For 300 trading days, the daily closing price of a stock (in $) is well modeled by a Normal model with mean $195.89 and standard deviation $7.18
According to this model, what cutoff value of price would separate the
a) lowest 15% of the days?
b) highest 0.61%?
c) middle 80%?
d) highest 50%?
c) Select the correct answer below and fill in the answer box(es) within your choice.
A.The cutoff points are _________ and _________.
(Use ascending order. Round to two decimal places as needed.)
B.The cutoff point is _________. (Round to two decimal places as needed.)
In: Math
You push a block of mass 24kg up a frictionless ramp that makes an angle 0.4189rad as measured from horizontal. Immediately after the block leaves your hands it has a velocity of magnitude 11m/s. The block travels up the ramp, reaches its highest point, and heads back down the ramp.
When is the magnitude of the block's acceleration the greatest?
a) On its way up the ramp.
b) On its way down the ramp.
c) At its highest point.
d) The magnitude of the net force on the block is constant throughout its motion.
In: Physics
A passenger with mass 85 kg rides in a Ferris wheel. The seats travel in a circle of radius 35 m. The Ferris wheel rotates at constant speed and makes one complete revolution every 25 s.
A)
Calculate the magnitude of the net force exerted on the passenger by the seat when she is one-quarter revolution past her highest point.
B)
Calculate the direction of the net force exerted on the passenger by the seat when she is one-quarter revolution past her highest point.
I need exact answers. Thank you
In: Physics
Section A1: All staff members receive an annual bonus of $200 plus an additional percentage of their Annual Income. Each staff member has been allocated their own percentage bonus rate (column C). In D6:D16 calculate each staff members Bonus
Section A2: Using the % Superannuation given to all staff. Calculate the annual Super amount paid to each staff member using the value in 9.5%. Copy the formula down to the last cell (to get full marks, a named range or an absolute cell reference must be used).
Section A3: In 'total package' calculate the total package for each staff member (Annual Salary, Bonus and Super). Adjust the spreadsheet so that the “#####” problem is addressed.
Section A4: Find total labour cost
Section A5: Inserest a row under the total labour cost and find the average total package
Section A6: use a formula to calculate the highest Total Package paid to an individual staff member (i.e. the biggest total package)
PLEASE PROVIDE AN EXPLAINATIONS during each step
Table is provided below.
| Employee Number | Annual | Bonus Rate | Bonus Amount | Annual Super | Total Package |
| 10026 | $ 49,283.00 | 1% | |||
| 10027 | $ 33,968.00 | 2% | |||
| 10030 | $ 32,158.00 | 2% | |||
| 10032 | $ 45,435.00 | 0% | |||
| 10033 | $ 51,722.00 | 0% | |||
| 10034 | $ 42,040.00 | 2% | |||
| 10035 | $ 44,161.00 | 1% | |||
| 10036 | $ 41,368.00 | 3% | |||
| 10037 | $ 57,029.00 | 2% | |||
| 10038 | $ 33,193.00 | 1% | |||
| 10039 | $ 37,410.00 | 0% | |||
| Total | |||||
| Super% | 9.50% | Highest |
In: Accounting
The National Football League (NFL) records a variety of performance data for individuals and teams. To investigate the importance of passing on the percentage of games won by a team, the following data show the conference (Conf), average number of passing yards per attempt (Yds/Att), the number of interceptions thrown per attempt (Int/Att), and the percentage of games won (Win%) for a random sample of 16 NFL teams for the 2011 season (NFL web site, February 12, 2012)
| Team | Conference | Yds/Att | Int/Att | Win% |
| Arizona Cardinals | NFC | 6.5 | 0.042 | 50.0 |
| Atlanta Falcons | NFC | 7.1 | 0.022 | 62.5 |
| Carolina Panthers | NFC | 7.4 | 0.033 | 37.5 |
| Cincinnati Bengals | AFC | 6.2 | 0.026 | 56.3 |
| Detroit Lions | NFC | 7.2 | 0.024 | 62.5 |
| Green Bay Packers | NFC | 8.9 | 0.014 | 93.8 |
| Houstan Texans | AFC | 7.5 | 0.019 | 62.5 |
| Indianapolis Colts | AFC | 5.6 | 0.026 | 12.5 |
| Jacksonville Jaguars | AFC | 4.6 | 0.032 | 31.3 |
| Minnesota Vikings | NFC | 5.8 | 0.033 | 18.8 |
| New England Patriots | AFC | 8.3 | 0.020 | 81.3 |
| New Orleans Saints | NFC | 8.1 | 0.021 | 81.3 |
| Oakland Raiders | AFC | 7.6 | 0.044 | 50.0 |
| San Francisco 49ers | NFC | 6.5 | 0.011 | 81.3 |
| Tennessee Titans | AFC | 6.7 | 0.024 | 56.3 |
| Washington Redskins | NFC | 6.4 | 0.041 | 31.3 |
Estimated Regression: Y= -5.763 + 12.949 + -1083.788
The average number of passing yards per attempt for the
Baltimore Ravens during the 2011 season was 6.7, and the team’s
number of interceptions thrown per attempt was 0.022. Use the
estimated regression equation developed in part (c) to predict the
percentage of games won by the Baltimore Ravens during the 2011
season. (Note: For the 2011 the 2011 season, the Baltimore
Ravens' record was 7 wins and 9 loses.)
| Compare your prediction to the actual percentage of games won
by the Baltimore Ravens. If required, round your answer to one
decimal digit. The Baltimore Ravens performed worse than predicted by ___% ? |
In: Statistics and Probability
Suppose that a survey of registered voters shows that 58% of people surveyed are in favor of changing the constitution to eliminate use of the electoral college in presidential elections.
