) India may win or may lose the next test series.

Hence, it can happen but not certain.

(vi) Tomorrow may or may not be a cloudy day.

Hence, it can happen but not certain.

(vii) The next traffic light seen may be red, yellow or green.

Hence, it can happen but not certain.

When a coin is flipped, the possible outcomes are Head (H) and Tail (T).

So, the total number of possible outcomes = 2

The event is that your team will start the game, i.e. your team gets the side of the coin it has chosen.

∴ The number of outcomes favourable to the event = 1

By formula,

P(Event) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

∴ P(your team will start the game) = $\frac{1}{2}$

Hence, the probability that your team will start the game = $\bm{\frac{1}{2}}$.

There are 6 marbles in the box with numbers 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

By formula,

P(Event) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

(i) The event is drawing a marble with number 5.

∴ The number of outcomes favourable to the event = 1

∴ P(drawing a marble with number 5) = $\frac{1}{6}$

Hence, the probability of drawing a marble with number 5 = $\bm{\frac{1}{6}}$.

(ii) The event is drawing a marble with number 2.

∴ The number of outcomes favourable to the event = 1

∴ P(drawing a marble with number 2) = $\frac{1}{6}$

Hence, the probability of drawing a marble with number 2 = $\bm{\frac{1}{6}}$.

When a die is tossed once, the possible outcomes are the numbers 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

By formula,

P(Event) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

(i) The event is getting a number less than 3 and the outcomes less than 3 are 1 and 2.

So, the number of outcomes favourable to the event getting a number less than 3 = 2

∴ P(getting a number less than 3) = $\frac{2}{6} = \frac{1}{3}$

Hence, the probability of getting a number less than 3 = $\bm{\frac{1}{3}}$.

(ii) The event is getting a prime number and the prime numbers (outcomes) are 2, 3 and 5.

So, the number of outcomes favourable to the event getting a prime number = 3

∴ P(getting a prime number) = $\frac{3}{6} = \frac{1}{2}$

Hence, the probability of getting a prime number = $\bm{\frac{1}{2}}$.

(iii) The event is getting a number greater than 2 and the outcomes greater than 2 are 3, 4, 5 and 6.

So, the number of outcomes favourable to the event getting a number greater than 2 = 4

∴ P(getting a number greater than 2) = $\frac{4}{6} = \frac{2}{3}$

Hence, the probability of getting a number greater than 2 = $\bm{\frac{2}{3}}$.

Total number of mangoes in the box = 3 + 21 = 24

By formula,

P(Event) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

(i) The event is getting a defective mango.

The number of outcomes favourable to the event getting a defective mango = 3

∴ P(getting a defective mango) = $\frac{3}{24} = \frac{1}{8}$

Hence, the probability of getting a defective mango = $\bm{\frac{1}{8}}$.

(ii) The event is getting a good mango.

The number of outcomes favourable to the event getting a good mango = 21

∴ P(getting a good mango) = $\frac{21}{24} = \frac{7}{8}$

Hence, the probability of getting a good mango = $\bm{\frac{7}{8}}$.

Total number of cards in a pack = 52

So, the total number of possible outcomes = 52

By formula,

P(Event) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

(i) The event is getting a red card.

There are 26 red cards (13 hearts and 13 diamonds) in a pack.

∴ The number of outcomes favourable to the event getting a red card = 26

∴ P(getting a red card) = $\frac{26}{52} = \frac{1}{2}$

Hence, the probability of getting a red card = $\bm{\frac{1}{2}}$.

(ii) The event is getting a king.

There are 4 kings in a pack.

∴ The number of outcomes favourable to the event getting a king = 4

∴ P(getting a king) = $\frac{4}{52} = \frac{1}{13}$

Hence, the probability of getting a king = $\bm{\frac{1}{13}}$.

(iii) The event is getting a card of spades.

There are 13 cards of spades in a pack.

∴ The number of outcomes favourable to the event getting a card of spades = 13

∴ P(getting a card of spades) = $\frac{13}{52} = \frac{1}{4}$

Hence, the probability of getting a card of spades = $\bm{\frac{1}{4}}$.

