Question: solve f using standard normal table, 36 weeks or less. oss Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in

solve f using standard normal table, 36 weeks or less.

oss Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in the following table for uiliding new motels. a) The expected (estimated) time for activity C is weoks. (Round your response to two decimal places.) b) The variance for activity C is weeks. (Round your rosponse to two decimal places.) c) Based on the calculation of the estimated times, the critical path is ACFHJK d) The estimated time for the critical path is weeks. (Round your response to two decimal places.) e) The activity variance aiong the critical path is Weeks. (Round your msponse to two decimal places.) f) Using the slandard normal table, the probability that Hopkins Hospitality will finish the project in 36 weeks or less is probatiity and round your response to throe decimal places.) The table below shows tha tntal araa undar the nnrmal rume fnr a nnint that ie 7 etandard daviatione in the rinht of the mean. \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline 2.2 & 0.9861 & 0.9865 & 0.9868 & 0.9871 & 0.9875 & 0.9878 & 0.9881 & 0.9884 & 0.9887 & 0.9890 \\ \hline 2.3 & 0.9893 & 0.9896 & 0.9898 & 0.9901 & 0.9904 & 0.9906 & 0.9909 & 0.9911 & 0.9913 & 0.9916 \\ \hline 2.4 & 0.9918 & 0.9920 & 0.9922 & 0.9925 & 0.9927 & 0.9929 & 0.9931 & 0.9932 & 0.9934 & 0.9936 \\ \hline 2.5 & 0.9938 & 0.9940 & 0.9941 & 0.9943 & 0.9945 & 0.9946 & 0.9948 & 0.9949 & 0.9951 & 0.9952 \\ \hline 2.6 & 0.9953 & 0.9955 & 0.9956 & 0.9957 & 0.9959 & 0.9960 & 0.9961 & 0.9962 & 0.9963 & 0.9964 \\ \hline 2.7 & 0.9965 & 0.9966 & 0.9967 & 0.9968 & 0.9969 & 0.9970 & 0.9971 & 0.9972 & 0.9973 & 0.9974 \\ \hline 2.8 & 0.9974 & 0.9975 & 0.9976 & 0.9977 & 0.9977 & 0.9978 & 0.9979 & 0.9980 & 0.9980 & 0.9981 \\ \hline 2.9 & 0.9981 & 0.9982 & 0.9983 & 0.9983 & 0.9984 & 0.9984 & 0.9985 & 0.9985 & 0.9986 & 0.9986 \\ \hline 3.0 & 0.9987 & 0.9987 & 0.9987 & 0.9988 & 0.9988 & 0.9989 & 0.9990 & 0.9989 & 0.9990 & 0.9990 \\ \hline \end{tabular}

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