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Project 3 instructions Based on Brase & Brase: sections XXXXXXXXXX Note that you must do this project on your own—you may not work with other students. You are always welcome to ask your instructor...

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Project 3 instructions

Based on Brase & Brase: sections XXXXXXXXXX

Note that you must do this project on your own—you may not work with other students. You are always welcome to ask your instructor for help.

Visit the NASDAQ historical prices weblink. First, set the date range to be for exactly 1 year ending on the Monday that this course started. For example, if the current term started on April 1, 2018, then use April 1, 2017 – March 31, XXXXXXXXXXDo NOT use these dates. Use the dates that match up with the current term.) Do this by clicking on the blue dates after “Time Period”. Next, click the “Apply” button. Next, click the link on the right side of the page that says “Download Data” to save the file to your computer.

This project will only use the Close values. Assume that the closing prices of the stock form a normally distributed data set. This means that you need to use Excel to find the mean and standard deviation. Then, use those numbers and the methods you learned in sections XXXXXXXXXXof the course textbook for normal distributions to answer the questions. Do NOT count the number of data points.

Complete this portion of the assignment within a single Excel file. Show your work or explain how you obtained each of your answers. Answers with no work and no explanation will receive no credit.

1. a) Submit a copy of your dataset along with a file that contains your answers to all of the following questions.

b) What the mean and Standard Deviation (SD) of the Close column in your data set?

c) If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at less than the mean for that year? Hint: You do not want to calculate the mean to answer this one. The probability would be the same for any normal distribution. (5 points)

2. If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at more than $1150? (5 points)

3. 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) (5 points)

4. If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at less than $950 per share. Would this be considered unusal? Use the definition of unusual from the course textbook that is measured as a number of standard deviations (5 points)

5. At what prices would Google have to close in order for it to be considered statistically unusual? You will have a low and high value. Use the definition of unusual from the course textbook that is measured as a number of standard deviations. (5 points)

6. What are Quartile 1, Quartile 2, and Quartile 3 in this data set? Use Excel to find these values. This is the only question that you must answer without using anything about the normal distribution. (5 points)

7. 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. (5 points)

