Write a Python program to solve the 8-puzzle problem (and its natural generalizations) using the A* search algorithm.
The problem. The 8-puzzle problem is a puzzle invented and popularized by Noyes Palmer Chapman in the 1870s. It is played on a 3-by-3 grid with 8 square blocks labeled 1 through 8 and a blank square. Your goal is to rearrange the blocks so that they are in order, using as few moves as possible. You are permitted to slide blocks horizontally or vertically into the blank square. The following shows a sequence of legal moves from an initial board (left) to the goal board (right).
1 3 1 3 1 2 3 1 2 3 1 2 3
4 2 5 => 4 2 5 => 4 5 => 4 5 => 4 5 6
7 8 6 7 8 6 7 8 6 7 8 6 7 8
initial 1 left 2 up 5 left goal
Best-first search. Now, we describe a solution to the problem that illustrates a general artificial intelligence methodology known as the A* search algorithm. We define a search node of the game to be a board, the number of moves made to reach the board, and the previous search node. First, insert the initial search node (the initial board, 0 moves, and a null previous search node) into a priority queue. Then, delete from the priority queue the search node with the minimum priority, and insert onto the priority queue all neighboring search nodes (those that can be reached in one move from the dequeued search node). Repeat this procedure until the search node dequeued corresponds to a goal board. The success of this approach hinges on the choice of priority function for a search node. We consider two priority functions:
Manhattan priority function. The sum of the Manhattan distances (sum of the vertical and horizontal distance) from the blocks to their goal positions, plus the number of moves made so far to get to the search node
Input and output formats. The input and output format for a board is the board size N followed by the N-by-N initial board, using 0 to represent the blank square.
more puzzle04.txt
3
0 1 3
4 2 5
7 8 6
Solver puzzle04.txt
Minimum number of moves = 4
3
0 1 3
4 2 5
7 8 6
3
1 0 3
4 2 5
7 8 6
3
1 2 3
4 0 5
7 8 6
3
1 2 3
4 5 0
7 8 6
3
1 2 3
4 5 6
7 8 0
more puzzle-unsolvable3x3.txt
3
1 2 3
4 5 6
8 7 0
Solver puzzle3x3-unsolvable.txt
Unsolvable puzzle
Your program should work correctly for arbitrary N-by-N boards (for any 1 ≤ N < 128), even if it is too slow to solve some of them in a reasonable amount of time.
In: Computer Science
Patient is 88 year old female admitted to the hospital with a two day history of feeling SOB, lightheaded, dizzy, and chest pain. The patient lives in a single family house, and the bedroom and bathroom are located on the second floor. The patient was brought to the ED by the family. The nurse starts to perform the nursing assessment and finds that the patient is only oriented to person. The patient does not report any pain. The vital signs and the laboratory results are recorded as the following:
Temp. 38.1 pulse 98 resp. 32 Blood pressure 122/80 SpO2 with RA at Pulse Ox 89% , weight at 120 pounds (54.54 Kilograms), height 5.4
Na 148
Cl 96
Potassium 3.5
Bicarb 39
BUN 50
Creat 1.78
Glucose 124
Hgb 6.6
Hct 21.5
WBC 24.9
Platelets 145,000
Albumin 3.5
ROS:
Negative
PMH: CAD, CKD, HTN, COPD
PSH: Bilateral knee replacement
Social hx: lives alone, drinks one glass of wine once a day, denies smoking and drug abuse
Allergies:
NKDA
Medications at home: Lisinopril 40 mg once a day, ASA 81 mg once a day, Lasix 40 mg once a day, Advair one puff every 12 hours
Physical Exam:
Awake, confused to time and place, oriented to self
Mucous membranes dry, tongue pink
HEENT: AT/NC, PERRLA, neck supple, no JVD
CV: RRR, Normal S1 S2, no m/r/c heard
Lungs: Breath sounds vesicular with crackles in the posterior bases
ABD: distended with faint bowel sounds, tender to touch
Extremities: upper and lower extremities with strength at 3/5 for all four extremities
Has 3+ edema in lower legs bilaterally
Skin: small open redden area on left thigh
Diagnostic Exam:
ECG: Normal
2D echocardiogram: LV function at 20%
Answer the following questions:
Identify the abnormal lab tests.
