Mean\(\bar{x}=\cfrac{\Sigma fx}{\Sigma f}=\cfrac{2795}{50}=55.9\)
Median = \(\left(\cfrac{N}{2}\right)^{th} \) term is called the median class
N = 50
\(\cfrac{N}{2}=\cfrac{50}{2}=25\)
\(\therefore\) Median class = 25^th value = 54.5 - 64.5
l = 54.5, m = 22, f = 14, c = 10
Median = \( l+\cfrac { \left( \cfrac { N }{ 2 } -m \right) }{ f } \times c\)
= 54.5 + \(\cfrac{25-22}{14}\times10\)
= 54.5 + \(\cfrac{3}{14}\times10\)
= 54.5 + \(\cfrac{30}{14}\)
= 54.5 + 2.14
= 56.64
Model class is 54.5 - 64.5 since it has the maximum frequency.
l = 54.5, f = 14, f₁ = 10, f₂ = 8, c = 10
\(\therefore\) Mode =\(l+\left[\cfrac{f-{f}_{1}}{2f-{f}_{1}{f}_{2}}\right]\times c\)
= 54.5+\(\left[\cfrac{14-10}{2\times 14-10-8}\right]\times 8\)
= 54.5+ \(\left(\cfrac{4}{28-18}\right)\times 10\)
= 54.5+ \(\cfrac{4}{10}\times 10=58.5\)
\(\therefore \) Mean = 55.9
Median = 56.64
Mode = 58.5

Mode =\( l+\left[ \cfrac { f-{ f }_{ 1 } }{ 2f-{ f }_{ 1 }-{ f }_{ 2 } } \right] \times c\)
Here Model class 20-30 since ,it has the maximum frequency
l=20, f=46, f₁=38. f₂= 34, c = 30-20 =10

\(\therefore\) Mode = \( l+\left[ \cfrac { f-{ f }_{ 1 } }{ 2f-{ f }_{ 1 }-{ f }_{ 2 } } \right] \times c\)

= \(20+\left[\cfrac{46-38}{2\times46-38-34}\right]\times40\)

= 24

Total of the data = 11 + 15 + 17 + x + 1 + 19 + x -2 + 3
= 2x + 64
Mean of the data = \(\frac{2 x+64}{7}\)
= 14 (given)
\(\frac{2 x+64}{7}=\frac{14}{1}\)
2x + 64 = 98
2x = 98 - 64
2x = 34
\(\therefore x=\frac{34}{2}=17\)
The data = 11, 15, 17, 17+1, 19, 17- 2, 3
= 3, 11, 15, 15, 17, 18, 19
15 occurs two times
\(\therefore\) The Mode of data = 15

The monthly salary of 10 employees are Rs. 5000, Rs. 7000, Rs. 5000, Rs. 7000, Rs. 8000, Rs. 7000, Rs. 7000, Rs. 8000, Rs.7000, Rs. 5000
Writing in ascending order Rs. 5000, Rs. 5000, Rs. 5000, Rs. 7000, Rs. 7000, Rs. 7000, Rs. 8000, Rs. 8000
Number of values = 10 which is an even number.
Median = Average of \(\cfrac{{10}^{th}}{2} \) and \(\left(\cfrac{10}{2}+1\right)^{th}\)
\( \text {Mean } =\frac{(5000 \times 3)+(7000 \times 5)+(8000 \times 2)}{10} \)
\( =\frac{15000+35000+16000}{10} =\frac{66000}{10} \)
= Rs. 6600 
\( \text { Median } =\text { Average of }\left(\frac{10}{2}\right)^{\text {th }} \text { and }\left(\frac{10}{2}+1\right)^{t h} \text { term } \)
\( =\text { Average of } 5^{\text {th }} \text { and } 6^{\text {th }} \text { term } \)
\(=\frac{7000+7000}{10}=\frac{14000}{2} = 7000\) 
Median = Rs. 7000 
Mode of salary = Rs. 7000

Data in ascending order 27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63 
Number of terms = 13
If n is odd rhen Median = \(\left[\frac{\mathrm{N}+1}{2}\right]^{t h}\) observation
\(=\left(\frac{13+1}{2}\right)^{t h}\)observation
= 7^th observation
7^th observation = 32

Values in ascending order 
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
Number of terms = 10 which is even
Median = Average of \(\left[\frac{N}{2}\right]^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\)
= Average of [5^th term + 6^th term]
\(=\frac{44+44}{2}=44\)
Median = 44