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 \(\bar X={correct\sum x\over n}={1987\over 25}=79.48\)

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 \(\bar X={\sum^6_{i=1}(x_i+4)\over6} \)
\(={(x_1+4)+(x_2+4)+(x_3+4)+(x_4+4)+(x_5+4)+(x_6+4)\over6}\)
\(={\sum^6_{i=1}(x_i+24)\over6}={\sum_{i=1}^6x_i\over6}+4\)
\(\bar X=45+4=49\)

\(\bar{X}=\frac{\sum_{i=1}^{8} x_{i}}{n}=\frac{21+30+22+16+24+28+18+17}{8}=\frac{176}{8}=22\)
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.

Let Assumed mean A = 28, class width c = 8

Mean
\(\bar X=A+{Σfd\over Σf}\times c\)
\(=28+\left(-22\over100\right)\times8\)
= 28 -1.76 = 26.24

Let Assumed mean A =170 

Mean \(\bar X = A+{Σfd\over Σf}\)
\(=170+\left(-780\over32\right)\)
Therefore \(\bar X\) = 170 - 24.375
= 145.625

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