Identify two nursing diagnosis for this patient.
Identify two nursing intervention for each nursing diagnosis with rationale.
Identify one short term goal for each nursing diagnosis.
Identify one outcome for each nursing diagnosis.
List the medications for the patient.
List the diagnosis test ordered for the patient and the results.
In: Nursing
Need Excel Format and screenshot
The attached printout of an Excel spreadsheet shows the use of six financial formulas related to the time-value-of-money concepts discussed in Chapter 5. Your task is to reproduce the spreadsheet using Excel financial formulas in the red cells, which have the names shown in blue in the adjacent cells. You can find the financial formulas in Excel by clicking on Formulas at the top of the spreadsheet, and then clicking on Financial.
You will submit your spreadsheet through D2L, and I will check your work by changing one of the input values for each formula to see if your spreadsheet calculates the correct answer.
Note that interest rates in Excel are entered in decimal form, not as a percent as with the TI calculator. For example, in Excel 9.5% is entered in a cell as 0.095. Also, the type variable for each formula defines when the cash flows occur. Setting type equal to 0 means the cash flows occur at the end of each period. Setting type equal to 1 means the cash flows occur at the beginning of each period.
|
A |
B |
C |
|
|
1 |
Present value |
||
|
2 |
Rate |
0.11 |
|
|
3 |
Nper |
8 |
|
|
4 |
PMT |
10 |
|
|
5 |
FV |
100 |
|
|
6 |
Type |
0 |
|
|
7 |
PV |
‐94.85 |
|
|
8 |
|||
|
9 |
Number of periods |
||
|
10 |
Rate |
0.11 |
|
|
11 |
PMT |
10 |
|
|
12 |
PV |
‐94.85 |
|
|
13 |
FV |
100 |
|
|
14 |
Type |
0 |
|
|
15 |
Nper |
8.01 |
|
|
16 |
|||
|
17 |
Payment |
||
|
18 |
Rate |
0.11 |
|
|
19 |
Nper |
8 |
|
|
20 |
PV |
‐94.85 |
|
|
21 |
FV |
100 |
|
|
22 |
Type |
0 |
|
|
23 |
PMT |
10.00 |
|
|
24 |
|||
|
25 |
Interest rate |
||
|
26 |
Nper |
8 |
|
|
27 |
PMT |
10 |
|
|
28 |
PV |
‐94.85 |
|
|
29 |
FV |
100 |
|
|
30 |
Type |
0 |
|
|
31 |
Rate |
0.11 |
|
|
32 |
|||
|
33 |
Future value |
||
|
34 |
Rate |
0.11 |
|
|
35 |
Nper |
8 |
|
|
36 |
PMT |
10 |
|
|
37 |
PV |
‐94.85 |
|
|
38 |
Type |
0 |
|
|
39 |
FV |
99.99 |
|
|
40 |
|||
|
41 |
Net present value |
||
|
42 |
Rate |
0.11 |
|
|
43 |
Value 1 |
100 |
|
|
44 |
Value 2 |
200 |
|
|
45 |
Value 3 |
300 |
|
|
46 |
Value 4 |
400 |
|
|
47 |
Value 5 |
500 |
|
|
48 |
NPV |
1031.99 |
In: Finance
• A MAJOR POTENTIOMETER MANUFACTURER IS CONSIDERING TWO ALTERNATIVES FOR NEW PRODUCTION MACHINES WITH CAPACITY TO PRODUCE 20 000 UNITS PER DAY. ONE ALTERNATIVE IS FOR A HIGH-CAPACITY AUTOMATED PRODUCTION MACHINE CAPABLE OF PRODUCING 20 000 UNITS PER DAY WHEN OPERATED FOR THREE SHIFTS PER DAY. A QUARTER-TIME EMPLOYEE WOULD BE ASSIGNED TO MONITOR THE MACHINE (EMPLOYEE WOULD MONITOR OTHER MACHINES AT THE SAME TIME). WITH THE THREE-SHIFT SCHEDULE THIS WOULD BE EQUIVALENT TO A THREE-QUARTER-TIME EMPLOYEE.
