4. Consider the random variable Z from problem 1, and the random variable X from problem 2.
Also let f(X,Z)represent the joint probability distribution of X and Z. f is defined as follows:
f(1,-2) = 1/6
f(2,-2) = 2/15
f(3,-2) = 0
f(4,-2) = 0
f(5,-2) = 0
f(6,-2) = 0
f(1,3) = 0
f(2,3) = 1/30
f(3,3) = 1/6
f(4,3) = 0
f(5,3) = 0
f(6,3) = 0
f(1,5) = 0
f(2,5) = 0
f(3,5) = 0
f(4,5) = 1/6
f(5,5) = 1/6
f(6,5) = 1/6
Compute the covariance of X and Z.
Then, compute the correlation coefficient of X and Z. (Note: You will need values that you computed in problems 1 and 2.)
These are questions 1 and 2.
1. Let Z be a random variable with the following probability distribution f:
f(-2) = 0.3
f(3) = 0.2
f(5) = 0.5
Compute the E(Z), Var(Z) and the standard deviation of Z.
2. Tossing a fair die is an experiment that can result in any integer number from 1 to 6 with equal probabilities. Let X be the number of dots on the top face of a die. Compute E(X) and Var(X).
In: Statistics and Probability
|
Use the given data set to answer parts (a) and (b). a. Find the regression equation for the data points.b. Graph the regression equation and the data points. |
|||||||
|
x |
5 |
4 |
33 |
1 |
2 |
||
|---|---|---|---|---|---|---|---|
|
y |
0 |
2 |
0 |
4 |
6 |
a.
Find the regression equation for the data points.
In: Statistics and Probability
Journalize transactions from account data and prepare a trial balance.
(LO 2, 4)
The T-accounts below summarize the ledger of Daggett Landscaping Company at the end of the first month of operations.
| Cash | No. 101 | Unearned Service Revenue | No. 209 | ||||
|---|---|---|---|---|---|---|---|
| 4/1 | 12,000 | 4/15 | 1,300 | 4/30 | 1,000 | ||
| 4/12 | 900 | 4/25 | 1,500 | ||||
| 4/29 | 400 | ||||||
| 4/30 | 1,000 | ||||||
| Accounts Receivable | No. 112 | Owner's Capital | No. 301 | ||||
| 4/7 | 3,200 | 4/29 | 400 | 4/1 | 12,000 | ||
| Supplies | No. 126 | Service Revenue | No. 400 | ||||
| 4/4 | 1,800 | 4/7 | 3,200 | ||||
| 4/12 | 900 | ||||||
| Accounts Payable | No. 201 | Salaries and Wages Expense | No. 726 | ||||
| 4/25 | 1,500 | 4/4 | 1,800 | 4/15 | 1,300 | ||
In: Accounting
I don't know why I keep getting the following error:
AttributeError: 'tuple' object has no attribute 'size'
I am using python in Anaconda.
import numpy as np
def information_gain(x_array, y_array):
parent_entropy = entropy(x_array)
split_dict = split(y_array)
for val in split_dict.values():
freq = val.size / x_array.size
child_entropy = entropy([x_array[i] for i in val])
parent_entropy -= child_entropy* freq
return parent_entropy
x = np.array([0, 1, 0, 1, 0, 1])
y = np.array([0, 1, 0, 1, 1, 1])
print(round(information_gain(x, y), 4))
x = np.array([0, 0, 1, 1, 2, 2])
y = np.array([0, 1, 0, 1, 1, 1])
print(round(information_gain(x, y), 4))
In: Computer Science
From Recording Transactions (Including Adjusting Journal Entries) to Preparing Financial Statements and Closing Journal Entries (Chapters 2, 3, and 4) [LO 2-3, LO 3-3, LO 4-1, LO 4-2, LO 4-3, LO 4-4, LO 4-5, LO 4-6]
|
[The following information applies to the questions displayed below.] |
| Brothers Harry and Herman Hausyerday began operations of their machine shop (H & H Tool, Inc.) on January 1, 2013. The annual reporting period ends December 31. The trial balance on January 1, 2015, follows (the amounts are rounded to thousands of dollars to simplify): |
| Account Titles | Debit | Credit | ||||
| Cash | $ | 2 | ||||
| Accounts Receivable | 6 | |||||
| Supplies | 13 | |||||
| Land | 0 | |||||
| Equipment | 67 | |||||
| Accumulated Depreciation | $ | 5 | ||||
| Software | 21 | |||||
| Accumulated Amortization | 7 | |||||
| Accounts Payable | 4 | |||||
| Notes Payable (short-term) | 0 | |||||
| Salaries and Wages Payable | 0 | |||||
| Interest Payable | 0 | |||||
