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# Probability and Statistics Class

**Need help with this question or any other
Statistics assignment help
task?**

See attached homework sets 1,2,3,4,5,6, which are also quizzes I need to complete (60 minutes limit). If this goes well then I will need help for an exam coming up in four weeks (120 - 180 minutes limit).

##### Additional Instructions:

Sheet1
Company Group Ratio
1 h 1.50
2 h 0.10
3 h 1.76
4 h 1.14
5 h 1.84
6 h 2.21
7 h 2.08
8 h 1.43
9 h 0.68
10 h 3.15
11 h 1.24
12 h 2.03
13 h 2.23
14 h 2.50
15 h 2.02
16 h 1.44
17 h 1.39
18 h 1.64
19 h 0.89
20 h 0.23
21 h 1.20
22 h 2.16
23 h 1.80
24 h 1.87
25 h 1.91
26 h 1.67
27 h 1.87
28 h 1.21
29 h 2.05
30 h 1.06
31 h 0.93
32 h 2.17
33 h 2.61
34 h 3.05
35 h 1.52
36 h 1.93
37 h 1.95
38 h 2.61
39 h 1.11
40 h 0.95
41 h 0.96
42 h 2.25
43 h 2.73
44 h 1.56
45 h 2.73
46 h 0.90
47 h 2.12
48 h 1.42
49 h 1.62
50 h 1.76
51 h 2.22
52 h 2.80
53 h 1.85
54 h 0.96
55 h 1.71
56 h 1.02
57 h 2.50
58 h 1.55
59 h 1.69
60 h 1.64
61 h 1.03
62 h 1.80
63 h 0.67
64 h 2.44
65 h 2.30
66 h 2.21
67 h 1.96
68 h 1.81
69 f 0.82
70 f 0.89
71 f 1.31
72 f 0.05
73 f 0.83
74 f 0.90
75 f 1.68
76 f 0.99
77 f 0.62
78 f 0.91
79 f 0.52
80 f 1.45
81 f 1.16
82 f 1.32
83 f 1.17
84 f 0.42
85 f 0.48
86 f 0.93
87 f 0.88
88 f 1.10
89 f 0.23
90 f 1.11
91 f 0.19
92 f 0.13
93 f 2.03
94 f 0.51
95 f 1.12
96 f 0.92
97 f 0.26
98 f 1.15
99 f 0.13
100 f 0.88
101 f 0.09

30-year mortgage rate
30-year mortgage rate (Aug 2020)
2.89
3.36
3.28
2.79
3.08
3.18
2.96
2.94
3.02

Covid-19
12-Aug-20 Deaths per 1M Tests per 1M
Belgium 852 162,882
UK 688 277,778
Peru 651 79,115
Spain 611 159,806
Italy 583 121,909
Sweden 571 90,738
Chile 533 99,764
USA 509 202,846
Brazil 486 62,198
France 465 84,239
Mexico 418 8,615
Netherlands 359 69,751
Ecuador 337 15,206
Bolivia 322 16,952
Colombia 265 38,922
Canada 238 120,217
Switzerland 230 98,696
Iran 226 32,852
South Africa 185 55,486
Portugal 173 173,297
Honduras 153 11,934
Romania 146 74,269
Argentina 112 19,712
Germany 111 102,448
Russia 105 214,522
Turkey 70 64,601
Israel 69 217,118
Kazakhstan 67 116,123
Belarus 63 146,474
Hungary 63 37,935
Azerbaijan 49 79,266
Egypt 49 1,317
Poland 48 59,657
https://www.worldometers.info/coronavirus/country/belgium/https://www.worldometers.info/coronavirus/country/france/https://www.worldometers.info/coronavirus/country/mexico/https://www.worldometers.info/coronavirus/country/netherlands/https://www.worldometers.info/coronavirus/country/ecuador/https://www.worldometers.info/coronavirus/country/bolivia/https://www.worldometers.info/coronavirus/country/colombia/https://www.worldometers.info/coronavirus/country/canada/https://www.worldometers.info/coronavirus/country/switzerland/https://www.worldometers.info/coronavirus/country/iran/https://www.worldometers.info/coronavirus/country/south-africa/https://www.worldometers.info/coronavirus/country/uk/https://www.worldometers.info/coronavirus/country/portugal/https://www.worldometers.info/coronavirus/country/honduras/https://www.worldometers.info/coronavirus/country/romania/https://www.worldometers.info/coronavirus/country/argentina/https://www.worldometers.info/coronavirus/country/germany/https://www.worldometers.info/coronavirus/country/russia/https://www.worldometers.info/coronavirus/country/turkey/https://www.worldometers.info/coronavirus/country/israel/https://www.worldometers.info/coronavirus/country/kazakhstan/https://www.worldometers.info/coronavirus/country/belarus/https://www.worldometers.info/coronavirus/country/peru/https://www.worldometers.info/coronavirus/country/hungary/https://www.worldometers.info/coronavirus/country/azerbaijan/https://www.worldometers.info/coronavirus/country/egypt/https://www.worldometers.info/coronavirus/country/poland/https://www.worldometers.info/coronavirus/country/spain/https://www.worldometers.info/coronavirus/country/italy/https://www.worldometers.info/coronavirus/country/sweden/https://www.worldometers.info/coronavirus/country/chile/https://www.worldometers.info/coronavirus/country/us/https://www.worldometers.info/coronavirus/country/brazil/
Wages
wage IQ
3845 93
4040 119
4125 108
3250 96
2810 74
7000 116
3000 91
5405 114
5770 111
5000 95
4650 132
4605 102
4500 125
6590 119
8960 118
4790 105
6800 109
4250 72
4150 105
2355 101
6375 123
8075 113
4365 95
10685 145
5265 114
8010 124
5940 93
4000 115
7085 125
3175 128
5000 103
7120 98
13340 108
3330 129
8895 132
3910 92
7860 108
6370 106
3570 105
5405 123
3460 108
6590 122
6195 109
5135 100
8740 125
4905 122
3850 105
5770 94
5775 102
4040 109
5500 105
5770 134
8745 108
5000 104
2310 112
3845 120
4375 124
6875 103
7260 115
4000 96
8740 123
5755 98
4200 96
4890 89
4815 109
3095 93
2210 82
3000 120
6830 122
8215 117
7275 109
11550 114
8410 126
6175 82
4275 119
5360 104
5200 115
5000 97
3375 105
5500 100
4980 114
3660 73
6000 96
8470 113
3430 106
3770 104
4285 80
4160 104
2895 122
3360 96
12500 95
5380 105
3750 94
5930 91
4165 96
3250 69
6250 110
5610 111
4325 110
4040 97
6495 125
4515 91
4500 86
3125 110
7930 92
4810 85
7695 120
5550 106
6410 112
3850 91
5000 90
4475 86
6025 86
3750 113
3270 111
3005 111
3000 106
2165 98
5940 105
3175 105
6125 118
5755 90
4325 95
5155 112
5245 120
5000 123
5525 103
9620 121
6730 90
4045 125
7475 109
6730 128
6000 97
2500 96
6625 97
4500 78
4000 112
4000 88
5170 97
4900 101
4420 106
2400 59
4615 105
2565 119
5525 93
5965 82
13855 134
3895 84
4750 98
6970 118
7475 113
3250 100
3350 93
5630 119
2250 67
5140 111
12020 106
9495 127