• A SECOND ALTERNATIVE WOULD BE TO USE TWO MANUALLY OPERATED MACHINES, EACH CAPABLE OF 10 000 UNITS PER DAY ASSUMING THREE-SHIFT OPERATION. HERE, A TOTAL OF SIX EMPLOYEES (2 PER SHIFT, 3 SHIFTS) WOULD BE NEEDED.
• IF LABOR COSTS (INCLUDING WAGES, BENEFITS, ETC.) ARE $40 000 PER EMPLOYEE PER YEAR, RECOMMEND WHICH ALTERNATIVE IS BEST USING THE EQUIVALENT UNIFORM ANNUAL COST METHOD
| Alternative 1 | Alternative 2 | |
| Cost to purchase | $500 000 | $100 000 |
| Number of machines required | 1 | 2 |
| Number of employees required | 2 | 6 |
| Expected life of machine | 10 yr | 10 yr |
| Interest rate | 8% | 8% |
| Annual maintenance cost per machine | $30 000 | $10 000 |
| Salvage value at 10 years per machine | $100 000 | $20 000 |
In: Accounting
A travel agent wants to estimate the proportion of vacationers who plan to travel outside the United States in the next 12 months. A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. Construct a 95% confidence interval estimate of the population proportion. Make a statement about this in context of the problem
In: Statistics and Probability
The records of ABC Company showed the following:
| Units | Unit Cost | Total Cost | ||
| January 1 | Beginning | 10,000 | 60 | 600,000 |
| April 1 | Purchase | 18,000 | 50 | 900,000 |
| October 1 | Purchase | 22,000 | 40 | 880,000 |
The physical inventory reveals 15,000 units on hand on December
31.
Compute the cost of ending inventory and cost of sales using:
| Inventory Cost Flow | Ending Inventory | Cost of Goods Sold (COGS) |
| First in, first out (FIFO) | ||
| Weighted Average | ||
| Last in, first out (LIFO) |
In: Accounting
1.-The following data were obtained for a randomized block design involving five treatments and three blocks: SST = 570, SSTR = 390, SSBL = 95. Set up the ANOVA table. (Round your value for F to two decimal places, and your p-value to three decimal places.)
| Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F | p-value |
|---|---|---|---|---|---|
| Treatments | 390 | 4 | 97.5 | ||
| Blocks | 95 | 2 | 47.5 | ||
| Error | 85 | 8 | 10.625 | ||
| Total | 570 | 14 |
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ___
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow.
| Analyzer | |||
|---|---|---|---|
| Computerized | Electronic | ||
| Car | Compact | 51 | 43 |
| Intermediate | 55 | 44 | |
| Full-sized | 62 | 45 | |
Use α = 0.05 to test for any significant differences.
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ___
In: Statistics and Probability
2. You deposit $1000. How much will you have under each of the following conditions?
a) 8 percent compounded semi-annually for two years
b) 8 percent compounded quarterly for two years
c) 8 percent compounded monthly for two years
In: Finance
Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean μ = 50 and estimated standard deviation σ = 12. A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed.
(a) What is the probability that, on a single test, x
< 40? (Round your answer to four decimal places.)
(b) Suppose a doctor uses the average x for two tests
taken about a week apart. What can we say about the probability
distribution of x? Hint: See Theorem 6.1.
The probability distribution of x is not normal. The probability distribution of x is approximately normal with μx = 50 and σx = 8.49. The probability distribution of x is approximately normal with μx = 50 and σx = 12. The probability distribution of x is approximately normal with μx = 50 and σx = 6.00.
What is the probability that x < 40? (Round your answer
to four decimal places.)
(c) Repeat part (b) for n = 3 tests taken a week apart.
(Round your answer to four decimal places.)
(d) Repeat part (b) for n = 5 tests taken a week apart.
(Round your answer to four decimal places.)
(e) Compare your answers to parts (a), (b), (c), and (d). Did the
probabilities decrease as n increased?
Yes No
Explain what this might imply if you were a doctor or a nurse.
