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probability

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Question

How would I calculate this :The personnel office at a large electronics firm regularly schedules job interviews

and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 36 minutes and standard deviation σ = 6 minutes. (Round your answers to four decimal places.) (a) What is the probability that a first interview will last 40 minutes or longer? Seven first interviews are usually scheduled per day. What is the probability that the average length of time for the seven interviews will be 40 minutes or longer?

 
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histogram

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Question

How do I get calculate the class, frequency and midpoint to make a histogram for the distribution of this data?

Each time I tried to put in the data it does not come out correctly. I need assistance. Is the normality assumption that was made at the beginning valid? Why or why not? Hint: Does this distribution have the properties of a normal distribution as described in the course textbook? Real data sets are never perfect; however, it should be close. One option would be to construct a histogram like you did in Project 1 to see if it has the right shape. Something in the range of 10 to 12 classes is a good number. The max is 1268.33 and min is 913.81. I just can’t seem to get this one.

Close 913.81 921.29 929.57 939.33 937.34 928.45 927.81 935.95 926.5 929.08 932.07 935.09 925.11 920.29 915 921.81 931.58 932.45 928.53 920.97 924.86 944.49 949.5 959.11 953.27 957.79 951.68 969.96 978.89 977 972.6 989.25 987.83 989.68 992 992.18 992.81 984.45 988.2 968.45 970.54 973.33 972.56 1019.27 1017.11 1016.64 1025.5 1025.58 1032.48 1025.9 1033.33 1039.85 1031.26 1028.07 1025.75 1026 1020.91 1032.5 1019.09 1018.38 1034.49 1035.96 1040.61 1054.21 1047.41 1021.66 1021.41 1010.17 998.68 1005.15 1018.38 1030.93 1037.05 1041.1 1040.48 1040.61 1049.15 1064.19 1077.14 1070.68 1064.95 1063.63 1060.12 1056.74 1049.37 1048.14 1046.4 1065 1082.48 1086.4 1102.23 1106.94 1106.26 1102.61 1105.52 1122.26 1121.76 1131.98 1129.79 1137.51 1155.81 1169.97 1164.24 1170.37 1175.84 1175.58 1163.69 1169.94 1167.7 1111.9 1055.8 1080.6 1048.58 1001.52 1037.78 1051.94 1052.1 1069.7 1089.52 1094.8 1102.46 1111.34 1106.63 1126.79 1143.75 1118.29 1104.73 1069.52 1078.92 1090.93 1095.06 1109.64 1126 1160.04 1164.5 1138.17 1149.49 1149.58 1135.73 1099.82 1097.71 1090.88 1049.08 1021.57 1053.21 1005.1 1004.56 1031.79 1006.47 1013.41 1025.14 1027.81 1007.04 1015.45 1031.64 1019.97 1032.51 1029.27 1037.98 1074.16 1072.08 1087.7 1072.96 1067.45 1019.98 1021.18 1040.04 1030.05 1017.33 1037.31 1024.38 1023.72 1048.21 1054.79 1053.91 1082.76 1097.57 1098.26 1100.2 1079.23 1081.77 1078.59 1066.36 1079.58 1069.73 1079.69 1079.24 1075.66 1060.32 1067.8 1084.99 1119.5 1139.29 1139.66 1136.88 1123.86 1120.87 1129.99 1139.32 1134.79 1152.12 1152.26 1173.46 1168.06 1169.84 1157.66 1155.48 1124.81 1118.46 1103.98 1114.22 1115.65 1127.46 1102.89 1124.27 1140.17 1154.05 1152.84 1153.9 1183.48 1188.82 1183.86 1198.8 1195.88 1186.96 1184.91 1205.5 1248.08 1263.7 1268.33 1238.5 1219.74 1217.26 1220.01 1226.15 1223.71 1224.77 1242.22 1245.61 1249.1 1237.61 1235.01 1242.1 1214.38 1206.49 1200.96 1207.77 1201.62 1207.33 1205.38 1220.65 1241.82

 
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share

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Question

How would I calculate this: If a person bought 1 share of Google stock within the

last year, what is the probability that the stock on that day closed within $50 of the mean for that year? (between 50 below and 50 above the mean). The mean is 85.028 and the SD is 1076.025

 
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McMahon group

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Question

If you were McMahon group, how would you present the findings of your study to the NCRCC board? Explain the

rationale for your chosen method.

 
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