Median:
Arranging the given data in ascending order, we have: 110, 110, 120, 120, 120, 120, l30, l30, 140,140.
Number of data = 10, which is an even number.
\(\therefore \)Median = Mean of two middle values
\(\frac { 120+120 }{ 2 } \)=120
Mode: In the given data, we observe that the observation 120 occurs maximum number of times, i.e. 4 times.
\(\therefore \)Mode = 120

(i) If we are collecting the shoe SIze suitable for a particular age, and
(ii) A shopkeeper selling shirts needs to know suitable size to enrich the stock, then mode is more useful.

(a)

(b) Tabulation of data.
(c) In India, children need sufficient protein food for nutrition.

We have,
\(11=\frac{x+(x+2)+(x+4)+(x+6)+(x+8)}{5}\)
\(\Rightarrow 11=\frac{5x+20}{5}\)
\(\Rightarrow\) 55 = 5x + 20
\(\Rightarrow\) 5x=35
\(\Rightarrow\) x=7
\(\therefore\) Mean of first three observationsv
\(=\frac{x+(x+2)+(x+4)}{3}=\frac{3x+6}{3}\)
=x+2=7+2=9   \([\because x=7]\)

\(Mean=\frac{\sum fx}{N}\)
Putting values from table,
Mean \(=\frac{8480}{182}=46.59.\)

We will have to add atleast three 17's to change the mode to 17. On adding three 17's, the frequency of 17 will be maximum i.e. 4.

(a) We can clearly see that all the 5 letters are vowels i.e. a, e, i, i, u. Hence, it is certain to draw a vowel i.e. probability of drawing a vowel is 1
(b) \(probaility=\frac { number\ ofcards\ marked\ U }{ Total\ numbe\ of\ cards } =\frac { 1 }{ 5 } \)
(c) \(probaility=\frac { number\ of\ cards\ marked\ AorI }{ Total\ number\ of\ cards } =\frac { 2 }{ 5 } \)
(d) \(probaility=\frac { number\ of\ cards\ marked\ with\ a\ consonant }{ Total\ number\ of\ cards } \)
= \(\frac { 0 }{ 5 } \)=0
Hence, it is impossible to draw a consonant

Clearly, from the bar graph, grapes is liked by 4 students. So, the least popular fruit among the students is grapes.

As per the given information,
Total number of trials =100
Number of heads = 59
Number of tails = 100 - 59 = 41
(a) P (getting a head) = \(\frac { number\ of\ heads\ in\ trails }{ Total\ number\ of\ trails } \)
= \(\frac { 59 }{ 100 } \)
(b) P (getting a tail) = \(\frac { number\ of\ heads\ in\ trails }{ Total\ number\ of\ trails } \) = \(\frac { 41 }{ 100 } \)

On the Basic of above table

1972

C

Cricket is most popular sport

We start the scale from o. The greatest value of data is 135, so end the scale at a value greater than 135 such as 140. Use equal division along axes such as increment of 10. Here, we take 1 unit for 10 children.

From the given graph, it is clear that the bar representing cat is the highest i.e. 10.
Therefore, cat is the most popular pet.

(i) On arranging the data in ascending order, we get 12,12,13,13,14,14,14,16,19.
Here, 14 occurs more frequently i. e. 3 times.
\(\therefore\) Mode =14
(ii) Here, ascending order of the given data is 12,12,13,13,14,14,14,16,19
Since, the middle observation of the data = 14
\(\therefore\) Median of data = 14
Hence, mode = 14 and median = 14

Given, the enrolment of a school during six consecutive years
1555,1670,1750,2013,2540,2820
Now, sum of enrolments
= 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348
Number of years = 6
\(\therefore\)  Mean enrolment of the school for this period
 = \(\frac { sum\quad of\quad enrolments }{ number\quad of\quad years } =\frac { 12348 }{ 6 } =2058\)
Hence, the mean enrolment of the school for this period is 2058.

The frequency table is shwn below.

We have, sum of marks
=4+6+7+5+3+5+4+5+2+6+2+5+1
+ 9 + 6 + 5 + 8 + 4 + 6 + 7 = 100
Number of students = 20
\(\therefore\) Arithmetic mean = \(\frac { sum\ of\ marks }{ number\ of\ students } =\frac { 100 }{ 20 } =5\)
Hence, the arithmetic mean of the data is 5.

The frequency table is shown below.

It is clear from the table that, the highest number = 9

We know that, there are seven days in a week. Let the study hours for different days of the week are as follows

Total number of study hours = 6+5+6+7+6+5+7=42h
Number of days in a week = 7
\(\therefore\)  \(Average=\frac { Total\ study\ hours }{ total\ number\ of\ days } =\frac { 42 }{ 7 } =6h\)
Hence, average study hours is 6 h per day