\(\overset { - }{ x } =\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } \)
\(\Rightarrow 25=\frac { 570+45p }{ 26+p } \)
\(\Rightarrow 650+25p=570+45p\)
\(\Rightarrow 650-570=45p-25p\)
p=4

Mean = \(\frac { \Sigma c_{ i }f_{ i } }{ \Sigma f_{ i } } \)
\(\Rightarrow \quad x=\frac { 162 }{ 20 } =8.1\)
Mean number of plants per house is 8.1.

According to the question, Mode = 24.5
and Mean = 29.75
The relationship connecting measures of central tendencies is :
3 Median = Mode + 2 Mean
3 Median = 24.5 + 2 x 29.75
= 24.5 + 59.50
3 Median = 84.0
Median = \(\frac { 84 }{ 3 } =28\)

Change the above distribution as frequency distribution.

Mean = \(\frac { \Sigma x_{ i }f_{ i } }{ \Sigma f_{ i } } \)
\(\frac { 300 }{ 20 } =15\)

Modal Class = 80- 100
i=80 , f₁=18 , f₂=5,f₀=12,h=20
Model = i + \(\left( \frac { f_{ i }-f_{ 0 } }{ 2f_{ i }-f_{ 0 }-f_{ 2 } } \right) \times 20\)
\(=80+\left( \frac { 18-12 }{ 36-12-5 } \right) \times 20\)
\(=80+\frac { 6 }{ 19 } \times 20\)
=80+6.31
=86.31

Modal class is 30 - 35, f = 30,f₁ = 25, f₀ = 10,f₂ = 7, h =5
Mode = l+ \(\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\Rightarrow \)Mode = 30+ \(\frac { 25-10 }{ 50-10-7 } \times 5\)
= 30+2.27 or 32.27 approx.

Given ,
Mode = 50.5
Median = 45.5
3 Median = Mode + 2 Mean
3 x 45.5 = 50.5 + 2 Mean
\(\Rightarrow \) Mean = \(\frac { 136.5-50.5 }{ 2 } \)
= 43

Here, we have the cumulative frequency distribution of less than type. We observe that the number of students getting marks less than 10 is 14 and 22 students have marks less than 20.
Therefore, number of students getting marks between 10 and 20 is 22-14=8. Similarly, the number of students getting marks between 20 and 30 is 37-22=15 and so on.
Thus, we have the following frequency distribution table