Question

A random variable X has the following probability function:
Values of X, x: 0 1 2 3 4 5 6 7
p(x): 0 k 2k 2k 3k k2 2k2 7k2+ k


(i) Find k, (ii) Evaluate P(X < 6), (P X ≥ ),6 and 0(P < X < ),5 (iii) If ,
2
1
(P X ≤ )a >
Find the minimum value of a, and (iv) Determine the distribution function of X.

Answer 1

A random variable $X$ has the following probability function:  

$$\def\arraystretch{1.5} \begin{array}{c:c} Values\ of\ x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X=x) & 0 & k & 2k & 2k & 3k & k^2 & 2k^2 & 7k^2+k \end{array}$$

(i) Find $k$

For a discrete probability distribution of a random variable  $\displaystyle\sum_{i=1}^nP(X=x_i)=1.$ Then

$$0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1,\ k\geq0$$ $$10k^2+9k-1=0$$ $$(k+1)(10k-1)=0$$

Since $k\geq0,$ we take $k=\dfrac{1}{10}.$

(ii) Evaluate $P(X<6),\ P(X\geq6)$ and $P(0<X<5)$

$$P(X<6)=P(X=0)+P(X=1)+P(X=2)+$$ $$+P(X=3)+P(X=4)+P(X=5)=$$ $$1-P(X=6)-P(X=7)=$$ $$=1-2(0.1)^2-(7(0.1)^2+0.1)=0.81$$

$$P(X\geq6)=P(X=6)+P(X=7)=$$ $$=2(0.1)^2(7(0.1)^2+0.1)=0.19$$

$$P(0<X<5)=P(X=1)+P(X=2)+$$ $$+P(X=3)+P(X=4)=$$ $$=0.1+2(0,1)+2(0.1)+3(0.1)=0.8$$

(iii) If $P(X\leq a)>{1 \over 2},$ find the minimum value of $a$

$$\def\arraystretch{1.5} \begin{array}{c:c} Values\ of\ x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X=x) & 0 & 0.1 & 0.2 & 0.2 & 0.3 & 0.01 & 0.02 & 0.17 \end{array}$$

$$P(X\leq3)=0+0.1+0.2+0.2=0.5$$

$$P(X\leq4)=0+0.1+0.2+0.2+0.3=0.8>0.5$$

The minimum value of $a$ is $4.$

(iv) Determine the distribution function of $X.$

$F(0)=P(X\leq0)=0$

$F(1)=P(X\leq1)=0+k=k=0.1$

$F(2)=P(X\leq2)=0+k+2k=3k=0.3$

$F(3)=P(X\leq3)=0+k+2k+2k=5k=0.5$

$F(4)=P(X\leq4)=0+k+2k+2k+3k=8k=0.8$

$F(5)=P(X\leq5)=0+k+2k+2k+3k+k^2=$

$=8k+k^2=0.81$

$F(6)=P(X\leq6)=0+k+2k+2k+3k+k^2+2k^2=$

$=8k+3k^2=0.83$

$F(7)=P(X\leq7)=10k^2+9k=1$

$$F(x)=x = \begin{cases} 0 &\text{ } x<1 \\ 0.1 &\text{ } x<2 \\ 0.3 &\text{ } x<3 \\ 0.5 &\text{ } x<4 \\ 0.8 &\text{ } x<5 \\ 0.81 &\text{ } x<6 \\ 0.83 &\text{ } x<7 \\ 1 &\text{ } x\geq7 \end{cases}$$