| Income Tax Payable | 0 | |||||
| Common Stock | 84 | |||||
| Retained Earnings | 9 | |||||
| Service Revenue | 0 | |||||
| Salaries and Wages Expense | 0 | |||||
| Depreciation Expense | 0 | |||||
| Amortization Expense | 0 | |||||
| Income Tax Expense | 0 | |||||
| Interest Expense | 0 | |||||
| Supplies Expense | 0 | |||||
| Totals | $ | 109 | $ | 109 | ||
| Transactions during 2015 (summarized in thousands of dollars) follow: | |
| 1. | Borrowed $11 cash on a six-month note payable dated March 1, 2015. |
| 2. | Purchased land for future building site; paid cash, $8. |
| 3. | Earned revenues for 2015, $174, including $47 on credit and $127 collected in cash. |
| 4. | Issued additional shares of stock for $4. |
| 5. | Recognized salaries and wages expense for 2015, $92 paid in cash. |
| 6. | Collected accounts receivable, $31. |
| 7. | Purchased software, $11 cash. |
| 8. | Paid accounts payable, $12. |
| 9. | Purchased supplies on account for future use, $19. |
| 10. | Signed a $20 service contract to start February 1, 2016. |
| Data for adjusting journal entries: | |
| 11. | Unrecorded amortization for the year on software, $7. |
| 12. | Supplies counted on December 31, 2015, $12. |
| 13. | Depreciation for the year on the equipment, $5. |
| 14. | Accrued interest of $1 on notes payable. |
| 15. | Salaries and wages earned but not yet paid or recorded, $13. |
| 16. | Income tax for the year was $7. It will be paid in 2016. |
INCLUDE journal entires, adjusting journal entires, T-accounts,unadjusted trial balance, adjusted trial balance,income statement, statement of retained earnings, closing journal entry, post-closing trial balance
In: Accounting
Use SPSS® to check your mock data for the following:
| sex | film | arousal |
| 1 | 1 | 47 |
| 1 | 1 | 38 |
| 1 | 1 | 41 |
| 1 | 1 | 35 |
| 1 | 1 | 43 |
| 1 | 1 | 49 |
| 1 | 1 | 38 |
| 1 | 1 | 39 |
| 1 | 1 | 44 |
| 1 | 1 | 48 |
| 1 | 2 | 25 |
| 1 | 2 | 8 |
| 1 | 2 | 4 |
| 1 | 2 | 16 |
| 1 | 2 | 15 |
| 1 | 2 | 6 |
| 1 | 2 | 20 |
| 1 | 2 | 12 |
| 1 | 2 | 9 |
| 1 | 2 | 23 |
| 2 | 1 | 23 |
| 2 | 1 | 35 |
| 2 | 1 | 25 |
| 2 | 1 | 36 |
| 2 | 1 | 33 |
| 2 | 1 | 40 |
| 2 | 1 | 31 |
| 2 | 1 | 39 |
| 2 | 1 | 35 |
| 2 | 1 | 27 |
| 2 | 2 | 18 |
| 2 | 2 | 13 |
| 2 | 2 | 19 |
| 2 | 2 | 24 |
| 2 | 2 | 11 |
| 2 | 2 | 2 |
| 2 | 2 | 9 |
| 2 | 2 | 19 |
| 2 | 2 | 10 |
| 2 | 2 | 2 |
In: Statistics and Probability
Gatson manufacturing company produces 2 types of tires: Economy tire; Premium tire. The manufacturing time and the profit contribution per tire are given in the following table.
|
Operation |
Manufacturing Time (Hours) |
Time Available |
|
|
Economy tires |
Premium tires |
Hours |
|
| Material Preparation |
4/3 |
1/2 |
600 |
| Tire Building |
4/5 |
1 |
650 |
| Curing |
1/2 |
2/4 |
580 |
| Final Inspection |
1/5 |
1/3 |
120 |
| Profit/Tire |
$12 |
$10 |
|
Answer the following assuming that the company is interested in maximizing the total profit contribution.
a. Develop a spreadsheet model and find the optimal solution using Excel Solver. What is the total profit contribution Gatson can earn with the optimal production quantities? Enter your answer without a dollar sign and rounded to two decimal places.
b. Based on your answer to Question 1, how many Economy tires should Gatson manufacture to maximize profit contribution? Round your answer to one decimal place.
c. Based on your answer to Question 1, how many Premium tires should Gatson manufacture to maximize profit contribution? Round your answer to one decimal place.
In: Operations Management
I NEED AN NLOGN OR LINEAR SOLUTION TO THIS PROBLEM!
I NEED A FASTER SOLUTION THAN THE ONE GIVEN BELOW!! The solution below runs in quadratic time, I need one faster than this.
I REPEAT I NEED A FASTER SOLUTION!! THE SOLUTION GIVEN BELOW IS TOO SLOW!