3785 113
6250 115
5810 102
5125 85
5500 117
3570 125
6590 118
7055 121
10810 115
6365 120
5700 109
2890 109
4710 85
5290 85
3750 106
5000 101
4755 116
3175 91
6250 91
3375 110
2000 75
2885 94
2950 109
4615 122
6250 102
5500 122
5650 105
3260 113
3090 104
4810 124
2645 110
4085 95
4810 120
4200 105
4330 121
12020 96
5630 110
5800 92
3615 98
8890 109
9515 118
5050 97
4855 95
2625 99
2625 106
3350 117
2500 90
5290 98
2750 95
2500 115
3635 106
4325 114
5405 92
6520 119
2875 85
3115 98
2575 114
6365 84
4950 108
3000 107
5800 84
2500 86
3975 88
2500 97
3700 84
6250 117
5070 110
6250 104
4565 110
6730 99
2225 80
1325 101
6250 117
8035 108
7260 69
6955 109
4105 85
3970 114
2500 90
2600 107
8650 117
9620 112
5775 106
10810 116
4615 123
5575 118
2245 91
7500 103
4130 114
4685 102
4890 104
5155 112
6360 107
5680 117
4000 110
6695 96
5315 90
4675 97
4040 99
1875 99
2685 97
5410 95
4650 126
5775 109
2740 86
3110 119
4205 102
3845 105
2935 117
9620 112
5290 89
2085 131
6010 114
5770 98
5350 124
6010 125
3555 100
6010 84
4250 113
5000 102
2450 82
5000 112
4325 111
6875 96
7930 127
8010 97
15390 120
4490 101
4530 120
4760 99
2855 78
2225 110
1445 94
7220 120
4810 108
5375 103
4545 97
6250 106
3100 99
5080 72
4000 88
5250 119
5395 87
3270 121
3905 112
5190 103
9620 99
6010 101
3330 117
4525 109
4450 73
4085 116
4060 98
2885 104
3780 78
5055 104
5775 85
5125 104
6750 107
5005 109
3980 119
6150 96
3770 93
3570 90
5000 103
10335 106
4560 96
3000 114
4755 86
3555 110
5755 87
5000 89
4205 106
2000 100
5875 111
6010 100
7210 114
2690 98
3905 89
3750 74
4205 110
3500 130
6730 108
4000 98
6250 125
5525 104
2375 98
3810 104
4810 120
3605 134
4000 92
3290 129
6350 118
6565 107
4120 92
7210 118
7000 108
5190 119
3340 109
5500 91
5000 118
2615 119
5555 103
4810 120
3645 122
3450 119
5050 104
3000 96
2980 83
4250 115
3350 88
3965 96
7210 123
3350 108
4380 116
4205 109
4875 97
6115 107
4550 95
2665 109
3750 104
6030 110
3725 94
4500 103
5850 123
2700 113
2750 110
3075 100
4545 95
3845 100
4920 108
4165 101
4395 90
5135 115
5000 114
2325 84
5500 92
3205 125
5175 107
6060 129
4750 98
4690 108
6250 114
2930 111
3465 101
2810 99
1875 75
3365 91
3270 113
3460 120
5555 98
6840 107
6410 121
6250 115
6730 134
7120 96
4270 120
4440 105
5805 121
2915 113
6300 106
4735 104
9250 80
7875 117
3790 126
7210 112
2445 104
5630 116
5000 70
2500 83
6000 84
2825 82
9600 109
3420 111
3870 94
1165 96
4875 101
6830 90
10685 120
3500 120
6000 108
5805 99
3645 80
3750 93
5130 85
5555 118
3125 115
6000 119
7705 94
5770 109
1550 104
3050 89
8745 101
5000 96
1750 109
3825 94
3950 115
4090 92
2385 71
4690 130
10495 119
1750 74
4700 104
6010 124
2250 94
5290 120
5000 96
1590 67
2780 69
4790 98
4975 92
8000 109
3030 100
2555 101
7055 111
6730 106
7610 102
5375 115
6000 99
6885 118
4370 132
3125 108
6250 98
5410 111
3465 79
3635 97
3075 99
4565 101
4420 120
3490 97
4000 106
4060 131
5000 93
2745 79
6500 100
4615 107
7695 112
3605 105
4075 114
3000 116
5500 104
5290 85
4810 93
2165 100
4700 71
7085 88
5000 93
2500 98
4000 110
3830 109
3750 94
2750 77
3975 89
3615 102
5195 113
1000 103
2000 90
2875 96
4250 97
2540 87
5810 89
5995 70
6350 104
1750 116
4815 85
7315 104
4010 99
3210 93
3755 114
2440 103
7000 92
4700 98
4330 114
3375 77
1875 89
1625 93
1730 110
7210 132
2800 112
2750 99
4750 96
3420 106
6610 128
8170 95
2085 80
2360 71
3330 89
8395 127
5195 114
6250 102
3800 88
5770 88
8010 110
7210 110
4325 130
3525 100
5000 101
4260 89
3850 80
5000 72
6645 115
8410 115
2885 108
3425 93
2885 111
5200 101
6250 92
3000 115
2400 76
5000 93
4940 91
2390 90
2200 84
2425 120
3525 89
7750 104
4800 107
5000 100
3125 129
6730 112
4805 76
2395 97
6005 98
6730 60
5610 132
2715 90
2250 63
2535 110
2735 88
2930 98
2310 61
3525 76
7830 127
8170 94
6410 97
2780 127
6000 109
4275 89
5265 90
5000 105
4500 100
8010 111
2540 125
7500 124
8410 107
3750 110
4215 102
2045 93
5020 81
3250 99
4040 93
10685 102
3805 95
3845 98
1725 84
2475 88
4935 109
12500 82
5490 105
6060 95
2885 92
1950 74
7500 123
2915 86
2300 92
4725 87
7210 117
6665 112
7500 119
6665 103
3500 74
4865 81
10810 124
3985 89
2000 88
5075 67
8720 106
3150 103
2225 91
3300 101
3895 85
1885 83
2800 70
5610 102
2265 94
6930 97
7695 115
5770 117
4810 105
3610 113
2400 83
4040 101
7210 113
5455 88
1750 93
2500 83
5130 113
2885 106
6665 106
4575 114
5525 119
4550 82
5000 89
5800 74
5005 77
3565 75
4645 97
2000 68
6205 83
5325 97
2015 98
4700 102
4060 81
3500 85
2875 86
2250 79
3105 84
2205 74
3125 92
3630 93
2500 69
5000 113
1965 92
3000 127
4810 109
4810 97
4325 66
5770 109
6930 109
3660 104
4325 85
3500 81
4875 87
6500 78
4500 54
4145 98
5000 98
4135 105
2500 87
5775 117
4750 85
3500 82
8550 114
2665 65
10020 105
4450 92
15390 107
7695 111
2540 73
6770 119
5715 97
4810 110
6250 96
4950 67
4525 102
4630 64
7795 103
6560 118
4615 118
4395 118
4000 75
5245 112
2750 95
5950 119
2915 94
6000 103
3985 90
6855 128
1800 110
6350 84
9160 97
4545 102
8730 131
2600 126
4040 107
10685 108
3460 92
4655 108
4060 115
4330 111
7210 115
2500 101
6920 114
2640 116
4405 90
5130 125
6060 98
8100 111
9215 109
8010 124
5000 121
3685 103
8495 96
8495 100
5125 104
5535 112
3125 73
3600 101
4275 99
6250 95
5650 109
4500 110
9620 98
4810 112
9370 64
7865 90
4700 102
4500 119
4410 96
8550 118
6300 79
3755 96
5485 80
5005 137
5770 116
5000 69
6250 78
4285 99
6055 92
4685 113
4520 98
4360 115
6000 96
6305 117
4750 96
575 108
3365 92
5420 95
5290 94
4000 115
1845 117
9620 105
4275 109
4320 87
5460 110
5125 110
4250 109
3365 85
4000 105
5245 122
