p = 12

On arranging the data in ascending order, we get 128,132,135,139,141,143,146,149,150,151
It is clear that, we have mean height = 141.4 cm Hence, 5 girls have more height than the mean height i.e. 143cm, 146cm, 149cm, 150cm and 151 cm

Here, sum of rotal runs
= 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Number of innings = 8
\(\therefore\) mean score = \(\frac { total\ runs }{ number\ of\ innings } =\frac { 400 }{ 8 } =50\)
Hence, the mean score is 50 runs

Let the weight (in kg) of 20 children of your classare as follows:
35, 35, 40, 33, 36, 40, 50, 38, 45, 34,
40, 40, 37, 34, 32, 48, 42, 35, 36, 40
Now, arranging the weight of 20 children of your class in descending order, we have
50, 48, 45, 42, 40, 40, 40, 40, 40, 38
37, 36, 36, 35, 35, 35, 34, 34, 33, 32
The most common weight is 40 kg, because five children have 40 kg weight.

Given data are 35,32, 35, 42, 38, 32 and 34.
For median of the given data, we first arrange the given data in ascending order, we get
Here, n = 7, which is odd
\(\left( \frac { n+1 }{ 2 } \right) ^{ th }\) data will be the median i.e. \(\frac { 7+1 }{ 2 } =\frac { 8 }{ 2 } \)=4th observation
Hence, 4th observation, i.e. 35 is the median of the given data

About 475 books were sold in the year 1990 and about 225 books were sold in the year 1992.

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.

Number of students getting marks from 50-69
= Number of students getting marks from 50-59 + Number of students getting marks from 60-69
= 7+11=18

On arranging the adjacent data in descending order, we get

Here, total rainfall
= 0.0+ 12.2+ 2.1 + 0.0+ 20.5+ 5.5+ 1.0= 41.3 mm
Number of days = 7
\(\therefore\) Mean rainfall = \(\frac { total\quad rain\quad fall }{ number\quad of\quad days } =\frac { 41.3 }{ 7 } =5.9mm\)
Hence, mean rainfall for the week is 5.9 mm

Average number of points scared by A = 12.5 [as calculated in part (i)]
Average number of points scared by B = 4.5 [as calculated in part (iii)]
\(=\frac { sum\ of\ points\ in\ games }{ number\ of\ games } \)
= \(\frac { 8+11+13 }{ 3 } =\frac { 32 }{ 3 } \) =10.67
\(\because\)  12.5 > 10.67 > 4.5
Hence, A is the best performer.

23.42

After the study of double bar graph,
Cost of apple in city 1=82
Cost of apple in city II = 75
Cost of banana in city I = 45
Cost of banana in city II = 32
Cost of mango in city I = 75
Cost of mango in city II = 60
Cost of watermelon in city I = 20
Cost of watermelon in city II = 26
Cost of cherry in city I = 38
Cost of cherry in city II = 30
(a) Clearly, the double bar graph compares the cost of different fruits per kg in city I and II.
(b) Cost of apple in city I = 82, where in city II is = 75
Cost of Banana in city I = 45, where in city II = 32
Cost of Mango in city 1=75, where in city II = 60
Cost of Cherry in city I = 38, where in city II = 30
Hence, apple, banana, mango, and cherry
prices/costs in city I are greater than city II.
(c) Difference of rates for apples in cities I and II
= 82 -75 =7
(d) The ratio of the cost of mangoes per kg in city I to
the cost of mangoes per kg in city II
=75: 60 = 5: 4

\(Mean= \frac { sum\ of\ observations }{ number\ of\ observatins } \)
\(6=\frac { sum\ of\ observations }{ 12 } \)
Sum of observations = 12 x 6 = 72

number less than 3 are 1 and 2
p (getting a number less thn 3) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Range = maximum vale - minimum value 
= 97 - 17 = 80

Arithmetic mean 8, 4 ,x,6,2,7 is 5
Arithmetic mean = \(\frac { sum\ of\ observations }{ number\ of\ observations } \)
Sum of observations = 8 + 4 + x + 6 2 + 7 = 27 +x
5 =\(\frac { 27+x }{ 6 } \Rightarrow 5\times 6=27+x\)
30 = 27 +x \(\Rightarrow\) x = 30 -27 = 3

\(\underset { \overset { \downarrow }{ minimum\quad value } }{ 1 } ,2,4,\underset { \overset { \downarrow }{ median } }{ 6 } ,8,9,\underset { \overset { \downarrow }{ maximum\quad value } }{ 11 } \)