Question
PLEASE SHOW WORK(will upvote) The average weekly income of information workers in private industry is $777....
PLEASE SHOW WORK(will upvote)
The average weekly income of information workers in private
industry is $777. If the standard deviation is $77, what is the
probabiltity that a random sample of 50 information workers will
earn, on average, Between $500 and $800 per week?
Homework Answers
Answer #1
Not the answer you're looking for?
Similar Questions
According to a survey, the workers employed in a processing industry earn an average of $550...
According to a survey, the workers employed in a processing
industry earn an average of $550 per week. Assume that this mean is
based on a random sample of 1000 workers selected from the
processing industry and that the population standard deviation of
weekly earnings is known to be $75. Find a 99% confidence interval
for the mean weekly earnings of all workers employed in processing
industries.
Please solve all parts for upvote At work some business analysts estimate that the length of...
Please solve all parts for upvote
At work some business analysts estimate that the length of time
people work at a job has a mean of 6.2 years and a standard
deviation of 4.5 years
a) Explain why you suspect this distribution may be skewed to
the right.
b) If you take a random sample of 5 people, can you use Table Z
to find the probability they work at their current job for at least
10 years? Explain in...
The CEO of 3M believes that on the average employees work less than 50 hours per...
The CEO of 3M believes that on the average employees
work less than 50 hours per week. A small simple random sample of
employees provided the following weekly working hours.
Weekly
Hours
Xi
47
50
51
48
49
46
50
51
a. Compute the point estimates of population
average (sample average) and standard deviation (sample standard
deviation) of weekly working hours of all employees at 3M. You need
to show all of your computations in the table below.
b. Using...
1) Automobile Workers A worker in the automobile industry works an average of 42.9 hours per...
1) Automobile Workers A worker in the automobile industry works
an average of 42.9 hours per week. If the distribution
is approximately normal with a standard deviation of 1.4 hours,
what is the probability that a randomly selected automobile worker
works less than 41 hours per week? Round the answer to at least
four decimal places.
P(X<41)
Teachers' Salaries The average annual salary
for all U.S. teachers is 47750. Assume that the distribution is
normal and the standard deviation is 5680....
5. A recent news story reported the following information about a sample of workers in Germany...
5. A recent news story reported the following information about
a sample of workers in Germany and in the United States.
United States
Germany
Average length of work week ( in hours
)
42
38
Sample Standard
Deviation
5
6
Sample
Size
23
28
At a 5% level of significance, test to see if there is a
significant difference in the average work week between Germany and
the United States. What is your Conclusion
A random sample of 100 workers in one large plant took an average of 12 minutes...
A random sample of 100 workers in one large plant took an
average of 12 minutes to complete a task, with a standard deviation
of 2 minutes. A random sample of 50 workers in a second large plant
took an average of 11 minutes to complete the task, with a standard
deviation of 3 minutes. Construct a 95% confidence interval for the
difference between the two population mean completion times.
Suppose that the population average for weekly earnings for employees in general automotive repair shops is...
Suppose that the population average for weekly earnings for
employees in general automotive repair shops is $450, and that the
population standard deviation for the earnings for such employees
is $50. A sample of 100 such employees is selected at random.
16) What is the probability distribution of the average weekly
earnings of a sample of 100 for employees in general automotive
repair shops?
a) Normal
b) Not enough information to determine
c) t-distributed.
d) None of the above
17)...
Show your work. Carry out all calculations to at least 3 significant digits. Please show your...
Show your work. Carry out all calculations to at least 3
significant digits. Please show your work
Marketing strategists like to study the differences (in, e.g.,
age and income) between buyers and non-buyers of a product. In an
earlier study of the purchasers and non-purchasers of a product
sold by the AAA Company, demographic data were collected. Their age
profiles (in years) are summarized and reported as follows:
Purchasers
Sample size 900
Sample mean 43.8
Sample standard deviation 14.6
Non-Purchasers...
Show work please. 1) If the scores for a test have a mean of 70 and...
Show work please.
1) If the scores for a test have a mean of 70 and a standard
deviation of 12, find the percentage of scores that will fall above
50.
A) 95.25% B) 95.54% C) 45.07% D) 45.54%
2)
The average age of a vehicle registered in the U.S. is 8 years.
If a random vehicle is selected, find the probability that the age
is between 80 and 110 months. Assume the standard deviation is
15.
A) .6667 B)...
please show all work A random sample of 45 life insurance policy holders showed that the...
please show all work
A random sample of 45 life insurance policy holders showed that
the average premiums paid on their life insurance policies was $340
per year with a sample standard deviation of $62. Construct a 90%
confidence interval for the population mean. Make a statement about
this in context of the problem.
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 6 minutes ago
asked 10 minutes ago
asked 30 minutes ago
asked 50 minutes ago
asked 57 minutes ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 2 hours ago
asked 2 hours ago