4420 108
4000 122
6000 79
3320 105
5555 90
4250 111
5000 110
4375 92
6010 131
3330 104
4615 110
3940 99
4750 86
7210 105
4330 85
3125 72
3400 86
3465 68
4375 95
11540 116
5980 96
3500 85
3605 100
7500 101
5250 97
4005 100
3200 96
5990 117
1950 93
4445 116
5380 106
5635 98
3750 103
4275 118
2625 100
5520 96
2765 67
4000 96
6920 104
2125 106
2550 62
1850 73
2980 62
2010 76
2090 76
5665 106
3405 94
4750 81
3000 87
3305 96
4625 75
2750 95
3030 86
2125 86
1700 104
3460 95
2875 116
2855 66
4085 101
4935 96
3080 104
5130 105
4040 106
4040 101
3940 96
4250 105
4500 106
7090 102
1300 75
3310 99
2810 90
2810 104
1785 115
5045 118
7210 108
3255 76
3750 50
3770 55
3500 73
2515 84
4685 104
3120 92
3750 83
4500 98
2700 70
3210 87
2000 96
4500 80
2565 104
4470 104
6410 96
2425 89
1625 113
3845 105
3090 84
5200 108
3755 104
1900 106
1500 83
3765 70
5325 104
5350 91
7865 112
3250 97
3500 99
2470 74
4450 100
2600 86
4455 90
2850 101
7220 109
2405 98
2500 107
7365 110
4015 120
4810 97
5000 103
3000 114
2250 95
3145 93
2460 67
7810 97
1785 97
4800 83
2830 104
2405 82
7210 113
3225 93
3940 100
3220 101
2385 100
3320 82
2600 79
6010 102
2690 77
4365 109
5000 107
Amazon
Date AMZN SP500 10-Year TB
1-Aug-20 3,167.46 3,351.28 0.56%
1-Jul-20 3,164.68 3,271.12 0.54%
1-Jun-20 2,758.82 3,100.29 0.65%
1-May-20 2,442.37 3,044.31 0.65%
1-Apr-20 2,474.00 2,912.43 0.62%
1-Mar-20 1,949.72 2,584.59 0.70%
1-Feb-20 1,883.75 2,954.22 1.13%
1-Jan-20 2,008.72 3,225.52 1.52%
1-Dec-19 1,847.84 3,230.78 1.92%
1-Nov-19 1,800.80 3,140.98 1.78%
1-Oct-19 1,776.66 3,037.56 1.69%
1-Sep-19 1,735.91 2,976.74 1.68%
1-Aug-19 1,776.29 2,926.46 1.51%
1-Jul-19 1,866.78 2,980.38 2.02%
1-Jun-19 1,893.63 2,941.76 2%
1-May-19 1,775.07 2,752.06 2.14%
1-Apr-19 1,926.52 2,945.83 2.51%
1-Mar-19 1,780.75 2,834.40 2.41%
1-Feb-19 1,639.83 2,784.49 2.71%
1-Jan-19 1,718.73 2,704.10 2.64%
1-Dec-18 1,501.97 2,506.85 2.69%
1-Nov-18 1,690.17 2,760.17 3.01%
1-Oct-18 1,598.01 2,711.74 3.16%
1-Sep-18 2,003.00 2,913.98 3.06%
1-Aug-18 2012.709961 2901.52002 2.85%
1-Jul-18 1777.439941 2816.290039 2.96%
1-Jun-18 1699.800049 2718.370117 2.85%
1-May-18 1629.619995 2705.27002 2.82%
1-Apr-18 1566.130005 2648.050049 2.94%
1-Mar-18 1447.339966 2640.870117 2.74%
1-Feb-18 1512.449951 2713.830078 2.87%
1-Jan-18 1450.890015 2823.810059 2.72%
1-Dec-17 1169.469971 2673.610107 2.40%
1-Nov-17 1176.75 2584.840088 2.42%
1-Oct-17 1105.280029 2575.26001 2.38%
1-Sep-17 961.349976 2519.360107 2.33%
1-Aug-17 980.599976 2471.649902 2.12%
1-Jul-17 987.780029 2470.300049 2.29%
1-Jun-17 968 2423.409912 2.30%
1-May-17 994.619995 2411.800049 2.20%
1-Apr-17 924.98999 2384.199951 2.28%
1-Mar-17 886.539978 2362.719971 2.40%
1-Feb-17 845.039978 2363.639893 2.36%
1-Jan-17 823.47998 2278.870117 2.45%
1-Dec-16 768.659973 2238.830078 2.48%
1-Nov-16 750.570007 2198.810059 2.37%
1-Oct-16 789.820007 2126.149902 1.83%
1-Sep-16 837.309998 2168.27002 1.61%
1-Aug-16 769.16 2170.95 1.56%
1-Jul-16 758.81 2173.6 1.45%
1-Jun-16 715.62 2098.86 1.48%
1-May-16 722.79 2096.95 1.85%
1-Apr-16 659.59 2065.3 1.84%
1-Mar-16 593.64 2059.74 1.77%
1-Feb-16 552.52 1932.23 1.74%
1-Jan-16 580 1940.24 1.92%
1-Dec-15 675.89 2043.94 2.27%
1-Nov-15 664.8 2080.41 2.21%
1-Oct-15 625.9 2079.36 2.15%
1-Sep-15 511.89 1920.03 2.04%
1-Aug-15 512.89 1972.18 2.21%
1-Jul-15 536.15 2103.84 2.19%
1-Jun-15 434.09 2063.11 2.35%
1-May-15 429.23 2107.39 2.13%
1-Apr-15 421.78 2085.51 2.04%
1-Mar-15 372.1 2067.89 1.93%
1-Feb-15 380.16 2104.5 2.00%
1-Jan-15 354.53 1994.99 1.64%
1-Dec-14 310.35 2058.9 2.17%
1-Nov-14 338.64 2067.56 2.17%
1-Oct-14 305.46 2018.05 2.34%
1-Sep-14 322.44 1972.29 2.49%
1-Aug-14 339.04 2,003.37 2.42%
1-Jul-14 312.99 1,930.67 2.55%
2-Jun-14 324.78 1,960.23 2.60%
1-May-14 312.55 1,923.57 2.56%
1-Apr-14 304.13 1,883.95 2.70%
3-Mar-14 336.37 1,872.34 2.72%
3-Feb-14 362.1 1,859.45 2.71%
2-Jan-14 358.69 1,782.59 2.86%
2-Dec-13 398.79 1,848.36 2.90%
1-Nov-13 393.62 1,805.81 2.72%
1-Oct-13 364.03 1,756.54 2.61%
3-Sep-13 312.64 1,681.55 2.81%
1-Aug-13 280.98 1,632.97 2.74%
1-Jul-13 301.22 1,685.73 2.59%
3-Jun-13 277.69 1,606.28 2.30%
1-May-13 269.2 1,630.74 1.94%
1-Apr-13 253.81 1,597.57 1.75%
1-Mar-13 266.49 1,569.19 1.96%
1-Feb-13 264.27 1,514.68 1.98%
2-Jan-13 265.5 1,498.11 1.91%
3-Dec-12 250.87 1,426.19 1.72%
1-Nov-12 252.05 1,416.18 1.65%
1-Oct-12 232.89 1,412.16 1.75%
4-Sep-12 254.32 1,440.67 1.72%
1-Aug-12 248.27 1,406.58 1.68%
2-Jul-12 233.3 1,379.32 1.53%
1-Jun-12 228.35 1,362.16 1.63%
1-May-12 212.91 1,310.33 1.80%
2-Apr-12 231.9 1,397.91 2.05%
1-Mar-12 202.51 1,408.47 2.18%
1-Feb-12 179.69 1,365.68 1.97%
3-Jan-12 194.44 1,312.41 1.97%
1-Dec-11 173.1 1,257.60 1.97%
1-Nov-11 192.29 1,246.96 2.01%
3-Oct-11 213.51 1,253.30 2.15%
1-Sep-11 216.23 1,131.42 1.97%
1-Aug-11 215.23 1,218.89 2.28%
1-Jul-11 222.52 1,292.28 2.99%
1-Jun-11 204.49 1,320.64 3.00%
2-May-11 196.69 1,345.20 3.17%
1-Apr-11 195.81 1,363.61 3.45%
1-Mar-11 180.13 1,325.83 3.41%
1-Feb-11 173.29 1,327.22 3.58%
3-Jan-11 169.64 1,286.12 3.39%
1-Dec-10 180 1,257.64 3.31%
1-Nov-10 175.4 1,180.55 2.77%
1-Oct-10 165.23 1,183.26 2.54%
1-Sep-10 157.06 1,141.20 2.65%

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 1: BASIC PROBABILITY, DISCRETE RANDOM VARIABLES
This first assignment tests your understanding of the basic probability models, the various rules
for computing probabilities of events represented as sets, joint probability tables, conditional probabil-
ity and independence, the total probability rule, variance, correlation, and the binomial and Poisson
distributions.