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

(ii) The frequency of 9 is written symbolically as IIII IIII using tally marks.

(iii) In a bar graph, the height (or length) of a bar is equal to or proportional to the frequency of the corresponding observation.

(iv) Double bar graphs help us in comparing two collections of data at a glance.

(v) The most common representative value of a statistical data is the arithmetic mean of the data.

(vi) Mode of a data is the observation which occurs maximum number of times in the given data.

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

Mean = $\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10}{10}$

= $\frac{55}{10}$

= 5.5

The arithmetic mean of first 10 natural numbers is 5.5.

(viii) When a die is rolled, the total number of possible outcomes = 6

The only number more than 5 is 6.

∴ The number of favourable outcomes = 1

∴ P(getting a number more than 5) = $\frac{1}{6}$

When a die is rolled, the probability of getting a number more than 5 is $\bm{\frac{1}{6}}$.

(ix) The probability of an impossible event is 0.

(x) The probability of a sure event is 1.

(xi) Median is one of the observations in the data if the number of observations is odd.

(i) Mean is the sum of all observations divided by the number of observations. The value of observation that occurs most frequently is the mode and not the mean.

Hence, the given statement is false.

(ii) In the given data,

Highest observation = 4

Lowest observation = −5

∴ Range = 4 − (−5)

= 4 + 5

= 9

The range is 9 and not −9.

Hence, the given statement is false.

(iii) Mode is the observation which occurs maximum number of times in the data, so it is always one of the numbers in the data.

Hence, the given statement is true.

(iv) A data can have more than one mode. For example, in the data 2, 1, 2, 3, 5, 6, 2, 5, 8, 3, 5, 10, 8, both 2 and 5 occur three times, so both 2 and 5 are modes.

Hence, the given statement is false.

(v) Mean need not be one of the numbers in the data. For example, the mean of 1, 2, 3 and 4 is $\frac{1 + 2 + 3 + 4}{4} = \frac{10}{4}$ = 2.5, which is not one of the given numbers.

Hence, the given statement is false.

(vi) Sum of all observations = 6 + 4 + 3 + 8 + 9 + 12 + 13 + 9

= 64

and, number of observations = 8

∴ Mean = $\frac{64}{8}$

= 8

The mean is 8 and not 9.

Hence, the given statement is false.

(vii) If the number of observations is odd, the median is the middle observation, which is from the given data. If the number of observations is even, the median is the average of the two middle observations, which may not be from the given data.

Hence, the given statement is true.

(viii) The measures of central tendency (mean, median and mode) always lie between the lowest and the highest values of the data.

Hence, the given statement is false.

(ix) When the extreme observations on both ends of the arranged data are removed, the middle observation(s) remain the same.

∴ The median does not get affected.

Hence, the given statement is false.

(x) Probability of an event is always less than or equal to 1. So, the probability of getting an ace can never be greater than 1.

Actually, P(getting an ace) = $\frac{4}{52} = \frac{1}{13}$

Hence, the given statement is false.

(xi) On arranging the given data in ascending order of their magnitudes, we get :

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

Since, the number of terms $(n)$ is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 6\end{aligned}$$

The median is 6 and not 2.

Hence, the given statement is false.

Mean, median and mode are the three measures of central tendency (representative values) of a data.

Range is the difference between the highest and the lowest observations, which tells the spread of the data and not its central tendency.

Hence, option 4 is the correct option.

The difference between the highest and the lowest observations in a data is called its range.

Hence, option 3 is the correct option.

From the given data,

Highest runs scored = 100

Lowest runs scored = 0

By formula,

Range = Highest observation − Lowest observation

Substituting the values, we get :

Range = 100 − 0

= 100

Hence, option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{3 + 1 + 5 + x + 9}{5} = 4 \\[1em] \Rightarrow \frac{18 + x}{5} = 4 \\[1em] \Rightarrow 18 + x = 4 \times 5 \\[1em] \Rightarrow 18 + x = 20 \\[1em] \Rightarrow x = 20 - 18 \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, option 3 is the correct option.