The more tests a patient completes, the stronger is the evidence for excess insulin. The more tests a patient completes, the weaker is the evidence for excess insulin. The more tests a patient completes, the stronger is the evidence for lack of insulin. The more tests a patient completes, the weaker is the evidence for lack of insulin.
In: Statistics and Probability
| Date | Close |
| 7/3/2017 | 898.700012 |
| 7/5/2017 | 911.710022 |
| 7/6/2017 | 906.690002 |
| 7/7/2017 | 918.590027 |
| 7/10/2017 | 928.799988 |
| 7/11/2017 | 930.090027 |
| 7/12/2017 | 943.830017 |
| 7/13/2017 | 947.159973 |
| 7/14/2017 | 955.98999 |
| 7/17/2017 | 953.419983 |
| 7/18/2017 | 965.400024 |
| 7/19/2017 | 970.890015 |
| 7/20/2017 | 968.150024 |
| 7/21/2017 | 972.919983 |
| 7/24/2017 | 980.340027 |
| 7/25/2017 | 950.700012 |
| 7/26/2017 | 947.799988 |
| 7/27/2017 | 934.090027 |
| 7/28/2017 | 941.530029 |
| 7/31/2017 | 930.5 |
| 8/1/2017 | 930.830017 |
| 8/2/2017 | 930.390015 |
| 8/3/2017 | 923.650024 |
| 8/4/2017 | 927.960022 |
| 8/7/2017 | 929.359985 |
| 8/8/2017 | 926.789978 |
| 8/9/2017 | 922.900024 |
| 8/10/2017 | 907.23999 |
| 8/11/2017 | 914.390015 |
| 8/14/2017 | 922.669983 |
| 8/15/2017 | 922.219971 |
| 8/16/2017 | 926.960022 |
| 8/17/2017 | 910.97998 |
| 8/18/2017 | 910.669983 |
| 8/21/2017 | 906.659973 |
| 8/22/2017 | 924.690002 |
| 8/23/2017 | 927 |
| 8/24/2017 | 921.280029 |
| 8/25/2017 | 915.890015 |
| 8/28/2017 | 913.809998 |
| 8/29/2017 | 921.289978 |
| 8/30/2017 | 929.570007 |
| 8/31/2017 | 939.330017 |
| 9/1/2017 | 937.340027 |
| 9/5/2017 | 928.450012 |
| 9/6/2017 | 927.809998 |
| 9/7/2017 | 935.950012 |
| 9/8/2017 | 926.5 |
| 9/11/2017 | 929.080017 |
| 9/12/2017 | 932.070007 |
| 9/13/2017 | 935.090027 |
| 9/14/2017 | 925.109985 |
| 9/15/2017 | 920.289978 |
| 9/18/2017 | 915 |
| 9/19/2017 | 921.809998 |
| 9/20/2017 | 931.580017 |
| 9/21/2017 | 932.450012 |
| 9/22/2017 | 928.530029 |
| 9/25/2017 | 920.969971 |
| 9/26/2017 | 924.859985 |
| 9/27/2017 | 944.48999 |
| 9/28/2017 | 949.5 |
| 9/29/2017 | 959.109985 |
| 10/2/2017 | 953.27002 |
| 10/3/2017 | 957.789978 |
| 10/4/2017 | 951.679993 |
| 10/5/2017 | 969.960022 |
| 10/6/2017 | 978.890015 |
| 10/9/2017 | 977 |
| 10/10/2017 | 972.599976 |
| 10/11/2017 | 989.25 |
| 10/12/2017 | 987.830017 |
| 10/13/2017 | 989.679993 |
| 10/16/2017 | 992 |
| 10/17/2017 | 992.179993 |
| 10/18/2017 | 992.809998 |
| 10/19/2017 | 984.450012 |
| 10/20/2017 | 988.200012 |
| 10/23/2017 | 968.450012 |
| 10/24/2017 | 970.539978 |
| 10/25/2017 | 973.330017 |
| 10/26/2017 | 972.559998 |
| 10/27/2017 | 1019.27002 |
| 10/30/2017 | 1017.109985 |
| 10/31/2017 | 1016.640015 |
| 11/1/2017 | 1025.5 |