Problem 1 (Probability rules). In the USA, the result of presidential elections often depends on a
small number of swing states in which the winner takes all the electoral votes in that state. Can you
compute the chances that presidential candidate X will win the elections in the following hypothetical
situation: Candidate X needs 37 electoral votes from 4 swing states to win the presidential election.
Swing State Electoral votes
F (Florida) 29
M (Michigan) 16
P (Pennsylvania) 20
W (Wisconsin) 10
(1.1) Easy. The various combinations of swing states that could be won by candidate X represent
the basic outcomes in the sample space. For example FP represents a basic outcome where X
wins Florida and Pennsylvania, and looses Michigan and Wisconsin. As an other example, ∅
represents a basic outcome where candidate X looses all four swing states. What is the size of
the sample space?
(1.2) Moderate. Represent the event ”X wins the election” as the union of all the basic outcomes
contributing to this event.
(1.3) Moderate. The probability that candidate X wins a swing state is 0.5, and the four events
”X wins F”, ”X wins M”, ”X wins P”, ”X wins W”, are pairwise independent. What is the
probability that candidate X wins the elections?
(1.4) Challenging. Polls provide the following independent probabilities for each of the four swing
states.
State Probability that X wins the state
F 0.6
M 0.8
P 0.7
W 0.6
What is the probability that candidate X will win the presidential elections?
1
2 PROBLEM SET 1: BASIC PROBABILITY, DISCRETE RANDOM VARIABLES
Problem 2 (Probability table, conditional probability and independence). Beth flies from
Miami to Pittsburgh with a 25-minute layover in Atlanta. If Beth misses her connection, she will have
to spend the night in Atlanta. The probability that Beth’s flight from Miami to Atlanta arrives on time
is 0.7; the probability that it is delayed is 0.3. If Beth arrives on time in Atlanta, the probability that
she makes the connection to Pittsburgh is 1, If she is delayed arriving in Atlanta, the probability that
she makes her connection to Pittsburgh is 0.4.
(2.1) Easy. Consider the following events: O = Flight from Miami arrives ON TIME in Atlanta; D
= Flight from Miami is DELAYED arriving in Atlanta; M = Beth MAKES the connection in
Atlanta, N = Beth does NOT make the connection and therefore spends the night in Atlanta.
Fill in the probability table.
M N
O 0.7
D 0.3
1
(2.2) Easy. What is the probability that Beth’s flight is delayed arriving in Atlanta AND she makes
her connection to Pittsburgh.
(2.3) Easy. What is the probability that Beth makes her connection in Atlanta?
If Beth arrives on time in Altanta, the probability that her luggage makes the connection
with her is 0.8. If Beth’s flight is delayed arriving in Atlanta, the probability that both Beth
and her luggage make the connection is 0.1.
(2.4) Moderate. What is the probability that both Beth and her luggage make it to Pittsburgh on
her scheduled flight?
(2.5) Moderate. Are the events O = ”Beth’s flight from Miami arrives ON TIME in Atlanta” and
L = ”Beth makes the connection and her LUGGAGE is there at her arrival in Pittsburgh”
statistically independent?
(2.6) Challenging. Given that Beth made it to Pittsburgh on her scheduled flight, what is the
probability that her luggage is there at her arrival in Pittsburgh?
(2.7) Challenging. Beth made it to Pittsburgh on her scheduled flight and her luggage was not there
at her arrival. What is the probability that Beth’s flight from Miami was delayed arriving in
Atlanta? [Hint: Let K = ”Beth made it to Pittsburgh on her scheduled flight and her luggage
was not there at her arrival”. You need to compute P (D|K). Note that M = L ∪K and that
L and K are mutually exclusive.]
Problem 3 (Correlation). Many companies such as Walmart, Costco, and Amazon take advantage of
”risk pooling” to reduce their inventories. To illustrate how risk pooling works, consider a company with
warehouses in Pittsburgh and Philadelphia to serve its customers in these two cities. Many different
products are held in inventory. The idea of risk pooling will become clear by focusing on one such
product. Weekly demand for one of the company’s products is 40 units on average in Pittsburgh with
a standard deviation of 20 units, and 80 units on average in Philadelphia with a standard deviation
of 30 units. The correlation between these two demands is ρ = 0.1. To protect against stockouts, the
company maintains an inventory at each warehouse equal to the expected weekly demand locally plus
two standard deviations of this weekly demand. The company is considering replacing its warehouses in
Pittsburgh and Philadelphia by a more centrally located warehouse in Harrisburg. The inventory level
at the Harrisburg warehouse will be based on the total demand in Pittsburgh and Philadelphia. Let us
call D this total weekly (random) demand. Inventory at the Harrisburg warehouse will be equal to the
expected value of D plus two standard deviations of D.
(3.1) Easy. What are the current inventory levels at the Pittsburgh and Philadelphia warehouses?
(3.2) Challenging. What would be the inventory needed at the Harrisburg warehouse? What
would be the decrease in inventory achieved by switching to a single warehouse in Harrisburg
as compared to (3.1)?
45-750 PROBABILITY AND STATISTICS 3
Problem 4 (Independence?). A fair coin is tossed 10 times. Let X denote the random variable
representing the number of heads and Y denote the number of tails.
(4.1) Easy. Compute the variance of X.
(4.2) Easy. Note that Y = 10−X. Use this formula to compute the variance of Y .
(4.3) Moderate. Compute the variance of X + Y . [Hint: Recall that Y = 10−X.]
(4.4) Challenging. Use (4.3) to compute the correlation between X and Y . Are X and Y indepen-
dent random variables?
(4.5) Challenging. Compute the variance of 2X − Y .
Problem 5 (Binomial Distribution). Airlines regularly overbook flights to compensate for no-show
passengers. In doing so, airlines are balancing the risk of having to compensate bumped passengers
against lost revenue associated with empty seats. In a USA Today analysis of airline statistics, it was
found that the average no-show rate was 12%. An airline books 110 passengers on a 100-seat plane.
Assume a 12% probability that a passenger will not show up for boarding.
(5.1) Easy. What is the probability that no passenger will be bumped?
(5.2) Easy. What is the probability that the plane will be full?
