Given p.d.f. is 
\(f(x)= \begin{cases}k & \text { for } 0< x \leq 4 \\ 0, & \text { otherwise }\end{cases}\)
Since f(x) is a p.d.f.\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ 4 }{ kdx=1\Rightarrow k\int _{ 0 }^{ 4 }{ dx=1\Rightarrow k{ [x] }_{ 0 }^{ 4 }=1 } } \)
\(\Rightarrow k(4-0)=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \)
Also P(2≤x≤4) =\(\int _{ 0 }^{ 4 }{ f(x)dx } \)
\(=2\int _{ 2 }^{ 4 }{ kdx=\int _{ 2 }^{ 4 }{ \frac { 1 }{ 4 } dx=\frac { 1 }{ 4 } \int _{ 2 }^{ 4 }{ dx } } } \)
\(=\frac { 1 }{ 4 } { [x] }_{ 2 }^{ 4 }=\frac { 1 }{ 4 } (4-2)=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\\ \therefore P(2\le X\le 4)=\frac { 1 }{ 2 } \)

Since the profit is Rs. 2000 and the loss is Rs. 1000
X can take values 2000 and -1000.
∴ The probability mass function is

∴ Expected profit E(X) = Σxp(x)
= 2000(0.4)-1000(0.6)
= 800-600
E(X) = Rs. 200
E(X²) = Σx²p(x)
= 2000²(0.4)-1000²(0.6)
= 1600000+600000
= 2200000
∴ Var(X) = E(X²)-[E(X)]²
= 2200000-(200)²
= 2200000-40000
= 2160000.
Standard deviation = \(\sqrt{variance}=\sqrt{2160000}\)
= 1469.69.

1) 0≤F(x)≤1, -∞
2) F(-∞) = 0 and F(∞) = 1
3) is a non-decreasing function, F(a) ≤F(b) for a
4) \(\underset { h\rightarrow 0 }{ lim } \) F(x+h) = F(x), since F(x) is continuous from the right
5) F'(x) = f(x)≥0
6) P(a≤x≤b) = F(b)-F(a)

Given probability distribution function is
\(= \begin{cases}kx & x =2, 4, 6 \\ k(x - 2), & x = 8 \\ 0, & \text { otherwise } \\ \end{cases}\)
where k is a constant

Since the given function is a probability distribution function, each ρ_i>0Σ ρ_i=1
⇒ 2k+4k+6k+6k = 1
⇒ 18 k = 1
⇒ k = \(\frac{1}{18}\)

If X is a continuous random variable and f(x) is the value of its probability density function at x, then the expected value of X is
\(E(X)=\int _{ -\infty }^{ \infty }{ x.(x)dx } \)

Let X be a discrete random variable with probability mass function p(x), then its expected value is defined by
E(X) =\(\sum _{ x }^{ }{ x.p(x) } \)

Given that in an investment profit is Rs. 5000 with probability of 0.62 or a loss of Rs.8000 with a probability of 0.38.
Hence, the probability mass function is

∴ Expected gain E(X) = 5000(0.62) - 8000 (0.32)
= 3100-3040
= Rs. 60
Hence, the expected gain is = Rs. 60

S = {1, 2, 3, 4, 5, 6}⇒ n(s) = 6
∴ X takes the values 1, 2, 3, 4, 5, 6
P(X = 1) = \(\frac{1}{6}\)
P(X = 2) = \(\frac{1}{6}\)
P(X = 3) = \(\frac{1}{6}\)
P(X = 4) = \(\frac{1}{6}\)
P(X = 5) = \(\frac{1}{6}\)
P(X = 6) = \(\frac{1}{6}\)
[Since in all the cases, only one favourable event and total no of events is 6]
∴ The probability mass function is

∴ Expected value for the random value of an unbiased die
\(E(X)=\sum _{ i=1 }^{ 6 }{ xp(x) } \)
\(=1(\frac { 1 }{ 6 } )+2(\frac { 1 }{ 6 } )+3(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 6 } )+5(\frac { 1 }{ 6 } )+6(\frac { 1 }{ 6 } )\)
\(=\frac { 1+2+3+4+5+6 }{ 6 } =\frac { 21 }{ 6 } =\frac { 7 }{ 2 } \)
\(\\ \therefore E(X)=3.5\)

When we perform any experiment, we expect an outcome. We associate a real numbers with each outcome of an experiment. In other words, we considering a function whose domain is the set of possible outcomes and whose range is subset of the set of real numbers such a function is called random variable.

Continuous random variable :
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable. For eg., height of students in a school.

The given probability function is

Since the given function is a probability function, each ρ_i>0 and Σρ_i = 1
⇒ k + 2k + 3k + 4k = 1
⇒10k = 1 ⇒ k = \(\frac{1}{10}\)
⇒ k = 0.1

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(0≤X≤10)=P(X=0)+P(X=10)
\(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X<0)=P(X=-2)
= 1/4

We have F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1  f(x) = 0 for x ≤ 1 
Again F(x) = 1 for x > 3 
f(x) = d/dx (1) = 0 for x > 3 
In 1 < x ≤ 3, F(x) = k(x – 1)4 
f(x) = d/dx (k(x – 1)4) = 4k(x – 1)3 
\(\therefore f(x)=4k{ (x-1) }^{ 3 }for\quad 1\le x\le 3\)
i) Since f(x) is a probability density function,
\(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ { 4k(x-1) }^{ 3 }dx=1 } \)

\(\Rightarrow k[{ (3-1) }^{ 4 }-{ (0) }^{ 4 }]=1\)
\(\Rightarrow k({ 2 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
ii) \(\therefore\)p.d.f
\(f(x)=\frac { 4\times 1 }{ 16 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)
\(f(x)=\frac { 1 }{ 4 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)