| 11/2/2017 | 1025.579956 |
| 11/3/2017 | 1032.47998 |
| 11/6/2017 | 1025.900024 |
| 11/7/2017 | 1033.329956 |
| 11/8/2017 | 1039.849976 |
| 11/9/2017 | 1031.26001 |
| 11/10/2017 | 1028.069946 |
| 11/13/2017 | 1025.75 |
| 11/14/2017 | 1026 |
| 11/15/2017 | 1020.909973 |
| 11/16/2017 | 1032.5 |
| 11/17/2017 | 1019.090027 |
| 11/20/2017 | 1018.380005 |
| 11/21/2017 | 1034.48999 |
| 11/22/2017 | 1035.959961 |
| 11/24/2017 | 1040.609985 |
| 11/27/2017 | 1054.209961 |
| 11/28/2017 | 1047.410034 |
| 11/29/2017 | 1021.659973 |
| 11/30/2017 | 1021.409973 |
| 12/1/2017 | 1010.169983 |
| 12/4/2017 | 998.679993 |
| 12/5/2017 | 1005.150024 |
| 12/6/2017 | 1018.380005 |
| 12/7/2017 | 1030.930054 |
| 12/8/2017 | 1037.050049 |
| 12/11/2017 | 1041.099976 |
| 12/12/2017 | 1040.47998 |
| 12/13/2017 | 1040.609985 |
| 12/14/2017 | 1049.150024 |
| 12/15/2017 | 1064.189941 |
| 12/18/2017 | 1077.140015 |
| 12/19/2017 | 1070.680054 |
| 12/20/2017 | 1064.949951 |
| 12/21/2017 | 1063.630005 |
| 12/22/2017 | 1060.119995 |
| 12/26/2017 | 1056.73999 |
| 12/27/2017 | 1049.369995 |
| 12/28/2017 | 1048.140015 |
| 12/29/2017 | 1046.400024 |
| 1/2/2018 | 1065 |
| 1/3/2018 | 1082.47998 |
| 1/4/2018 | 1086.400024 |
| 1/5/2018 | 1102.22998 |
| 1/8/2018 | 1106.939941 |
| 1/9/2018 | 1106.26001 |
| 1/10/2018 | 1102.609985 |
| 1/11/2018 | 1105.52002 |
| 1/12/2018 | 1122.26001 |
| 1/16/2018 | 1121.76001 |
| 1/17/2018 | 1131.97998 |
| 1/18/2018 | 1129.790039 |
| 1/19/2018 | 1137.51001 |
| 1/22/2018 | 1155.810059 |
| 1/23/2018 | 1169.969971 |
| 1/24/2018 | 1164.23999 |
| 1/25/2018 | 1170.369995 |
| 1/26/2018 | 1175.839966 |
| 1/29/2018 | 1175.579956 |
| 1/30/2018 | 1163.689941 |
| 1/31/2018 | 1169.939941 |
| 2/1/2018 | 1167.699951 |
| 2/2/2018 | 1111.900024 |
| 2/5/2018 | 1055.800049 |
| 2/6/2018 | 1080.599976 |
| 2/7/2018 | 1048.579956 |
| 2/8/2018 | 1001.52002 |
| 2/9/2018 | 1037.780029 |
| 2/12/2018 | 1051.939941 |
| 2/13/2018 | 1052.099976 |
| 2/14/2018 | 1069.699951 |
| 2/15/2018 | 1089.52002 |
| 2/16/2018 | 1094.800049 |
| 2/20/2018 | 1102.459961 |
| 2/21/2018 | 1111.339966 |
| 2/22/2018 | 1106.630005 |
| 2/23/2018 | 1126.790039 |
| 2/26/2018 | 1143.75 |
| 2/27/2018 | 1118.290039 |
| 2/28/2018 | 1104.72998 |
| 3/1/2018 | 1069.52002 |
| 3/2/2018 | 1078.920044 |
| 3/5/2018 | 1090.930054 |
| 3/6/2018 | 1095.060059 |
| 3/7/2018 | 1109.640015 |
| 3/8/2018 | 1126 |
| 3/9/2018 | 1160.040039 |
| 3/12/2018 | 1164.5 |
| 3/13/2018 | 1138.170044 |
| 3/14/2018 | 1149.48999 |
| 3/15/2018 | 1149.579956 |
| 3/16/2018 | 1135.72998 |
| 3/19/2018 | 1099.819946 |
| 3/20/2018 | 1097.709961 |
| 3/21/2018 | 1090.880005 |
| 3/22/2018 | 1049.079956 |
| 3/23/2018 | 1021.570007 |