(5.3) Challenging. The airline makes a profit of $100 per passenger on this flight, and it estimates
the cost of a bumped passenger to be $1000 (this includes cash awards to bumped passengers,
free air vouchers, loss of goodwill, etc.). For this reason, the airline would like a probability of
at least 0.9 that no passenger will be bumped. How many tickets should the airline sell on this
100-seat plane?
Problem 6 (Poisson Distribution). The neighborhood coffee shop Moonbucks sells three types of
coffees: regular, Americano and caffe latte and charges $2, $2.50 and $4 respectively. Orders for regular
coffees occur according to a Poisson distribution with a mean of 20 per hour during the 9-10am time
period. Orders for Americanos occur according to a Poisson distribution with a mean of 8 per hour
during the same period. Orders for caffe lattes occur according to a Poisson distribution with a mean
of 10 per hour during this period.
(6.1) Easy. What is the expected number of regular coffees that Moonbucks sells between 9AM to
9:15AM.
(6.2) Easy. What is the probability that Moonbucks sells at least one regular coffee between 9AM
and 9:05AM?
(6.3) Easy. It is 9:05AM. Moonbucks had no client for regular coffees from 9AM to 9:05AM. What
is the probability that it has no client for regular coffee between 9:05AM and 9:10AM?
(6.4) Moderate. What is Moonbucks’ expected revenue between 9AM and 10AM?
(6.5) Moderate. What is the probability that Moonbucks sells 35 coffees or more (all three types
combined) between 9AM and 10AM?

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 2: DECISION TREES, CONTINUOUS RANDOM VARIABLES
This assignment tests your understanding of decision theory and continuous random variables. The
learning objectives are: constructing decision trees, calculating the value of information when making
decisions under uncertainty, working with common continuous distributions (uniform, exponential),
combining continuous and discrete distributions.
Problem 1 (Decision Trees and Value of Information). NovelTech is a high-tech company
dedicated to technological innovations. The company is trying to decide whether or not to develop a
new product. It is estimated that, with a probability of 40%, this product will be a major success with
an estimated profit of $400 million; with a probability of 30% the product will break even; and with a
probability of 30% it will be a total failure and will cost the company $600 million.
(1.1) Easy. Draw a decision tree for the problem of deciding whether to develop the new product or
not. Solve the decision tree assuming the goal is to maximize the expected payoff. Should the
company develop the new product?
(1.2) Easy. What is the expected value of perfect information (EVPI)? Explain what it means.
The company can hire an expert that will design an experiment to test the new product. Based on
previous experience with the expert, the following conditional probabilities are realistic estimates of the
expert’s evaluation accuracy.
Pr(experiment is positive | major success) = 0.8
Pr(experiment is positive | break even) = 0.6
Pr(experiment is positive | failure) = 0.3
(1.3) Easy. Use the law of total probability to compute the probability that the experiment is positive.
(1.4) Moderate. What is the probability of a major success given that the experiment is positive?
What is the probability of breaking even given that the experiment is positive? What is the
probability of a failure given that the experiment is positive? What is the probability of a major
success given that the experiment is negative? What is the probability of breaking even given
that the experiment is negative? What is the probability of a failure given that the experiment
is negative?
(1.5) Moderate. Construct a decision tree for this problem. Solve the decision tree assuming that
the goal is to maximize the expected payoff. What is the expected value of sample information
(EVSI)? What is the optimal decision strategy? The expert charges $1 million for her services.
Should you hire the expert?
(1.6) Challenging. What is the standard deviation of the payoff of the optimal decision strategy?
Solve with and without including the expert’s fee. What is the difference?
Problem 2 (pdfs and cdfs). Let X be a continuous random variable with pdf f(x) and cdf F (x).
(2.1) Easy Can f(x) ever be negative?
(2.2) Easy Can F (x) ever be negative?
(2.3) Easy Can F (x) ever be greater than 1?
(2.4) Moderate Can f(x) ever be greater than 1?
(2.5) Moderate Suppose that f(x) = 0 when x ≤ 0, f(x) = x when 0 < x < a, and f(x) = 0 when
x ≥ a. Compute a.
(2.6) Moderate Consider the pdf of question (2.5). Compute F (1).
1
2 PROBLEM SET 2: DECISION TREES, CONTINUOUS RANDOM VARIABLES
(2.7) Challenging Suppose that f(x) = 0 when x ≤ 0, f(x) = x when 0 < x ≤ 1, f(x) = 2−x when
1 < x < 2, and f(x) = 0 when x ≥ 2. Compute F (x) in each of the ranges x ≤ 0, 0 < x ≤ 1,
1 < x < 2, and x ≥ 2.
Problem 3 (Uniform Distribution). You arrive at a bus stop at 8 o’ clock, knowing that the bus
will arrive at some time uniformly distributed between 8am and 8:20am.
(3.1) Easy What is the expected value of your waiting time at the bus stop?
(3.2) Easy What is the probability that you will have to wait at least 15 minutes?
(3.3) Easy What is the median of your waiting time at the bus stop?
(3.4) Moderate If at 8:05am, the bus has not yet arrived, what is the probability that you will have
to wait at least an additional 10 minutes?
(3.5) Challenging You arrive at a bus stop at 8 o’ clock, but suppose now that there are three
different buses that can bring you to your destination. Each bus will arrive at the bus stop at
some time uniformly distributed between 8 and 8:20am and these arrivals are independent. You
take the first bus that shows up. What is the probability that you will have to wait no more
than 10 minutes?
Problem 4 (Exponential Distribution). A small private ambulance service in Western Pennsylvania
has determined that the time between emergency calls is exponentially distributed with a mean of 120
minutes. When a unit goes on call, it is busy for 50 minutes. If a unit is busy when an emergency call is
received, the call is immediately routed to another service. The company is considering buying a second
ambulance. However, before doing so, the owner would like to have answers to the following questions.
(4.1) Easy What is the probability that an emergency call is received when the unit is busy?
(4.2) Easy What is the probability that at least two hours elapse between an emergency call and the
following one?
(4.3) Easy The unit just went on call. What is the probability that the next emergency call will
arrive between 50 minutes and two hours?
(4.4) Moderate What is the distribution of the number of emergency calls received in 24 hours?
What is the expected number of emergency calls received in 24 hours?
Problem 5 (Combining Continuous and Discrete Distributions). The life of a certain type of
automobile tire, called MX tire, is uniformly distributed in the range 36, 000 to 48, 000 miles.
(5.1) Easy Compute the standard deviation of the life of an MX tire.
Consider a car equipped with 4 MX tires. The lives of the 4 tires are assumed to be independent.
(5.2) Moderate What is the probability that the car can drive at least 40, 000 miles without changing
any tire?
(5.3) Moderate What is the probability that at least one of the tires has to be changed before the
car reaches 42, 000 miles?
(5.4) Challenging What is the probability that at least 2 tires have to be changed before the car
reaches 42, 000 miles?

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 3: NORMAL RANDOM VARIABLES, SAMPLING DISTRIBUTIONS
This assignment tests your understanding of normal random variables, and of sampling distributions.
The learning objectives are: working with the probability distribution of a normal random variable, com-
bining independent normal random variables, the central limit theorem, and the sampling distribution
of the sample mean.
Problem 1 (Value at Risk). In finance, VaR does not stand for variance but for “value at risk.”
It is another measure of risk. It was introduced by J.P. Morgan in the 1980s as a way to answer a
question that was often asked by investors: “How much might I loose?” Let X be a random variable
representing the profit on an investment at the end of some time horizon, such as a year. If X is negative
the investment results in a loss. Choose a confidence level α (α = 95% is a commonly used value). The
α-VaR is the value x for which P (X ≤ x) = 1 − α. This means that the profit on the investment will
be greater than x with probability α, and it will be x or smaller with probability 1-α. In particular, if
x is negative and α = 95%, a loss of |x| or greater will occur with probability 5%.
