Q.A random variable $x$ has the following probability distribution : $X = x$: $0, 1, 2, 3, 4, 5, 6, 7$ with $P(X = x)$: $0,\ k,\ 2k,\ 2k,\ 3k,\ k^2,\ 2k^2,\ 7k^2 + k$. Find

$k=0.1$, the mean is $3.66$, and $P(0<X<5)=0.8$.

(i) Find $k$. Since $\sum P(X=x)=1$:

$0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$

$\Rightarrow 10k^2+9k-1=0\Rightarrow(10k-1)(k+1)=0.$

Rejecting $k=-1$, we get $k=\dfrac{1}{10}.$

(ii) Mean. With $k=0.1$ the probabilities are

$0,\ 0.1,\ 0.2,\ 0.2,\ 0.3,\ 0.01,\ 0.02,\ 0.17$ for $x=0,\dots,7.$ …