| 3/26/2018 | 1053.209961 |
| 3/27/2018 | 1005.099976 |
| 3/28/2018 | 1004.559998 |
| 3/29/2018 | 1031.790039 |
| 4/2/2018 | 1006.469971 |
| 4/3/2018 | 1013.409973 |
| 4/4/2018 | 1025.140015 |
| 4/5/2018 | 1027.810059 |
| 4/6/2018 | 1007.039978 |
| 4/9/2018 | 1015.450012 |
| 4/10/2018 | 1031.640015 |
| 4/11/2018 | 1019.969971 |
| 4/12/2018 | 1032.51001 |
| 4/13/2018 | 1029.27002 |
| 4/16/2018 | 1037.97998 |
| 4/17/2018 | 1074.160034 |
| 4/18/2018 | 1072.079956 |
| 4/19/2018 | 1087.699951 |
| 4/20/2018 | 1072.959961 |
| 4/23/2018 | 1067.449951 |
| 4/24/2018 | 1019.97998 |
| 4/25/2018 | 1021.179993 |
| 4/26/2018 | 1040.040039 |
| 4/27/2018 | 1030.050049 |
| 4/30/2018 | 1017.330017 |
| 5/1/2018 | 1037.310059 |
| 5/2/2018 | 1024.380005 |
| 5/3/2018 | 1023.719971 |
| 5/4/2018 | 1048.209961 |
| 5/7/2018 | 1054.790039 |
| 5/8/2018 | 1053.910034 |
| 5/9/2018 | 1082.76001 |
| 5/10/2018 | 1097.569946 |
| 5/11/2018 | 1098.26001 |
| 5/14/2018 | 1100.199951 |
| 5/15/2018 | 1079.22998 |
| 5/16/2018 | 1081.77002 |
| 5/17/2018 | 1078.589966 |
| 5/18/2018 | 1066.359985 |
| 5/21/2018 | 1079.579956 |
| 5/22/2018 | 1069.72998 |
| 5/23/2018 | 1079.689941 |
| 5/24/2018 | 1079.23999 |
| 5/25/2018 | 1075.660034 |
| 5/29/2018 | 1060.319946 |
| 5/30/2018 | 1067.800049 |
| 5/31/2018 | 1084.98999 |
| 6/1/2018 | 1119.5 |
| 6/4/2018 | 1139.290039 |
| 6/5/2018 | 1139.660034 |
| 6/6/2018 | 1136.880005 |
| 6/7/2018 | 1123.859985 |
| 6/8/2018 | 1120.869995 |
| 6/11/2018 | 1129.98999 |
| 6/12/2018 | 1139.319946 |
| 6/13/2018 | 1134.790039 |
| 6/14/2018 | 1152.119995 |
| 6/15/2018 | 1152.26001 |
| 6/18/2018 | 1173.459961 |
| 6/19/2018 | 1168.060059 |
| 6/20/2018 | 1169.839966 |
| 6/21/2018 | 1157.660034 |
| 6/22/2018 | 1155.47998 |
| 6/25/2018 | 1124.810059 |
| 6/26/2018 | 1118.459961 |
| 6/27/2018 | 1103.97998 |
| 6/28/2018 | 1114.219971 |
| 6/29/2018 | 1115.650024 |
| 7/2/2018 | 1127.459961 |
Project 3 instructions
Based on Larson & Farber: sections 5.2–5.3
Using the provided data. 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 5.2–5.3 of 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.
Show all work
1) 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 $825? (5 points)
3a) 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? (5 points)
3b) 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 $700 per share. (5 points)
3c) 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)
4) 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)
5) 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)
There are also 5 points for miscellaneous items like correct date range, correct mean, correct SD, etc.
In: Statistics and Probability