(1.1) Easy Investment A has a profit at the end of a year that is normally distributed with mean $2
million and standard deviation $1.5 million. What is the 95%-VaR of investment A?
(1.2) Moderate The investor owning A considers the potential loss calculated in (1.1) to be excessive.
Risk might be reduced using a hedging strategy. Also available is investment B, with a profit
at the end of a year that has mean -0.3 million dollars and standard deviation $ 1 million. The
correlation between the profits on investments A and B is -0.8. What is the expected profit
of investment A+B at the end of a year? What is the standard deviation of the profit on
investment A+B at the end of a year? Assuming that the profit on investment A+B is normally
distributed, compute the 95%-VaR of investment A+B.
(1.3) Challenging The investor is considering hedged portfolios of the form A + t B for t > 0. Can
you find the value of t that minimizes the variance of this portfolio? Assuming that the profit
on investment A + t B is normally distributed, compute the 95%-VaR of investment A + t B
for the value of t found above.
Problem 2 (Difference of Independent Normal Random Variables). The nation of Somonga,
located in the South Pacific, has asked you to analyze its trade balance (the trade balance is the
difference between the total revenue from exports and the total cost of imports in a year). Somonga’s
only export is coconut oil. It exports 18,000 metric tons of coconut oil per year. The price of coconut
oil in the world market is normally distributed with mean $920 per metric ton and standard deviation
$160. Somonga’s total cost of imports in a year is also normally distributed, with mean $16,500,000 and
standard deviation $1,600,000. Total cost of imports is independent of the price of coconut oil in the
world market.
(2.1) Easy What is the mean of the trade balance?
(2.2) Easy What is the standard deviation of the trade balance?
(2.3) Moderate What is the probability that the trade balance is negative?
(2.4) Challenging What is the probability that the price of coconut oil in the world market is greater
than $1000 given that it is greater than $900 ?
1
2 PROBLEM SET 3: NORMAL RANDOM VARIABLES, SAMPLING DISTRIBUTIONS
Problem 3 (Independence versus Dependence). Electronix produces several products, one of
which is called ElecPlus. The production cost C of one unit of ElecPlus is a normally distributed
random variable with mean 32 and standard deviation 6. The sales price P for a unit of ElecPlus is a
normally distributed random variable with mean 40 and standard deviation 8. The random variables C
and P are independent of each other.
(3.1) Easy What is the distribution of the profit P − C made on one unit of ElecPlus?
(3.2) Easy What is the probability of a positive profit on one unit of ElecPlus?
(3.3) Challenging What is the distribution of the profit made on ten units of ElecPlus? Assume
independence of the random variables associated with different units. What is the probability
of a positive profit on ten units of ElecPlus?
(3.4) Challenging What is the distribution of the profit made on ten units of ElecPlus? Do not
assume independence of the random variables associated with different units. Instead assume
that all ten units have the same production cost C, which is a normally distributed random
variable with mean 32 and standard deviation 6, and all ten units have the same sales price P ,
which is a normally distributed random variable with mean 40 and standard deviation 8. As
earlier the random variables C and P are independent of each other. What is the probability
of a positive profit on ten units of ElecPlus?
Problem 4 (Sample mean). The operations manager at a cookie factory is responsible for process
adherence. Each cookie has an average weight of 15 g. Because of the irregularity in the number of
chocolate chips in each cookie, the weight of an individual cookie varies, with a standard deviation of 1
g. The weights of the cookies are independent of each other. The cookies come in standard packs of 30
and jumbo packs of 60. In both packs, the label states that the cookies have an average weight of 15 g
each. Federal guidelines require that the weight stated on the packaging be consistent with the actual
weight.
(4.1) Easy. Based on the information given above is it possible to compute the probability that a
randomly selected cookie has a weight less than 14 g?
(4.2) Moderate. What is the probability that the average cookie weight in a pack of 30 is less than
14.8 g?
(4.3) Moderate. What is the probability that the average cookie weight in a pack of 60 is less than
14.8 g? Explain how increasing pack size affects this probability.
(4.4) Moderate. What is the probability that the average cookie weight in a pack of 60 is between
14.8g and 15.2 g?
(4.5) Challenging. Find an interval symmetrically distributed around the population mean that will
include 90% of the sample means, for the packs of 30 cookies.
Problem 5 (Central Limit Theorem). Are the following statements true or false?
(5.1) Easy. Based on the Central Limit Theorem, the sample mean can be used as a good estimator
of the population mean, assuming that the sample size, n, is sufficiently large.
(5.2) Easy. When applying the Central Limit Theorem, one usually considers that a sample size
n ≥ 30 is sufficiently large.
(5.3) Easy. The Central Limit Theorem can be applied to both discrete and continuous random
variables.
(5.4) Easy. Assuming that the sample size, n, is sufficiently large, the Central Limit Theorem permits
to draw conclusions about the population based strictly on sample data, and without having
any knowledge about the distribution of the underlying population.
(5.5) Moderate. Assuming that the population is normally distributed, the sampling distribution
of the sample mean is normally distributed for samples of all sizes.

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 4
This assignment tests your understanding of Interval Estimation and Hypothesis Testing. The assign-
ment aims to walk you through the following concepts: computing confidence intervals for population
means with known and unknown variance, and for population proportions with sufficiently large sam-
ples; phrasing hypotheses tests that rely on these confidence interval computations, and understanding
the difference between using z-stats and t-stats.
Problem 1 ( Confidence Interval Estimation with Known Variance). According to USA Today,
customers are not settling for automobiles straight off the production lines. They typically get accessories
that add a significant amount over the base sticker price. For BMW 3 Series with a sticker price of
$40,250, the cost of accessories has an unknown mean µ and a known standard deviation σ = $3,450.
A recent sample of 75 purchases yielded a sample mean of $8,200 above the sticker price as the cost of
the accessories.
(1.1) Easy Calculate the 90%, 95% and 99% confidence intervals for the average cost of accessories
customers get on 3 Series.
(1.2) Easy For the 95% confidence interval, determine the margin of error in estimating the average
cost of accessories customers get on 3 Series.
(1.3) Moderate What sample size would be required to reduce the margin of error by 50% (i.e. to
half of its current size)?
(1.4) Challenging Based on the above sample of 75 purchases, a statistician calculates a confidence
interval of 8200 ± 400. What is the confidence level 1 − α associated with this interval.
Problem 2 (Confidence Interval Estimation with Unknown Variance). Executive compensation
has risen dramatically beyond the rising levels of an average worker’s wage over the years. Sarah is an
MBA student who decides to use her statistical skills to estimate the mean CEO compensation for all
large companies in the United States. She takes a random sample of eight CEO compensations in 2019.
Firm Compensation ($ Million)
eBay 34.8
Walt Disney 65.6
Wyndham 21.5
T-Mobile 66.5
Amex 24.2
AT&T 29.1
Williams-Sonoma 50.8
JPMorgan Chase 30.0
(2.1) Easy How will Sarah use the above information to provide a 95% confidence interval of the
mean CEO compensation at large companies in the United States.
(2.2) Easy What assumptions did Sarah make for deriving the interval estimate in (3.1)?
(2.3) Easy How can Sarah reduce the margin of error?
Problem 3 (Confidence Interval Estimation for Population Proportion).
The coronavirus pandemic is altering the way students attend school. The admission’s office at
Carnegie Mellon University would like to have a sense of how this will affect its graduate programs. A
sample of 500 students was surveyed. Among them 145 enrolled in an online program.
