Question

An article reported that patients under care for HIV have CD4 tests every 4 months on...

An article reported that patients under care for HIV have CD4 tests every 4 months on average with a standard deviation of 1.5 months. What is the probability that in a sample of 60 patients, the mean time between tests exceeds 4.5 months?

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 4

standard deviation = = 1.5

n = 60

= 4

= / n = 1.5 / 60 = 0.19365

P( >4.5 ) = 1 - P( <4.5 )

= 1 - P[( - ) / < (4.5-4) / 0.19365]

= 1 - P(z <2.58 )

Using z table

= 1 - 0.9951

= 0.0049

probability= 0.0049

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
At an urgent care facility, patients arrive at an average rate of one patient every five...
At an urgent care facility, patients arrive at an average rate of one patient every five minutes. Assume that the duration between arrivals is exponentially distributed. (Round your answers to four decimal places.) a. Find the probability that the time between two successive visits to the urgent care facility is less than four minutes. b. Find the probability that the time between two successive visits to the urgent care facility is more than 16 minutes. c. If 10 minutes have...
In an article about the cost of health care, Money magazine reported that a visit to...
In an article about the cost of health care, Money magazine reported that a visit to the hospital emergency room for something as simple as sore throat has a mean cost of $328. Assume that the cost of this type of hospital emergency room visit is normally distributed with a standard deviation of $92. ) a. What is the probability that the cost will be more than $500? b. What is the probability that the cost will be less than...
A database contains information on the wait times patients have reported in the emergency room to...
A database contains information on the wait times patients have reported in the emergency room to see a physician since the hospital opened 5 years ago. Suppose the wait times vary according to a normal distribution with mean wait time of 1.9 hours and a standard deviation of 0.7 hours. Based on a simple random sample of 23 patients at the hospital in the waiting room, what is the probability that they will have to wait for more than 2.5...
A college has 260 ​full-time employees that are currently covered under the​ school's health care plan....
A college has 260 ​full-time employees that are currently covered under the​ school's health care plan. The average​ out-of-pocket cost for the employees on the plan is 1,820.20 with a standard deviation of ​$510. The college is performing an audit of its health care plan and has randomly selected 30 employees to analyze their​ out-of-pocket costs. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be less than 1,755.00 c. What...
A college has 240 full-time employees that are currently covered under the school's health care plan....
A college has 240 full-time employees that are currently covered under the school's health care plan. The average out-of-pocket cost for the employees on the plan is $1,910 with a standard deviation of 520. The college is performing an audit of its health care plan and has randomly selected 35 employees to analyze their out-of-pocket costs. a. Calculate the standard error of the mean (round to 2 decimal points). b. What is the probability that the sample mean will be...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.3 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.0 day per employee with a standard deviation of 1.8 days. Suppose...
The effectiveness of a new treatment was tested against the existing procedure for Alzheimer's patients. A...
The effectiveness of a new treatment was tested against the existing procedure for Alzheimer's patients. A study randomized 13 patients to the new treatment and 18 patients to the existing. After 3 years, all patients relapsed, and Alzheimer's symptoms re-occurred. The average time-to-relapse was X_bar1=18.6 months under the new treatment and X_bar_2=10.1 months under the existing procedure, with observed standard deviation s_1=3 and s_2=3.2. To test the difference between the mean relapse time of the new treatment and the standard...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.3 days per employee with a standard deviation of 1.9 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.4 day per employee with a standard deviation of 1.8 days. Suppose...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.4 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.4 day per employee with a standard deviation of 1.6 days. Suppose...
It has been reported that the percentage of American households with children under 18 living at...
It has been reported that the percentage of American households with children under 18 living at home has hit its lowest point at 46% in half a century. How many households do you expect will have children under 18 living at home in a sample mean of 1000? Calculate the standard deviation of the same proportions What is the probability that in a sample of 1000 households there are between 450 and 470 observations that have children under 18 living...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT