(a) $10k^2+9k=1\Rightarrow k=0.1$; then probabilities $0.81,\ 0.19,\ 0.8$ and minimum $x=4$. (b) $Z=1.23<1.96\Rightarrow$ accept $H_0$.

Part (a) — probability distribution. (The variable is discrete; "continuous" in the stem is a printing slip — the table is a probability mass function.)

(i) Find $k$. Total probability $=1$:

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

$10k^{2}+9k=1\ \Rightarrow\ 10k^{2}+9k-1=0\ \Rightarrow\ k=\frac{-9\pm\sqrt{81+40}}{20}=\frac{-9\pm11}{20}.$

So $k=0.1$ or $k=-1$; probability cannot be negative, hence $k=0.1$.

(ii) Required probabilities (with $k=0.1,\ k^{2}=0.01$):

$P(x<6)=P(0)+\dots+P(5)=0+k+2k+2k+3k+k^{2}=8k+k^{2}=0.8+0.01=0.81.$

$P(x\ge6)=P(6)+P(7)=2k^{2}+(7k^{2}+k)=9k^{2}+k=0.09+0.1=0.19.$

(Check: $0.81+0.19=1$ ✓.)

$P(0<x<5)=P(1)+P(2)+P(3)+P(4)=k+2k+2k+3k=8k=0.8.$

(iii) Minimum $x$ with $P(X\le x)>\tfrac12$. Cumulative values:

$P(X\le0)=0,\ P(X\le1)=0.1,\ P(X\le2)=0.3,\ P(X\le3)=0.5,\ P(X\le4)=0.8.$

$P(X\le3)=0.5$ is not $>\tfrac12$, but $P(X\le4)=0.8>\tfrac12$. So the minimum value is $x=4$.

Part (b) — test of a single mean (large sample $Z$-test). …