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

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

Arrange the values in ascending order.
10, 10, 12, 13, 16, 17, 18, 19, 21, 22.
The number of values = 10
Median = Average of \(\left(10\over2\right)^{th}\) and \(\left({10\over2}+1\right)^{th}\) values
= Average of 5^th and 6^th values
\(={16+17\over2}={33\over2}=16.5\)

56.96 (or) 57 (approximately)

Given that the total number of students n = 25, \(\bar X\)= 784
So, Incorrect \(\sum x=\bar X\times n=78.4\times25=1960\)
Correct Σx = incorrect Σx - wrong entry + correct entry
= 1960 - 69 + 96 = 1987
Correct \(\bar X={correct\sum x\over n}={1987\over 25}=79.48\)

\(\bar{X}=\frac{\sum_{i=1}^{8} x_{i}}{n}=\frac{21+30+22+16+24+28+18+17}{8}=\frac{176}{8}=22\)
Deviation of an entry x_i from the arithmetic mean \(\bar X\) is \(x_i-\bar X\), i = 1, 2,........8
Sum of the deviations = (21-22)+(30-22)+(22-22)+(16-22)+(24-22)+(28-22)+(18-22)+(17-22)
= 16–16 = 0. or equivalently, \(\sum_{i=1}^{8}\left(x_{i}-\bar{X}\right)=0\)
Hence, we conclude that sum of the deviations from the Arithmetic Mean is zero.

Mean \(\bar X={Σfx\over Σf}={860\over 30}=28.67\)
Hence the average age = 28.67

Modal class is 16 -20 since it has the maximum frequency.
l = 15.5, f = 32, f₁ = 16, f₂ = 24, c = 20.5–15.5 = 5
Mode \(=l+{\left(f-f_1\over 2f-f_1-f_2\right)}\times c\)
\(=15.5+\left(32-16\over64-16-24\right)\times 5\)
\(=15.5+\left(16\over24\right)\times = 15.5 + 3.33 =18.83\)

Let us arrange the marks in ascending order and prepare the following data:

Here N = 41
Median = size of \(\left(N+1\over 2\right)^{th}\) value = size of \(\left(41+1\over2\right)^{th}\) value = size of 21^st value.
If the 41 students were arranged in order (of height), the 21^st student would be the middle most one, since there are 20 students on either side of him/her. We therefore need to find the height against the 21^st student. 15 students (see cumulative frequency) have height less than or equal to 154 cm. 22 students have height less than or equal to 155 cm. This means that the 21^st student has a height 155 cm.
Therefore, Median = 155 cm

3.1, 3.2, 3.3, 2.1,1.3, 3.3, 3.1
In this given data 3.1, 3.3 occurs twice
\(\therefore\) mode = 3.1 and 3.3(bimodal)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 145+148+142+141+139+140 }{ 6 } =\frac { 855 }{ 6 } \)
x = 142.5 mm²

\(\bar { x } =60kg\)  
\(\bar { x } =\cfrac { \Sigma x }{ n } =\cfrac { 56+68+72+x }{ 4 } =60\)
196 + x = 260
x = 240-196
\(\therefore \) The weight of the fourth number = 44 kg

6 has the maximum frequency 9. Therefore 6 is the mode.

In this example, three values 20, 21, 22 occur two times each. There are three modes for the given data!