Mean \(\overset { - }{ x } =\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \)
\(\Rightarrow 53=\frac { 3340+70p }{ 72+p } \)
\(\Rightarrow 3340+70p=53(72+p)\)
\(\Rightarrow 3340+70p=3816-3340\)
\(\Rightarrow 70p-53p=3816-3340\)
\(\Rightarrow 17p=476\)
\(p=\frac { 476 }{ 17 } =28\)

\(\Sigma f_{ i }=40\)
26+p+q=40
p+q=14
Mean x= \(\frac { \Sigma x_{ i }f_{ i } }{ \Sigma f_{ i } } \)
\(14.7=\frac { 408+9p+27q }{ 40 } \)
588=408+9p+27q
180=9p+27q
p+3q=20
Subtracting eq. (i) from eq. (ii),
2q = 6
q=3
Putting this value of q in eq. (i),
p = 14- q = 14- 3 = 11
P = 11,q = 3

Mean=42.2, Mode=43.1

167.13

Given, distribution is the more than type distribution.
Here, we observe that, all 30 students have obtained marks more than or equal to 10. So, 30-23=7 students lie in the class 5-10. Similarly, we can find the other classes and their corresponding frequencies. Now, we construct the continuous grouped frequency distribution as

Here, class intervals are not continuous. But mid-value x_i of each class interval would be same either class interval is continuous or not continuous.
So, we solve it without making it continuous.
Also, x_i are larger so we apply step-deviation method. Her, class width (h)=5. Table for the given data is

We have, a = 72, h = 5, \(\sum { f_{ i }=40 } \) and \(\sum { f_{ i }u_{ i } } =72\)
By step-deviation method,
Mean
\(\left( \overline { x } \right) =a+\frac { \sum { f_{ i } } }{ \sum { f_{ i }u_{ i } } } \times h=72+\frac { 22 }{ 40 } \times 5=72+2.75=74.75\)

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.

From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f₁) of modal class = 18
Frequency (f₀) of class preceding modal class = 4
Frequency (f₂) of class succeeding modal class = 9
Class size (h) = 1000
\(=4000+\left(\frac{18-4}{2(18)-4-9}\right) \times 1000 \)
\(\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \)
\(=4000+\left(\frac{14000}{23}\right)\)
= 4000+608.695
= 4608.695
Therefore, mode of the given data is 4608.7 runs.

To find the class mark (x_i) for each interval, the following relation is used.
Class size = 50
Taking 225 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows.

From the table, we obtain
\(\sum f_{i}=25 \)
\(\sum f_{i} u_{i}=-7 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{1} u_{i}}{\sum f_{i}}\right) x h \)
\(=225+\left(\frac{-7}{25}\right) \times(50)\)
= 225 - 14
= 211
Therefore, mean daily expenditure on food is Rs 211.

 Here, \(\frac { n }{ 2 } =\frac { 25 }{ 2 } =12.5\)
Since, cumulative frequency just greater than 12.5 and corresponding class is 60-70.
Median class=60-70
Since, highest frequency is 9 and corresponding class is 50-60
Modal class=50-60

The abscissa of the point of intersection of the "Less than type" and "More than type" cumulative frequency curve of a grouped data is median.

Median

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

Mean  \(= \frac { \Sigma fx }{ \Sigma f } \)
\(50=\frac { \Sigma fx }{ 100 } \)
\(\Sigma fx=5000\)
Correct \(\Sigma fx^{ ' }=5000-100+110\)
=5010
Correct Mean  \(= \frac { 5010 }{ 100 } \)
= 50.1
Median Will remain same median = 52

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

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

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

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

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.

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