Values in ascendiirg order 21, 41, 43, 47, 47, 53, 62, 71, 83. 
Number of terms = 9 which is odd
The middle term (Median) = \(=\left(\frac{l+1}{2}\right)^{t h} \text { term }\)
\(=\frac{9+1}{2} \text { term }=5^{\text {th }} \text { term }\)
Median = 47

Let x₁, x₂, x₃, x₄, x₅, x₆ be the given set of values then \({\sum^6_{i=1}\over6}=45\)
If each value is increased by 4, then the mean of new set of values is
New A.M \(\bar X={\sum^6_{i=1}(x_i+4)\over6} \)
\(={(x_1+4)+(x_2+4)+(x_3+4)+(x_4+4)+(x_5+4)+(x_6+4)\over6}\)
\(={\sum^6_{i=1}(x_i+24)\over6}={\sum_{i=1}^6x_i\over6}+4\)
\(\bar X=45+4=49\)

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

Given, Mean = 66 and Mode = 60.
Using, Mode ≈ 3Median – 2Mean
60 ≈ 3Median – 2(66)
3 Median ≈ 60 +132
Therefore, Median ≈ \({192\over 3}≈64\)

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

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!