Mean = E(X)=\(\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ -12 }^{ 12 }{ x.\left( \frac { 1 }{ 24 } \right) } dx } \)
\(=\frac { 1 }{ 24 } \int _{ -12 }^{ 12 }{ x.dx } \)
\(=0[\because \int _{ -a }^{ a }{ f(x)dx=0 } when\ f(x)\ is\ an\ odd\ function]\)
\(\therefore E(X)=0\)

Let X denote the amount
∴ X is a random variable. taking the values 10 and -5 when 3 coins are tossed, sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH,TTT}
∴ P(X = 10) = P (getting 3 heads or 3 tails)
\(=\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)
P(X = -5) = p(getting 1head or 2heads)
\(=\frac { 6 }{ 8 } =\frac { 3 }{ 4 } \)
∴ Probability distribution function is

∴ Expected gain E(X) = Σx_ip_i = 10(\(\frac{1}{4}\))-5(\(\frac{3}{4}\))
\(=\frac { 10 }{ 4 } -\frac { 15 }{ 4 } =-\frac { 5 }{ 4 } =-1.25\)
∴E(X) = -1.25 [A loss of Rs. 1.25]

Let X be a random variable. That denote the mark obtained by a student for answering a question.
∴ X can take values 1 and -\(\frac{1}{2}\)
∴ P(X = 1) = P (answering a question correctly) 
= \(\frac{1}{4}\)
P(X = -\(\frac{1}{2}\)) = P(answering a question wrongly)
=\(1-\frac{1}{4}=\frac{3}{4}\)
∴ Probability distribution function is

\(\therefore E(x)=\sum { xp(x)=1(\frac { 1 }{ 4 } )-\frac { 1 }{ 2 } \left( \frac { 3 }{ 4 } \right) =\frac { 1 }{ 4 } -\frac { 3 }{ 8 } } \)
\(=\frac { 2-3 }{ 8 } =-\frac { 1 }{ 8 } \)
∴ Expectation of mark for answering a single question is -\(\frac{1}{8}\)
∴ Expectation of mark for answering 120 questions = 120(-\(\frac{1}{8}\)) = -15.

Since f(x) is a probability density function,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ \infty }{ { ce }^{ -ax }dx=1 } \Rightarrow C\int _{ 0 }^{ \infty }{ { e }^{ -ax }dx=1 } \)
\(\Rightarrow c{ \left[ \frac { { e }^{ -ax } }{ -a } \right] }_{ 0 }^{ \infty }=1\Rightarrow \frac { -c }{ a } [{ e }^{ -\infty }-{ e }^{ 0 }]\)
\(\Rightarrow \frac { -c }{ a } [0-1]=1\quad [\because { e }^{ -\infty }=0,{ e }^{ 0 }=1]\)
\(\Rightarrow \frac { c }{ a } =1\Rightarrow C=a\quad \therefore C=a\)

Clearly f(x) ≥0 for all real values of x
\(\therefore \int _{ -\infty }^{ \infty }{ f(x)dx } =\int _{ 0 }^{ 3 }{ \frac { 2x }{ 9 } dx=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ xdx } } \)
\(=\frac { 2 }{ 9 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }=\frac { 2 }{ 9 } \left[ \frac { 9 }{ 2 } -0 \right] =\frac { 2 }{ 9 } \times \frac { 9 }{ 2 } =1\)
∴ f(x) is a probability density function.