Processor Scheduling (SJF / FCFS)
1. FCFS

Turnaround Time T T = 5.866 is correct.

2. Non-PreEmptive SJF

T T = 5.2 is correct.

3. Future-Knowledge (Non-PreEmptive SJF with Idle time)

T T = 6.2, I'm not sure about this.

We never get a future knowledge of how big the next process will be.

4. PreEmptive SJF (SRTF)

Here, your scheduling is wrong.

$P_2$ arrives at time 1.4, how can you schedule it at time 0.4?

Refer below scheduling:

\begin{array} {|r|r|}
\hline
P_1 &amp;P_2 &amp;P_3 &amp;P_2 &amp;P_1 \\
\hline
0-1.4 &amp;1.4-3 &amp;3-4 &amp;4-5.4 &amp;5.4-9 \\
\hline
\end{array}  

T T ($P_1$) = 9 - 0=9
T T ($P_1$) = 5.4 - 1.4=4
T T ($P_1$) = 4 - 3 =1
Avg T T = 4.6  

Dont you see the improved TT of one over other.

FCFS = 5.8 > NON-PRE SJF = 5.2 > PRE SJF =4.6