(3.1) Easy Construct a 95% confidence interval for the proportion of graduate students enrolled in
an online program.
1
2 PROBLEM SET 4
(3.2) Moderate What will the confidence level be if we want our confidence interval to have a margin
of error of 2%?
Problem 4 (Hypothesis Testing for Population Mean Using Two Approaches).
Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a
famous Elizabethan poet are known to contain an average of µ = 10.4 new words (words not used in
the poet’s other work). The number of new words in this poet’s sonnets is normally distributed with
standard deviation σ = 2.7.
Now a manuscript with 5 new sonnets has come to light. Several scholars claim that these sonnets
are the work of the famous Elizabethan poet. The new sonnets contain an average of X̄ = 12.1 words
not used in the poet’s known works. Is this sufficiently strong evidence to reject the scholars’ claim that
the manuscript is by the Elizabethan poet? We will carry out a one-tailed hypothesis test.
(4.1) Easy Clearly write out the null hypothesis and the alternative hypothesis.
(4.2) Easy Give the Z-test statistic and its p-value.
(4.3) Moderate Test the null hypothesis using both the critical value approach (also called the Z-
test) and the p-value method, and confirm that they lead to the same conclusion. Use α = 0.05.
What do you conclude about the authorship of the new poems? Would your conclusion change
if you were using a significance level α = 0.1?
Problem 5 (Hypothesis Testing for Population Proportion). Before a new drug enters the
market, it needs to get approval by the Food and Drug Administration (FDA) to guarantee that it is
safe. Only if a vaccine’s benefits are found to outweigh its potential risks does the FDA grant a license
for the vaccine, allowing it to be used by the public. The new vaccine Novatrax will get approved only if
at most 5% of the patients will have side effects after taking it. The pharmaceutical lab that developed
Novatrax claims that it meets this criterion. The FDA administered Novatrax to a test group of 1000
randomly selected patients, and found that 62 of them had side effects. Is this strong enough evidence
for the FDA to deny approval of the vaccine?
(5.1) Easy At a 5% level of significance, test whether at most 5% of the patients taking the new drug
will have side effects: clearly write out the null and alternative hypothesis (explain why a one-
sided test is appropriate here), construct the test statistic, compute its p-value, and compare
against the required level of significance.
(5.2) Easy Would your conclusion change if you repeated the analysis at the 1% significance level?
Problem 6. Hypothesis Testing for Mean with unknown variance (30-Year Fixed Mortgage
rate): Rates on 30-year fixed mortgages continue to be at historic lows. According to Freddie Mac,
the average rate for 30-year fixed loans in the last week of July 2020 was 2.99%. An economist wants
to test if there is any change in the mortgage rates since then. She searches for 30-year fixed loans on
google.com and reports the rates offered by 9 banks, which is given in Excel workbook titled ‘mortgage’
in the sheet ‘HW4data2020.xslx’. Assume that rates are normally distributed.
(6.1) Moderate. Does the average mortgage rate differ from 2.99%? Use a t-test and α = 0.10.
Clearly state your null hypothesis, the test statistic, the critical value, your decision rule and
conclusion.
(6.2) Challenging. Report the 90% confidence interval for the average mortgage rate. Use this
confidence interval to confirm your conclusion in part (6.1).
(6.3) Challenging. Finally, carry out a hypothesis test against the null that the average mortgage
rate is at most 2.99%. Clearly state your hypotheses, the test statistic, the critical value, your
decision rule and conclusion. Use a t-test and α = 0.10. Does your conclusion change if you use
α = 0.05?

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 5
This assignment tests your understanding of Simple Regression. The assignment aims to walk you
through the following concepts: doing a hypothesis test for zero correlation, carrying out a simple one-
variable regression and interpreting its results, conducting hypothesis tests for the value of the slope
and intercept coefficients in a simple regression, and computing prediction intervals for the value of the
dependent variable for a new data point.
Problem 1 (Testing for Covid-19). The data set in sheet Covid-19 in the Excel document
HW5data2020 contains information on the amount of testing for Covid-19 (column C) and the number
of deaths per million (column B) in various countries since the beginning of the pandemic. We would
like to know whether there is a relationship between these two variables.
(1.1) Easy. Compute the sample correlation between ”deaths per 1M” and ”tests per 1M”. Test the
hypothesis that there is no correlation between ”deaths per 1M” and ”tests per 1M” at the 5%
significance level.
(1.2) Moderate. Estimate the simple regression using ”tests per 1M” as the independent variable
and ”deaths per 1M” as the ”dependent” variable. Interpret the slope coefficient. What is the
R2 statistic for this regression?
deathsi = β0 + β1 testsi + �i
(1.3) Moderate. Compute a 95% confidence interval for the slope β1. Test the hypothesis that
β1 = 0 at the 5% significance level.
Problem 2 (Wages and IQ). For this problem use the data in sheet Wages in the Excel file
HW5data2020 to estimate a simple regression explaining monthly salary (wage) in terms of IQ score
(IQ).
(2.1) Easy. Using the data provided, find the sample mean and sample standard deviation for IQ.
(2.2) Moderate. IQ tests are standardized so that the average in the population is 100 with a
standard deviation 15. Is the sample mean found (2.1) fairly ”average” or somewhat ”extreme”?
To quantify this, compute the p-value of getting a sample mean as extreme or more extreme
than the one computed in (2.1). Assume a 2-sided test.
(2.3) Easy. Estimate the simple regression.
wagei = β0 + β1IQi + �i
(2.4) Easy. Calculate a 95% confidence interval for β1.
(2.5) Easy. What is your best estimate for the monthly salary of an individual with an IQ of 120?
(2.6) Challenging. Give a 95% prediction interval for the monthly salary of an individual with an
IQ of 120?
(2.7) Moderate. Does IQ explain most of the variation in monthly salary?
Problem 3 (CAPM Model in Finance). In this problem, you will investigate the static capital
asset pricing model (CAPM) of finance. This model holds that the excess return (i.e. the return in
excess of the risk-free rate) on an asset is made up of two components, a systematic component and
a diversifiable component. Let r̃m be a random variable representing the excess return on the market,
and r̃ be a random variable representing the excess return on some asset.
Then the CAPM says that:
(1) r̃ = α+ βr̃m + �
1
2 PROBLEM SET 5
In writing equation (1), I have followed the finance convention of denoting the intercept as α and
the slope as β. In this equation, � is the diversifiable component – a random variable that cannot be
predicted with information prior to date t. The CAPM presumes that the � for a stock is not correlated
with the � for any other stock and hence is not correlated with r̃m.
The variation in the return on the market, r̃m, is driven by changes in the economic environment. The
systematic component of the return to the asset is the component associated with the market return.
Coefficient β in equation (1) is the asset’s ”beta” in finance parlance. It is a measure of the systematic
risk of the asset. For a given total expected return, an asset with a high beta is considered riskier than
an asset with a low beta. Why? Suppose you buy equal amounts of shares in a large number of assets.
Since the idiosyncratic return (�’s) on these assets are uncorrelated, the variance of the idiosyncratic
component of your portfolio will go to zero as the number of assets held gets large – thanks to the law
of large numbers. In short, the risk on the � component of the asset return r̃ can be diversified away.
This implies that the variance of the non-diversifiable component of r̃ is just β2 times the variance of r̃m
(Recall that Var(α+ βr̃m) = β
2 Var(r̃m) always holds when α and β are constants and r̃m is a random
variable). As a consequence, assets with a larger beta have a larger non-diversifiable risk and should
therefore have a larger expected return.
We will compute alpha and beta for Amazon, using the data in the Excel spreadsheet Amazon in
HW5data2020.