Define a success as "a person is in favor of the change".
(a) What is p, the probability of success on a single
trial? _______ (show 2 decimal places)
(b) You decide to conduct a very small survey using 10
people to see whether you get similar results. For your survey,
what is n?
(c) If you conduct 10 trials, what is the probability
that you observe exactly 5 people in favor of the change?
P(X = 5) =________(round to 3 decimal places)
(d) What is the probability that you observe either 5
or 6 people who are in favor of the change?
P(5 or 6 people are in favor) =________ (round to 3
decimal places)
(e) What is the probability that everyone you survey
is in favor of the change?
P(all 10 people are in favor) =__________ (round to 3
decimal places)
(f) If you conducted your survey of 10 people multiple times, what is the long term average number of successes (people in favor of the change) you would expect to see in 10 trials? E(X) = mean =__________ (use one decimal place)
In: Statistics and Probability
It is advisable to have working smoke detectors in your
apartment. In fact, JJ Smoke Detectors Advisors (JJSDA) analyzes
the numbers of smoke detectors in apartments. Suppose X is the
number of smoke detectors in a randomly selected apartment. Then
according to JJSDA, x has the following probability mass function,
p(x), and cumulative distribution function F(x).
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| p(x) | .119 | .171 | .35 | .20 | b | .03 |
| F(x) | .119 | a | .640 | .840 | .970 | 1.00 |
a. If F is the cumulative distribution function of x, then what is
the value of a?
b. If p is the probability mass function for x, then evaluate b. c)
What is the expected value of x <_>
c. What is the expected value of x?
d. Calculate the standard deviation of x.
e. What is the variance of .5x?
f. JJSDA checks 9 apartments on a Tuesday. What is the probability
that there are exactly 3 smoke detectors in exactly 3 of the
checked apartments?
g. JJSDA checks 5 apartments on a Wednesday. What is the
probability that all 5 of those apartments have at least 2 smoke
detectors?
h.What is the expected value of
x3
Add any comments below.
In: Statistics and Probability
We have a binomial experiment with n = 18 trials, each with probability p = 0.15 of a success.
A success occurs if a student withdraws from a class, so the number of successes, x, will take on the values 0, 1, and 2. The probability of each x value, denoted f(x), can be found using a table like the one below. Note that these values are rounded to four decimal places.
| n | x | p | |||
|---|---|---|---|---|---|
| 0.10 | 0.15 | 0.20 | 0.25 | ||
| 18 | 0 | 0.1501 | 0.0536 | 0.0180 | 0.0056 |
| 1 | 0.3002 | 0.1704 | 0.0811 | 0.0338 | |
| 2 | 0.2835 | 0.2556 | 0.1723 | 0.0958 | |
| 3 | 0.1680 | 0.2406 | 0.2297 | 0.1704 | |
| 4 | 0.0700 | 0.1592 | 0.2153 | 0.2130 | |
| 5 | 0.0218 | 0.0787 | 0.1507 | 0.1988 | |
For an experiment with n = 18 trials, the probability of exactly x = 0 successes where the probability of a success on a trial is p = 0.15 can be found by going along the row for x = 0 within the n = 18 grouping until you get to the column for p = 0.15. Doing so gives f(0) = .
Use the above table to find the probabilities for x = 1 success, f(1), and x = 2 successes, f(2).
| f(1) | = | |
| f(2) | = |
In: Statistics and Probability
a. Suppose that 5 % of the items produced by a factory are defective. If 5 items are chosen at random, what is the probability that none of the items are defective? Write your answer as a decimal accurate to three decimal places.
b. Suppose that 7.9 % of the items produced by a second factory are defective. If 5 items are chosen at random from the second factory, what is the probability that exactly one of the items is defective? Write your answer as a decimal accurate to three decimal place
c. Suppose that 10.8 % of the items produced by a third factory are defective. If 5 items are chosen at random from the third factory, what is the probability that exactly two of the items are defective? Write your answer as a decimal accurate to three decimal places.
d. Suppose that 7.3 % of the items produced by a fourth factory are defective. If 5 items are chosen at random from the fourth factory, what is the probability that at least two of the items are defective? Write your answer as a decimal accurate to three decimal places.
e. Suppose that 14.2 % of the items produced by a fifth factory are defective. If 4 items are chosen at random from the fifth factory, what is the expected value (or mean value) for the number of defective items? Write your answer as a decimal accurate to three decimal places.
In: Math
Renal Disease:
The presence of bacteria in a urine sample (bacteriuria) is
sometimes associated with symptoms of kidney disease in women.
Suppose a determination of bacteriuria has been made over a large
population of women at one point in time and 5% of those sampled
are positive for bacteriuria.
1. If a sample size of 5 is selected from this population, what is the probability that 1 or more women are positive for bacteriuria?
2. Suppose 100 women from this population are sampled. What is
the probability that 3 or more of them are positive for
bacteriuria?
One interesting phenomenon of bacteriuria is that there is a
“turnover”; that is, if bacteriuria is measured on the same woman
at two different points in time, the results are not necessarily
the same. Assume that 20% of all women who are bacteriuric at time
0 are again bacteriuric at time 1 (1 year later), whereas only 4.2%
of women who were not bacteriuric at time 0 are bacteriuric at time
1. Let X be the random variable representing the number of
bacteriuric events over the two time periods for 1 woman and still
assume that the probability that a woman will be positive for
bacteriuria at any one exam is 5%.
3. what is the probability distribution of X?
4. what is the mean of X?
5. what is the variance of X?
In: Math