On arranging the given data in ascending order of their magnitudes, we get :

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

Since, the number of terms $(n)$ is 9 and 9 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{9 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 5^{\text{th}}\text{ term} \\[1em] = 2\end{aligned}$$

Hence, option 2 is the correct option.

First 49 natural numbers are 1, 2, 3, ......., 49, which are already in ascending order.

Since, the number of terms $(n)$ is 49 and 49 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{49 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 25^{\text{th}}\text{ term} \\[1em] = 25\end{aligned}$$

Hence, option 2 is the correct option.

Mean of three numbers = 40

⇒ Sum of three numbers = 40 × 3 = 120

Lowest number = 19

⇒ Sum of remaining two numbers = 120 − 19 = 101

Since, the three numbers are different natural numbers and 19 is the lowest, the smallest possible value of the middle number = 20

∴ Highest possible value of the largest number = 101 − 20

= 81

Hence, option 3 is the correct option.

On arranging the given data in ascending order, we get :

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

In the given data, the number 2 appears maximum times, i.e. 3 times.

∴ Mode = 2

Hence, option 3 is the correct option.

A sure event always happens, so its probability is 1.

Hence, option 4 is the correct option.

An impossible event never happens, so its probability is 0.

Hence, option 2 is the correct option.

A newly born baby can be either a boy or a girl.

So, the total number of possible outcomes = 2

The number of outcomes favourable to the event of the baby being a girl = 1

∴ P(baby is a girl) = $\frac{1}{2}$

Hence, option 3 is the correct option.

There are 4 letters in the word GIRL.

So, the total number of possible outcomes = 4

The event is selecting the letter G.

The number of outcomes favourable to the event = 1

∴ P(selecting letter G) = $\frac{1}{4}$

Hence, option 3 is the correct option.

There are 8 letters in the word ALPHABET.

So, the total number of possible outcomes = 8

The event is selecting a vowel.

As there are three vowels (A, A and E) in the word ALPHABET, the number of outcomes favourable to the event = 3

∴ P(selecting a vowel) = $\frac{3}{8}$

Hence, option 4 is the correct option.

When a die is thrown, the possible outcomes are the numbers 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

The event is getting a composite number and the composite numbers (outcomes) are 4 and 6.

So, the number of outcomes favourable to the event = 2

∴ P(getting a composite number) = $\frac{2}{6} = \frac{1}{3}$

Hence, option 2 is the correct option.

Total number of balls in the bag = 5 + 10 = 15

The event is drawing a white ball.

The number of outcomes favourable to the event = 5

∴ P(drawing a white ball) = $\frac{5}{15}$

Hence, option 2 is the correct option.

Out of 5 students, only 2 can participate in the quiz.

So, the total number of possible outcomes = 5

and, the number of outcomes favourable to the event = 2

∴ P(student makes it to the competition) = $\frac{2}{5}$

= $\frac{2}{5} \times 100$%

= 40%

Hence, option 2 is the correct option.

The difference between the highest observation and the lowest observation is called the range of the data.

So, Statement II is true.

Given heights : 160 cm, 163 cm, 167 cm, 150 cm and 170 cm.

Highest height = 170 cm

Lowest height = 150 cm

∴ Range = 170 − 150

= 20 cm

So, Statement I is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Sum of all observations = 5 + 3 + 3 + 2 + 4 + 1 + 5 + 1 + 3 + 3

= 30

and, number of observations = 10

∴ Mean = $\frac{30}{10}$

= 3

On arranging the given data in ascending order of their magnitudes, we get :

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

As the number of terms $(n)$ is 10 and 10 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(3 + 3) \\[1em] = \frac{1}{2} \times 6 \\[1em] = 3\end{aligned}$$

In the given data, the number 3 appears maximum times, i.e. 4 times.

∴ Mode = 3

Thus, mean = median = mode = 3.