(3.1) Challenging. Using the data provided, compute the monthly rates of return on Amazon’s
stock. Note that Amazon does not pay dividends, so the monthly rates of return can be computed
directly from the stock prices in consecutive months (column B in the spreadsheet). For example,
the monthly rate of return on 1-Aug-20 is =(B2-B3)/B3. Enter this formula in cell F2.
Complete the computations by copying this formula down column F. You should now have
119 monthly rates of return (from 1-Oct-10 to 1-Aug-20).
In the next column, compute the monthly excess rates of return on Amazon’s stock. Use the
10-year Treasury Bill given in column D as the yearly risk-free rate. Divide by 12 to get the
monthly risk-free rate. Thus Amazon’s excess rate of return on 1-Aug-20 is =F2-D2/12. Enter
this in cell G2. Complete the computations by copying this formula down column G.
Repeat the same steps for the market’s rates of return. Use the SP500 as the market; the
monthly values of this index are given in column C. In cell I2, write the formula =(C2-C3)/C3
and copy the formula down column I; in cell J2, write the formula =I2-D2/12 and copy it down
column J.
Run a linear regression where the independent variable is the market’s excess rate of return
and the dependent variable is Amazon’s excess rate of return. What is the beta of Amazon?
(3.2) Moderate. Provide a 95% confidence interval for the beta of Amazon. Yahoo.finance claims
that Amazon’s beta is 1.33. Test this hypothesis at the 5% significance level.
(3.3) Moderate. A prediction about the CAPM that will be developed in your finance class is that
the intercept α = 0 for any asset. What value of α did you find in the regression performed
in (3.1)? Test the hypothesis that α = 0 at a 5% significance level. Is the prediction α = 0
supported?
(3.4) Moderate. Interpret the coefficient of determination R2 in terms of the systematic and diver-
sifiable components of the risk.

45-750 PROBABILITY AND STATISTICS
PROBLEM SET 6: ADVANCED HYPOTHESIS TESTS AND REVIEW OF EARLIER MATERIAL
The first part of this assignment tests your understanding of hypothesis testing for the difference
between two populations. The second part of the assignment tests your understanding of statistical
inference concerning variance. The third part is a review of earlier material from the course.
Problem 1 (Difference in means). Healthy Companies versus Failed Companies: In what
ways are companies that fail different from those that continue to do business? To answer this question,
one study compared various characteristics of 68 healthy and 33 failed firms. One of the variables was
the ratio of current assets to current liabilities. Roughly speaking, this is the amount that the firm is
worth divided by what it owes. The data is given in excel file ‘companies.xlsx’.
Take Group 1 to be the firms that are healthy and Group 2 to be those that fail. The question of
interest is whether or not the mean ratio of current assets to current liabilities is different for the two
groups.
(1.1) Easy. Write down the appropriate null and alternative hypotheses.
(1.2) Challenging. Assuming equal population standard deviations for the two groups, give a 95%
confidence interval for the difference in mean current assets to current liabilities ratios for healthy
versus failed firms.
(1.3) Moderate. Compute the test statistic to test the null hypothesis in (1.1). Test your hypothesis
at the significance level α = 0.05. Assume equal population standard deviations for the two
groups.
Problem 2 (Testing for Difference in Proportion). “No Sweat” Garment Labels: A survey
of U.S. residents aged 18 or older asked a series of questions about how likely they would be to purchase
a garment under various conditions. For some conditions, it was stated that the garment had a “No
Sweat” label; for others there was no mention of such a label. On the basis of the responses, each person
was classified as a “label user” or a “label nonuser”. One purpose of the study was to describe the
demographic characteristics of label users and nonusers.
Here is a summary of the data. We let X denote the number of label users.
Population n X p̂ = X/n
Women 296 63 0.213
Men 251 27 0.108
(2.1) Easy. What is the fraction of ”label users” overall (aggregating over men and women).
(2.2) Easy. Estimate the gender difference in label use.
(2.3) Moderate. Calculate the 95% margin of error for the estimated gender difference in label usage
as well as the 95% confidence interval for gender difference in label usage.
(2.4) Challenging. We would like to test the hypothesis that men and women are equally likely to
be label users, using the p-value. State the null hypothesis. Compute the Z-statistic under the
null hypothesis. Compute the p-value. Test the null hypothesis at a significance level of 1%.
Test the null hypothesis at a significance level of 0.1%.
Problem 3 (Hypothesis testing for Variance). Rentals: We would like to determine whether the
standard deviation on monthly rents for one-bedroom apartments is smaller than $200 in Pittsburgh.
That is, we hope to reject the null hypothesis that the standard deviation is $200 or greater. The
monthly rents on a random sample of 7 one-bedroom apartments are listed below (in $’s). Assume these
monthly rents are normally distributed. 725 652 662 723 926 867 901
1
2 PROBLEM SET 6: ADVANCED HYPOTHESIS TESTS AND REVIEW OF EARLIER MATERIAL
(3.1) Easy. Obtain an estimate of the standard deviation given the data.
(3.2) Moderate. State the appropriate null and alternative hypotheses for the test and compute the
value of the test statistic.
(3.3) Moderate. Test the null hypothesis using critical value at α = 0.01.
(3.4) Moderate. Repeat the test at α = 0.10.
Problem 4 (Testing for equality of spread). Healthy vs. Failed Firms contd. Recall that
in Problem 1 part (1.2) above you ran the hypothesis test for difference in means assuming that the
population standard deviation for both the groups is identical. Let’s test that assumption.
(4.1) Easy. What statistic would you use to test that the population standard deviations for the two
types of firms is equal? Clearly write out the hypothesis and calculate the relevant statistic.
(4.2) Moderate. What are the exact degrees of freedom associated with the F-statistic in this case?
(4.3) Challenging. Test the hypothesis in part (4.1) at the 5% level. Does your conclusion justify
the assumption made in Problem 1?
Problem 5 (Decision Analysis). You are considering investing $500,000 in a real estate project.
There is an 80% chance that you will make a profit of $300,000. But there is also a 20% chance that
you will loose $100,000.
(5.1) Easy. What is your expected profit if you invest?
(5.2) Moderate. Suppose you have the option of hiring the research firm RELCO, that can accurately
predict if the project will be successful, for a fee of $10,000. Should you hire RELCO?
Suppose RELCO is not always able to predict accurately. Based on historical performance,
RELCO predicts correctly 90% of the time when a project is successful; RELCO also predicts
correctly 90% of the time that a project is unsuccessful when this is the case.
(5.3) Moderate. Compute the probability that RELCO predicts a successful project.
Compute the probability that the project will be successful, given that RELCO predicts it
will be successful.
Compute the probability that the project will be successful, given that RELCO predicts it
will be unsuccessful.
(5.4) Challenging. Would you hire RELCO for $10,000?
Problem 6 (Normal Distribution). Mike is in charge of producing a sophisticated electronic device
and the deadline is coming up. The specifications for the device are stringent. In particular, the device
requires 60 capacitors which must operate within ±0.50 pF from a specified standard of 28 pFs.
The problem is that the supplier can provide capacitors that operate according to a normal dis-
tribution, with a mean of 28 pFs and a standard deviation of 1 pF (the capacitor characteristics are
independent random variables). Therefore, to have 60 capacitors that Mike can use in the electronic
device, he will have to order more than 60 from the supplier. These items are expensive, and Mike wants
to order the smallest batch that will have at least a 95% chance of containing enough usable capacitors.
Mike needs your help in figuring out the size of the order he should place.
(6.1) Easy. What is the probability that a capacitor provided by the supplier will meet the specifi-
cation needed for the electronic device?
(6.2) Challenging. What is the size of the order that Mike should place?

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