draw network diagram and identify critical path ysinh expected duration

is image 2 correct

\begin{tabular}{clcccccc} & \multicolumn{9}{c}{ Most } & Most & \multicolumn{2}{c}{ Most } \\ Task & \multicolumn{1}{c}{ Description } & Predecessor & Optimistic Likely & Pessimistic Expected & Variance \\ A & Requirements & - & 2 & 3 & 4 & 3 & 0.111111111 \\ B & Market Assesment & A & 4 & 7 & 10 & 7 & 1 \\ C & Design & A & 5 & 6 & 7 & 6 & 0.111111111 \\ D & Development & C & 6 & 7 & 14 & 8 & 1.777777778 \\ E & Testing & D & 7 & 9 & 11 & 9 & 0.444444444 \\ F & Revisions & B,E & 4 & 5 & 6 & 5 & 0.111111111 \\ G & Documentation & D & 3 & 6 & 9 & 6 & 1 \\ H & Quality Assurance & C,E & 1 & 4 & 7 & 4 & 1 \\ I & Pricing & B & 2 & 2 & 2 & 2 & 0 \\ J & Production & F,G,H,I & 3 & 6 & 9 & 6 & 1 \\ K & Distribution & J & 2 & 3 & 4 & 3 & 0.111111111 \end{tabular} CPM \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline ACTIVIT & DUFATION & ES & EF & IS & UF & SLACX & Cancen \\ \hline A & 3. & 0 & 0+3=3 & 33=0 & \begin{tabular}{l} RNLSIB \\ 8cl=3 \end{tabular} & 33=0 & res: \\ \hline a & 7 & 3MAXEF(A)= & 3+7=10 & \begin{tabular}{c} 267= \\ 19 \end{tabular} & & \begin{tabular}{c} 2610= \\ 16 \end{tabular} & \\ \hline c & 6 & & 3+6=9 & 96=3 & \begin{tabular}{l} MiNLSID \\ g H=9 \end{tabular} & 99=0 & yes: \\ \hline D & 8 & 9MAXEF(C)= & 9+8=17 & 178=9 & \begin{tabular}{l} Mentsie \\ 8.07=17 \end{tabular} & \begin{tabular}{c} 17+17= \\ 0 \end{tabular} & VES: \\ \hline E & 9 & \begin{tabular}{c} MAX EF (D)= \\ in \end{tabular} & \begin{tabular}{c} 17+9= \\ 20 \end{tabular} & \begin{tabular}{c} 269= \\ 17 \end{tabular} & \begin{tabular}{c} MNLS[F \\ =20 \end{tabular} & \begin{tabular}{c} 2026= \\ 0 \end{tabular} & YES: \\ \hline F & 5. & \begin{tabular}{c} MAXEFE B \\ Ela 26 \end{tabular} & \begin{tabular}{c} 28+5= \\ 31 \end{tabular} & \begin{tabular}{c} 315= \\ 26 \end{tabular} & \begin{tabular}{c} MNLSPY \\ =31 \end{tabular} & \begin{tabular}{c} 3131= \\ 0 \end{tabular} & YES \\ \hline 6 & 6 & \begin{tabular}{c} MAX EF (D)= \\ 17 \end{tabular} & \begin{tabular}{c} 17+6= \\ 23 \end{tabular} & \begin{tabular}{c} 316= \\ 25 \end{tabular} & \begin{tabular}{c} MNLS(H) \\ an \end{tabular} & \begin{tabular}{c} 3123= \\ 8 \end{tabular} & \\ \hline H & 4. & \begin{tabular}{c} MAX EFIC 8 \\ E) =26 \end{tabular} & \begin{tabular}{c} 26+4= \\ 30 \end{tabular} & \begin{tabular}{c} 314= \\ 27 \end{tabular} & =31M & \begin{tabular}{c} 3130= \\ 1 \end{tabular} & \\ \hline 1 & 2. & \begin{tabular}{c} MAX EF (B)= \\ 10 \end{tabular} & \begin{tabular}{c} 10+2= \\ 12 \end{tabular} & \begin{tabular}{c} 312= \\ 29 \end{tabular} & \begin{tabular}{c} MNLST(J) \\ =31 \end{tabular} & \begin{tabular}{c} 31+12= \\ 19 \end{tabular} & \\ \hline J & 6 & & \begin{tabular}{c} 31+6= \\ 37 \end{tabular} & \begin{tabular}{c} 376= \\ 31 \end{tabular} & 37Lin(K)= & \begin{tabular}{c} 3737= \\ 0 \end{tabular} & YES: \\ \hlinex & 3. & \begin{tabular}{c} MAxEF(D)= \\ 37 \end{tabular} & \begin{tabular}{c} 37+3= \\ 40 \end{tabular} & \begin{tabular}{c} 403= \\ 37 \end{tabular} & 40 & \begin{tabular}{c} 4040= \\ 0 \end{tabular} & Yes \\ \hline \end{tabular} FS = MAX(EF OF ALL. PREDECESSORS); O FOR THE FIRST ACTIVTY EF ES * DURATION SLACK =1FEF CRTICAL PATH = LONCEST PATH WITH O SLACK CATICAL PATH I ACDEFJK DURATION OF CAITICAL PATH =40 3.66 2. PROBABLUTY OF COMPLETION ON OR BEFORE 40 EXPECTED DURATION =40 DUE DURATION =40 VARANCE =3.66, STANDARD CEVIATION OF COMPICTICN = SEPTNARIANCE = SOPT(366) z= \&DUE DURATION - EXPECTED DURATION//STANDARD DEVATIOW OF COMPLETION =(4040)/ SONT (366)=0 FOR A Z VALUE OF O PAOBABUTY DISTRBUTION =0.5 OR EON