So, Statement I is true.

The mean, median and mode of a data set need not be equal. For example, for the data 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15, mean = 39, median = 15 and mode = 15.

So, Statement II is false.

∴ Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

When a dice is rolled, the possible outcomes are the numbers 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

The event is getting an even number and the even numbers (outcomes) are 2, 4 and 6.

So, the number of outcomes favourable to the event = 3

∴ P(getting an even number) = $\frac{3}{6} = \frac{1}{2}$

So, Statement I is true.

The probability of a sure event is 1 and the probability of any event can never be greater than 1.

So, Statement II is false.

∴ Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Total number of balls in the bag = 4 + 2 + 1 = 7

Number of outcomes favourable to the event drawing a white ball = 1

∴ P(drawing a white ball) = $\frac{1}{7}$

Number of outcomes favourable to the event drawing a blue ball = 2

∴ P(drawing a blue ball) = $\frac{2}{7}$

Half of P(drawing a blue ball) = $\frac{1}{2} \times \frac{2}{7} = \frac{1}{7}$ = P(drawing a white ball)

So, Statement I is true.

An impossible event never happens, so its probability is 0.

So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

If the number of observations is odd, the median is the middle observation, which is one of the numbers in the data.

If the number of observations is even, the median is the average of the two middle observations, which may not be one of the numbers in the data.

For example, in the data 3, 4, 5, 7 :

Number of observations = 4, which is even.

∴ Median = $\frac{4 + 5}{2} = \frac{9}{2}$ = 4.5,

which is not a number in the given data.

Hence, the given statement is false.

Take the students along X-axis and the marks obtained along Y-axis.

Scale : Take 1 unit height along Y-axis = 4 marks.

The double bar graph representing the given data is shown below :

(i) From the bar graph, the bars representing half-yearly marks are higher than the bars representing quarterly marks for Ashish, Arun, Kavish and Rita. Only for Maya, the bar for half-yearly marks is lower.

Hence, yes, there is an improvement as all bars representing half-yearly marks are higher in all cases except one.

(ii) Maya scored 21 marks in quarterly and 20 marks in half-yearly examination.

Since, 20 < 21, her marks have decreased.

Hence, Maya has not done better.

All the factors of 12 are 1, 2, 3, 4, 6 and 12.

Sum of all observations = 1 + 2 + 3 + 4 + 6 + 12

= 28

and, number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{28}{6}$

= $\frac{14}{3}$

Hence, the mean of the factors of 12 is $\bm{\frac{14}{3}}$.

On arranging the given data in ascending order of their magnitudes, we get :

2, 9, 10, 12, 13, 13, 14, 14, 14, 16, 16, 18

As the number of terms $(n)$ is 12 and 12 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{12}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{12}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[6^{\text{th}}\text{ term} + 7^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(13 + 14) \\[1em] = \frac{1}{2} \times 27 \\[1em] = 13.5\end{aligned}$$

In the given data, the number 14 appears maximum times, i.e. 3 times.

∴ Mode = 14

Hence, the median of the given data is 13.5 and the mode is 14.

As the given data has large number of observations, we put it in tabular form :

Heights (in cm)	Tally marks	Number of students (Frequency)
148	I	1
152	I	1
155	I	1
157	I	1
160	III	3
161	I	1
162	II	2
163	IIII III	8
164	III	3
165	III	3
168	I	1
Total		25

Looking at the table, we find that the observation 163 has maximum frequency, i.e. 8.

∴ Mode = 163 cm

Hence, the mode of their heights is 163 cm.

When a die is thrown, the possible outcomes are the numbers 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

The event is getting an odd prime number and the odd prime numbers (outcomes) are 3 and 5.

So, the number of outcomes favourable to the event getting an odd prime number = 2

∴ P(getting an odd prime number) = $\frac{2}{6} = \frac{1}{3}$

Hence, the probability of getting an odd prime number = $\bm{\frac{1}{3}}$.

The letters of the word MATHEMATICS are M, A, T, H, E, M, A, T, I, C and S.