Answered Same Day Dec 08, 2021

Solution

Pooja answered on Dec 09 2021
136 Votes
GOOG (8)
    1) a)                                1) b)
    Date    Open    High    Low    Close    Adj Close    Volume        mean =     1137.316360876
    10/22/18    1103.060059    1112.22998    1091    1101.160034    1101.160034    1514200        sd =    68.1697666007
    10/23/18    1080.890015    1107.890015    1070    1103.689941    1103.689941    1848700
    10/24/18    1104.25    1106.119995    1048.73999    1050.709961    1050.709961    1982400        c)
    10/25/18    1071.790039    1110.97998    1069.550049    1095.569946    1095.569946    2545800        P(X    10/26/18    1037.030029    1106.530029    1034.089966    1071.469971    1071.469971    4187600        0.5
    10/29/18    1082.469971    1097.040039    995.830017    1020.080017    1020.080017    3880700
    10/30/18    1008.460022    1037.48999    1000.75    1036.209961    1036.209961    3212700        2)
    10/31/18    1059.810059    1091.939941    1057    1076.77002    1076.77002    2529800        P(X>1150)
    11/1/18    1075.800049    1083.974976    1062.459961    1070    1070    1482000         =1-NORMDIST(1150,1137.316,68.1698,TRUE)
    11/2/18    1073.72998    1082.974976    1054.609985    1057.790039    1057.790039    1839000        0.4261969739
    11/5/18    1055    1058.469971    1021.23999    1040.089966    1040.089966    2441400
    11/6/18    1039.47998    1064.344971    1038.069946    1055.810059    1055.810059    1233300        3)
    11/7/18    1069    1095.459961    1065.900024    1093.390015    1093.390015    2058400        P(mean-50     11/8/18    1091.380005    1093.27002    1072.204956    1082.400024    1082.400024    1488200        P(X < mean+50) - P(X< mean-50)
    11/9/18    1073.98999    1075.560059    1053.109985    1066.150024    1066.150024    1343200         =NORMDIST(1137.316+50,1137.316,68.1698,TRUE)-NORMDIST(1137.316-50,1137.316,68.1698,TRUE)
    11/12/18    1061.390015    1062.119995    1031    1038.630005    1038.630005    1471800        0.5367236806
    11/13/18    1043.290039    1056.60498    1031.150024    1036.050049    1036.050049    1513700
    11/14/18    1050    1054.563965    1031    1043.660034    1043.660034    1565900        4)
    11/15/18    1044.709961    1071.849976    1031.780029    1064.709961    1064.709961    1836100        P(X<950)
    11/16/18    1059.410034    1067    1048.97998    1061.48999    1061.48999    1658100         =NORMDIST(950,1137.316,68.1698,TRUE)
    11/19/18    1057.199951    1060.790039    1016.26001    1020    1020    1858600        0.0029999607
    11/20/18    1000    1031.73999    996.02002    1025.76001    1025.76001    2449100        This is unusual as probability is less than 0.05
    11/21/18    1036.76001    1048.560059    1033.469971    1037.609985    1037.609985    1534300
    11/23/18    1030    1037.589966    1022.398987    1023.880005    1023.880005    691500        5)
    11/26/18    1038.349976    1049.310059    1033.910034    1048.619995    1048.619995    1942800        P(Z    11/27/18    1041    1057.579956    1038.48999    1044.410034    1044.410034    1803200        z =     -1.644853627
    11/28/18    1048.76001    1086.839966    1035.76001    1086.22998    1086.22998    2475400        (X-mean)/sd =     -1.644853627
    11/29/18    1076.079956    1094.244995    1076    1088.300049    1088.300049    1468900        X=     1025.1870730344
    11/30/18    1089.069946    1095.569946    1077.880005    1094.430054    1094.430054    2580200
    12/3/18    1123.140015    1124.650024    1103.665039    1106.430054    1106.430054    1991200        6)
    12/4/18    1103.119995    1104.420044    1049.97998    1050.819946    1050.819946    2345200        Q1=    1079.945038
    12/6/18    1034.26001    1071.199951    1030.77002    1068.72998    1068.72998    2769200        Q2=    1139.664978
    12/7/18    1060.01001    1075.26001    1028.5    1036.579956    1036.579956    2101200        Q3=    1193.28994725
    12/10/18    1035.050049    1048.449951    1023.289978    1039.550049    1039.550049    1807700
    12/11/18    1056.48999    1060.599976    1039.839966    1051.75    1051.75    1394700        7)
    12/12/18    1068    1081.650024    1062.790039    1063.680054    1063.680054    1523800
    12/13/18    1068.069946    1079.76001    1053.930054    1061.900024    1061.900024    1329800
    12/14/18    1049.97998    1062.599976    1040.790039    1042.099976    1042.099976    1686600
    12/17/18    1037.51001    1053.150024    1007.900024    1016.530029    1016.530029    2385400
    12/18/18    1026.089966    1049.47998    1021.440002    1028.709961    1028.709961    2192500
    12/19/18    1033.98999    1062    1008.049988    1023.01001    1023.01001    2479300
    12/20/18    1018.130005    1034.219971    996.359985    1009.409973    1009.409973    2673500
    12/21/18    1015.299988    1024.02002    973.690002    979.539978    979.539978    4596000
    12/24/18    973.900024    1003.539978    970.109985    976.219971    976.219971    1590300
    12/26/18    989.01001    1040    983    1039.459961    1039.459961    2373300
    12/27/18    1017.150024    1043.890015    997    1043.880005    1043.880005    2109800
    12/28/18    1049.619995    1055.560059    1033.099976    1037.079956    1037.079956    1414800
    12/31/18    1050.959961    1052.699951    1023.590027    1035.609985    1035.609985    1493300
    1/2/19    1016.570007    1052.319946    1015.710022    1045.849976    1045.849976    1532600
    1/3/19    1041    1056.97998    1014.070007    1016.059998    1016.059998    1841100
    1/4/19    1032.589966    1070.839966    1027.417969    1070.709961    1070.709961    2093900
    1/7/19    1071.5    1074    1054.76001    1068.390015    1068.390015    1981900
    1/8/19    1076.109985    1084.560059    1060.530029    1076.280029    1076.280029    1764900
    1/9/19    1081.650024    1082.630005    1066.400024    1074.660034    1074.660034    1199300        Since the histogram is bell shaped, the distribution is approximately...
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