Mean = 8.15

Here, \(\Sigma\)f_i = 120 and \(\Sigma\)f_ix_i = 13280
\(\therefore \text { Mean }=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{13280}{120}=110.66\)
Hence, the mean of given data is 110.66.

Modal Class is 20 - 25, l = 20, f₁ = 75, f₀ = 45, f₂ = 35, h = 5
Mode = 20 + \(\left({ { 75-45}\over{ 2\times75-45-35} } \right)\times5\)
= 20 + \({ { 30 }\over{ 70 } }\times5=20+{ {15 }\over{ 7} }=22.142\) = 22.14

\(\therefore\)  Mode = 12
\(\because\)  it has highest frequency.

(i) Mode is 19.
(ii) If 19 is changed to 1 6. then frequency of 16 is 5 and frequency of 19 becomes 1i.e.,16 has maximum frequency
\(\Rightarrow\) 16 is the mode.

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

The multiples of 5, according to the problem are: 5, 15, 25, 35, 45
Mean
= \(\frac { 5+15+25+35+45 }{ 5 } \)
\(=\frac { 125 }{ 5 } =25\)

\(\Sigma fx=630\) , \(\Sigma f=40\)
Mean  = \(\frac { 630 }{ 40 } =15.75\)

Here , \(\frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
Hence, rnedian class is 30 - 40.

Modal class = 5-7
l = 5 , f₁=80,f₀=45,h=2,f₂=55
Mode = \(l+\frac { (f_{ 1 }f_{ 0 }) }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=5+\frac { 80-45 }{ 160-45-55 } \times 2=5+\frac { 35\times 2 }{ 60 } \)
= 6.17

Modal class = 35-40
l=35, f₁=50 , f₂=42 , f₀=34 , h=5
Mode = l+ \(\frac { (f_{ 1 }-f_{ 0 }) }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=35+\frac { 50-34 }{ 100-34-42 } \times 5\)
\(=35+\frac { 16\times 5 }{ 24 } =38.33\)

Modal class = 20 - 30
l=20,f₁=40 , f₀=24,f₂=36 , h=10
Mode = l + \(\frac { \left( f_{ 1 }-f_{ 0 } \right) }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=20+\frac { \left( 40-24 \right) }{ 80-24-36 } \times 10\)
\(=20+\frac { 16\times 10 }{ 20 } =28\)

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 = \(\frac { 84 }{ 3 } =28\)

Median class : 40 - 50 \(\Rightarrow \) Lower Limit =40
Modal class   : 40 - 50  \(\Rightarrow \) Upper Limit = 50
Their sum = 40+50=90

Here \(\frac { N }{ 2 } =\frac { 60 }{ 2 } =30\)
So , Median Class = 20-30

l=20, f=30 , c.f =20 , h=10
Median = l + \(\left( \frac { \frac { N }{ 2 } -c.f }{ f } \right) \times h\)
=20+ \(\frac { 30-20 }{ 30 } \times 10\)
\(=20+\frac { 100 }{ 30 } =20+\frac { 10 }{ 3 } \)
= 20+3.3
Median = 23.33

x₁=12
x₂=6
x₃=23
x₄=7

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

Mean \(=\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } \)
 \(= \frac { 200 }{ 30 } =6.67\)

Writing the data as discrete frequency distribution, we get

For 20 to be mode of the frequency distribution,  \(\lambda \) =20

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

p=23, q=13, median class=400-500, modal class=500-600

32

123.4 kg

14.6

10

20

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

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.

Table for the given data is

Here, \(\sum { f_{ i }=100 } \), \(\sum { f_{ i }x_{ i } } =1228+24k\) and mean=16.6
Mean\(=\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } \Rightarrow 16.6=\frac { 1228+24k }{ 100 } \)
\(\Rightarrow \quad 16.6\times 100=1228+24k\\ \Rightarrow \quad 24k=1660-1228=432\Rightarrow k=\frac { 432 }{ 24 } =18\)

The cumulative frequency table for given data is

Here, \(\frac { n }{ 2 } =\frac { 67 }{ 2 } =33.5\)
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125-145.
Thus, we have median class 125-145.
Also, the maximum frequency is of the class 125-145.
Therefore, the modal class is 125-145.
Difference of the upper limit of median class and the lower limit of modal class=145-125=20