So, the total number of possible outcomes = 11

(i) The event is getting the letter M.

There are two M's in the word MATHEMATICS.

∴ The number of outcomes favourable to the event getting letter M = 2

∴ P(getting letter M) = $\frac{2}{11}$

Hence, the probability of getting letter M = $\bm{\frac{2}{11}}$.

(ii) The event is getting a vowel.

The vowels in the word MATHEMATICS are A, E, A and I.

∴ The number of outcomes favourable to the event getting a vowel = 4

∴ P(getting a vowel) = $\frac{4}{11}$

Hence, the probability of getting a vowel = $\bm{\frac{4}{11}}$.

(iii) The event is getting a consonant.

Number of consonants = 11 − 4 = 7 (M, T, H, M, T, C and S)

∴ The number of outcomes favourable to the event getting a consonant = 7

∴ P(getting a consonant) = $\frac{7}{11}$

Hence, the probability of getting a consonant = $\bm{\frac{7}{11}}$.

Let the sixth observation be $x$.

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow 17.5 = \frac{14 + 9 + 23 + 25 + 10 + x}{6} \\[1em] \Rightarrow 17.5 = \frac{81 + x}{6} \\[1em] \Rightarrow 81 + x = 17.5 \times 6 \\[1em] \Rightarrow 81 + x = 105 \\[1em] \Rightarrow x = 105 - 81 \\[1em] \Rightarrow x = 24\end{aligned}$$

Hence, the sixth observation is 24.

Mean height of 10 students = 151.8 cm

⇒ Total height of 10 students = 151.8 × 10 = 1518 cm

Two more students of heights 157.6 cm and 154.4 cm join the group.

⇒ Total height of 12 students = 1518 + 157.6 + 154.4 = 1830 cm

∴ New mean height = $\frac{1830}{12}$ cm

= 152.5 cm

Hence, the new mean height is 152.5 cm.

The given data is already in ascending order :

3, 6, 7, 10, $x, x + 4$, 19, 20, 25, 28

As the number of terms $(n)$ is 10 and 10 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}[x + (x + 4)] \\[1em] = \frac{1}{2}(2x + 4) \\[1em] = x + 2\end{aligned}$$

Given, median = 13

$$\begin{aligned}\Rightarrow x + 2 = 13 \\[1em] \Rightarrow x = 13 - 2 \\[1em] \Rightarrow x = 11\end{aligned}$$

Hence, the value of $\bm{x = 11}$.

Given,

Mean = 4.5

$$\begin{aligned}\Rightarrow \frac{5 + 7 + 7 + 8 + x + 3 + 1 + 5 + 4 + 2}{10} = 4.5 \\[1em] \Rightarrow \frac{42 + x}{10} = 4.5 \\[1em] \Rightarrow 42 + x = 4.5 \times 10 \\[1em] \Rightarrow 42 + x = 45 \\[1em] \Rightarrow x = 45 - 42 \\[1em] \Rightarrow x = 3\end{aligned}$$

So, the data is 5, 7, 7, 8, 3, 3, 1, 5, 4, 2.

On arranging the data in ascending order of their magnitudes, we get :

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

As the number of terms $(n)$ is 10 and 10 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(4 + 5) \\[1em] = \frac{1}{2} \times 9 \\[1em] = 4.5\end{aligned}$$

Hence, the median of the data is 4.5.

A leap year has 366 days.

366 days = 52 weeks and 2 days

52 weeks contain 52 Sundays. The remaining 2 days can be :

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday) and (Saturday, Sunday).

So, the total number of possible outcomes = 7

A leap year will have 53 Sundays if one of the remaining 2 days is a Sunday, i.e. (Sunday, Monday) or (Saturday, Sunday).

∴ The number of outcomes favourable to the event getting 53 Sundays = 2

∴ P(getting 53 Sundays) = $\frac{2}{7}$

Hence, the probability of getting 53 Sundays in a leap year = $\bm{\frac{2}{7}}$.