Slow Solution:
def predictAnswer(stockData, queries):
stockData = [0] + stockData
length = len(stockData)
ans = []
for q in queries:
l = q-1
r = q+1
flag = True
while l > 0 or r < length:
if l > 0 and stockData[l] < stockData[q]:
ans.append(l)
flag = False
break
if r < length and stockData[r] < stockData[q]:
ans.append(r)
flag = False
break
l -= 1
r += 1
if flag:
ans.append(-1)
return ans
Question:
The function predictAnswer should be made based on the following.
In the prediction game, the first player gives the second player some stock market data for some consecutive days. The data contains a company's stock price on each day. The rules for the game are:
Example 1;
stockData size n =10;
stockData = [5,6,8,4,9,10,8,3,6,4]
queries = [6,5,4]
Result is [5,4,8]
On day 6, the stock price is 10. Both 9 and 8 are lower prices one day away. Choose 9 (day 5) because it is before day 6. On day 5, the stock price is 9. 4 is the closest lower price on day 4. On day 4, the stock price is 4. The only lower price is on day 8. The return array is [5,4,8]
Example - 2
stockData size n = 10
stockData = [5,6,8,4,9,10,8,3,6,4]
queries = [3,1,8]
Result is [2,4,-1]
If the day number is 3.both days 2 and 4 are smaller.choose the earlier day,day 2.
If the day number is 1,day 4 is the closet day with a smaller price.
If the day number is 8,there is no day where the price is less than 3.
The return array is [2,4,-1]
/*
* Complete the 'predictAnswer' function below.
*
* The function is expected to return an INTEGER_ARRAY.
* The function accepts following parameters:
* 1. INTEGER_ARRAY stockData
* 2. INTEGER_ARRAY queries
*/
def predictAnswer(stockData, queries):
In: Computer Science
Year Name MinPressure_before Gender_MF Category alldeaths 1950 Easy 958 1 3 2 1950 King 955 0 3 4 1952 Able 985 0 1 3 1953 Barbara 987 1 1 1 1953 Florence 985 1 1 0 1954 Carol 960 1 3 60 1954 Edna 954 1 3 20 1954 Hazel 938 1 4 20 1955 Connie 962 1 3 0 1955 Diane 987 1 1 200 1955 Ione 960 0 3 7 1956 Flossy 975 1 2 15 1958 Helene 946 1 3 1 1959 Debra 984 1 1 0 1959 Gracie 950 1 3 22 1960 Donna 930 1 4 50 1960 Ethel 981 1 1 0 1961 Carla 931 1 4 46 1963 Cindy 996 1 1 3 1964 Cleo 968 1 2 3 1964 Dora 966 1 2 5 1964 Hilda 950 1 3 37 1964 Isbell 974 1 2 3 1965 Betsy 948 1 3 75 1966 Alma 982 1 2 6 1966 Inez 983 1 1 3 1967 Beulah 950 1 3 15 1968 Gladys 977 1 2 3 1969 Camille 909 1 5 256 1970 Celia 945 1 3 22 1971 Edith 978 1 2 0 1971 Fern 979 1 1 2 1971 Ginger 995 1 1 0 1972 Agnes 980 1 1 117 1974 Carmen 952 1 3 1 1975 Eloise 955 1 3 21 1976 Belle 980 1 1 5 1977 Babe 995 1 1 0 1979 Bob 986 0 1 1 1979 David 970 0 2 15 1979 Frederic 946 0 3 5 1980 Allen 945 0 3 2 1983 Alicia 962 1 3 21 1984 Diana 949 1 2 3 1985 Bob 1002 0 1 0 1985 Danny 987 0 1 1 1985 Elena 959 1 3 4 1985 Gloria 942 1 3 8 1985 Juan 971 0 1 12 1985 Kate 967 1 2 5 1986 Bonnie 990 1 1 3 1986 Charley 990 0 1 5 1987 Floyd 993 0 1 0 1988 Florence 984 1 1 1 1989 Chantal 986 1 1 13 1989 Hugo 934 0 4 21 1989 Jerry 983 0 1 3 1991 Bob 962 0 2 15 1992 Andrew 922 0 5 62 1993 Emily 960 1 3 3 1995 Erin 973 1 2 6 1995 Opal 942 1 3 9 1996 Bertha 974 1 2 8 1996 Fran 954 1 3 26 1997 Danny 984 0 1 10 1998 Bonnie 964 1 2 3 1998 Earl 987 0 1 3 1998 Georges 964 0 2 1 1999 Bret 951 0 3 0 1999 Floyd 956 0 2 56 1999 Irene 987 1 1 8 2002 Lili 963 1 1 2 2003 Claudette 979 1 1 3 2003 Isabel 957 1 2 51 2004 Alex 972 0 1 1 2004 Charley 941 0 4 10 2004 Frances 960 1 2 7 2004 Gaston 985 0 1 8 2004 Ivan 946 0 3 25 2004 Jeanne 950 1 3 5 2005 Cindy 991 1 1 1 2005 Dennis 946 0 3 15 2005 Ophelia 982 1 1 1 2005 Rita 937 1 3 62 2005 Wilma 950 1 3 5 2005 Katrina 902 1 3 1833 2007 Humberto 985 0 1 1 2008 Dolly 963 1 1 1 2008 Gustav 951 0 2 52 2008 Ike 935 0 2 84 2011 Irene 952 1 1 41 2012 Isaac 965 0 1 5 2012 Sandy 945 1 2 159 Test if there is a significant difference in the death by Hurricanes and Min Pressure measured. Answer the questions for Assessment. (Pick the closest answer)
