Calculation:

Let the grade in the final test be x

Average of n numbers \(\displaystyle{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}\) is given by \(\displaystyle{A}={\frac{{{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}}}{{{n}}}}\)

Plugging the values,

\(\displaystyle{A}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\)

Given average is 75

Therefore,

\(\displaystyle{75}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\)

Multiplying both sides by 4

\(\displaystyle{75}\cdot{4}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\cdot{4}\)

\(\displaystyle{300}={219}+{x}\)

Subtracting 219 from both the sides,

\(\displaystyle{300}—{219}={219}+{x}-{219}\)

\(\displaystyle{81}={x}\)

\(\displaystyle{x}={81}\)

Therefore the score required in final test to get an average of 75 is 81.

Let the grade in the final test be x

Average of n numbers \(\displaystyle{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}\) is given by \(\displaystyle{A}={\frac{{{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}}}{{{n}}}}\)

Plugging the values,

\(\displaystyle{A}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\)

Given average is 75

Therefore,

\(\displaystyle{75}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\)

Multiplying both sides by 4

\(\displaystyle{75}\cdot{4}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\cdot{4}\)

\(\displaystyle{300}={219}+{x}\)

Subtracting 219 from both the sides,

\(\displaystyle{300}—{219}={219}+{x}-{219}\)

\(\displaystyle{81}={x}\)

\(\displaystyle{x}={81}\)

Therefore the score required in final test to get an average of 75 is 81.