Since, the given not continuous, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
The table for given data becomes

Here assumed mean, a=300.5
Class width, h=200 and total number of observations, N=50
By step deviation method,
Mean\(=a+\left\{ \frac { \sum _{ i=1 }^{ 50 }{ f_{ i }u_{ i } } }{ n } \right\} \times h\)
\(=300.5+\left\{ \frac { 14 }{ 50 } \right\} \times 200\) 
\(\\ =300.5+56=356.5\)

Given, median=137 units and mean=137.05 units
We know that,
Mode=3(Median-2(Mean)
=3(137)-2(137.05)
=411-274.10=136.90
Hence, the value of mode is 136.90 units.

The cumulative frequency table of given data is

Here, n=200 ∴\(\frac { n }{ 2 } =\frac { 200 }{ 2 } =100\) \(\)
Since, the cumulative frequency just greater than 100 is 160 and the corresponding class interval is 10-15.
∴ Median class=10-15, l=10, cf=80, h=5 ad f=80
Now, median\(\) \(=l+\left\{ \frac { \frac { n }{ 2 } -cf }{ f } \right\} \times h=10+\left\{ \frac { 100-80 }{ 80 } \right\} \times 5\)
\(=10+\left( \frac { 20 }{ 80 } \right) \times 5=10+1.25=11.25\) ​​​​

First ten prime numbers in ascending order are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Here, n=10                      [even]
∴ Median
\(=\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { n }{ 2 } \right) th+\left( \frac { n }{ 2 } +1 \right) th \right] observation\\ =\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { 10 }{ 2 } \right) th+\left( \frac { 10 }{ 2 } +1 \right) th \right] observation\)
\(\)\(=\frac { 1 }{ 2 } \) [Value of 5th observation+Value of 6th ob
servation]
\(=\frac { 1 }{ 2 } [11+13]\)
\(=\frac { 24 }{ 2 }=2\)
\(\)

Table for given data is

Here, \(\sum { f_{ i } } =39+p\) and \(\sum { f_{ i }x_{ i }=2370+30\quad p } \)
Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } \)
\(54=\frac { 2370+30p }{ 39+p } \quad \left[ \because \quad mean=54,\quad given \right] \)
\(\Rightarrow \quad 54(39+p)=2370+30p\\ \Rightarrow \quad 2106+54p=2370+30p\\ \Rightarrow \quad 24p=264\\ \Rightarrow \quad p=11\)
Hence, the value of p is 11.

a = assumed mean = 150
\(\overset { - }{ =a+ } \quad \frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
\(=150+\frac { -12 }{ 50 } \times 20\)
= 150 - 4.8 = 145.2
\(\frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
Median class = 120 -140
I = 120,/= 14, cf. = 12
Median = l + \(\left( \frac { N }{ 2 } -c.f \right) \times h\)
\(=120+\frac { 25-12 }{ 14 } \times 20\)
= 120 + 18.57 138.57
Mode = 3 Medain - 2 Mean
= 3 x 138.57 - 2 x 145.2
= 415.71 - 290.4 = 125.31
Hence, mean = 145.2, median = 138.57, mode = 125.31

It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula : Median \(=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) h, \text { we get }\)
\(525=500+\left(\frac{50-36-x}{20}\right) \times 100\)
i.e. 525 - 500 = (14 - x) \(\times\) 5
i.e. 25 = 70 - 5 x
i.e. 5 x = 70 - 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e. y = 15

Mode = \(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=45+\frac { 2 }{ 18 } x10\)
=46.1

Here class intervals are not in inclusive form. So, we first convert them in inclusive form by subtracting 1/2 from the lower limit and adding 1/2 to the upper limit of each cases, where h is the difference between the lower limit of a class and the upper limit of the preceding class. The given frequency distribution in inclusive form is as follows:

It is clear from table that the modal class is 34.5 - 44.5.
Here, l = 34.5, h = 10, f₁ = 23,f₀ = 21, f₂ = 14
Now,   Mode =  \(l+\frac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \times h\)
\(\Rightarrow\)      Mode = \(34.5+\frac { 23-21 }{ 46-21-14 } \times 10\)
= \(34.5+\frac { 2 }{ 11 }\times10\)
= 34.5 + 36.31 = 36.31 is the average age.
(ii) Mode of grouped data.
(iii) If we practise habit of cleanlines, we will be able to put disease at an arm's length.

35

Median

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 = \(\frac { 84 }{ 3 } =28\)

So, frequency of the class 30 - 40 is 3.
Class mark of the class: 10-25 = \(\frac { 10+25 }{ 2 } \)
 \(=\frac { 35 }{ 2 } =17.5\)

Median = \(\frac { 1 }{ 3 } \) Mode+ \(\frac { 2 }{ 3 } \) Mean
= \(\frac { 1 }{ 3 } (12.4)+\frac { 2 }{ 3 } (10.5)\)
\(=\frac { 12.4 }{ 3 } +\frac { 21 }{ 3 } \)
\(=\frac { 12.4+21 }{ 3 } =\frac { 33.4 }{ 3 } \)
\(\frac { 33.4 }{ 3 } =11.13\)

 N=100
 \(\frac { N }{ 2 } =\frac { 100 }{ 2 } =50\)
Hence Median Class is 155 - 160

Given, Median = Mean + 3
Also, we know that,
Mode = 3 Median - 2 Mean
= 3 (Mean + 3)-2 Mean
⇒ Mode = Mean + 9
Hence Mode exceeds Mean by 9

32

45

20

Table for given data is

Here, \(\sum { f_{ i }=30 } \) and \(\sum { f_{ i }x_{ i } } =780\)
Mean \(\overline { x } =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 780 }{ 30 } =26\)
Hence, the mean number of pages written per day is 26.