7. What is the P-value?
8. What is the Statistical interpretation?
9. What is the conclusion?
In: Statistics and Probability
Open Hurricanes data.
Test if there is a significant difference in the death by Hurricanes and Min Pressure measured. Answer the questions for Assessment. (Pick the closest answer)
7. What is the P-value?
8. What is the Statistical interpretation?
9. What is the conclusion?
Year Name MinPressure_before
Gender_MF Category alldeaths
1950 Easy 958 1
3 2
1950 King 955 0
3 4
1952 Able 985 0
1 3
1953 Barbara 987 1
1 1
1953 Florence 985 1
1 0
1954 Carol 960 1
3 60
1954 Edna 954 1
3 20
1954 Hazel 938 1
4 20
1955 Connie 962 1
3 0
1955 Diane 987 1
1 200
1955 Ione 960 0
3 7
1956 Flossy 975 1
2 15
1958 Helene 946 1
3 1
1959 Debra 984 1
1 0
1959 Gracie 950 1
3 22
1960 Donna 930 1
4 50
1960 Ethel 981 1
1 0
1961 Carla 931 1
4 46
1963 Cindy 996 1
1 3
1964 Cleo 968 1
2 3
1964 Dora 966 1
2 5
1964 Hilda 950 1
3 37
1964 Isbell 974 1
2 3
1965 Betsy 948 1
3 75
1966 Alma 982 1
2 6
1966 Inez 983 1
1 3
1967 Beulah 950 1
3 15
1968 Gladys 977 1
2 3
1969 Camille 909 1
5 256
1970 Celia 945 1
3 22
1971 Edith 978 1
2 0
1971 Fern 979 1
1 2
1971 Ginger 995 1
1 0
1972 Agnes 980 1
1 117
1974 Carmen 952 1
3 1
1975 Eloise 955 1
3 21
1976 Belle 980 1
1 5
1977 Babe 995 1
1 0
1979 Bob 986 0
1 1
1979 David 970 0
2 15
1979 Frederic 946 0
3 5
1980 Allen 945 0
3 2
1983 Alicia 962 1
3 21
1984 Diana 949 1
2 3
1985 Bob 1002 0
1 0
1985 Danny 987 0
1 1
1985 Elena 959 1
3 4
1985 Gloria 942 1
3 8
1985 Juan 971 0
1 12
1985 Kate 967 1
2 5
1986 Bonnie 990 1
1 3
1986 Charley 990 0
1 5
1987 Floyd 993 0
1 0
1988 Florence 984 1
1 1
1989 Chantal 986 1
1 13
1989 Hugo 934 0
4 21
1989 Jerry 983 0
1 3
1991 Bob 962 0
2 15
1992 Andrew 922 0
5 62
1993 Emily 960 1
3 3
1995 Erin 973 1
2 6
1995 Opal 942 1
3 9
1996 Bertha 974 1
2 8
1996 Fran 954 1
3 26
1997 Danny 984 0
1 10
1998 Bonnie 964 1
2 3
1998 Earl 987 0
1 3
1998 Georges 964 0
2 1
1999 Bret 951 0
3 0
1999 Floyd 956 0
2 56
1999 Irene 987 1
1 8
2002 Lili 963 1
1 2
2003 Claudette 979
1 1 3
2003 Isabel 957 1
2 51
2004 Alex 972 0
1 1
2004 Charley 941 0
4 10
2004 Frances 960 1
2 7
2004 Gaston 985 0
1 8
2004 Ivan 946 0
3 25
2004 Jeanne 950 1
3 5
2005 Cindy 991 1
1 1
2005 Dennis 946 0
3 15
2005 Ophelia 982 1
1 1
2005 Rita 937 1
3 62
2005 Wilma 950 1
3 5
2005 Katrina 902 1
3 1833
2007 Humberto 985 0
1 1
2008 Dolly 963 1
1 1
2008 Gustav 951 0
2 52
2008 Ike 935 0
2 84
2011 Irene 952 1
1 41
2012 Isaac 965 0
1 5
2012 Sandy 945 1
2 159
In: Statistics and Probability