Question

In: Statistics and Probability

A hospital has determined that the average duration of a stay at the hospital is 9.3...

A hospital has determined that the average duration of a stay at the hospital is 9.3 days. Suppose that the population standard deviation for the duration of stays at this hospital is 2.3 days. Researchers randomly select 36 former patients to include in further research on hospital stays. Use the central limit theorem to find the approximate probability that total amount of time these 36 patients spent at the hospital exceeds 180 days?

Solutions

Expert Solution

Let Xi denote the duration of stay (in days) of the ith patient. (i = 1,2,...,36)

Now, we are given that:

Let X denote the sum of duration of stays of these 36 patients.

Thus, X = X1+X2+...+X36 =

Now, consider:

Also:

Thus, from the central limit theorem, we can conclude that:

Now, the probability that total amount of time these 36 patients spent at the hospital exceeds 180 days is given by:

For any queries, feel free to comment and ask.

If the solution was helpful to you, don't forget to upvote it by clicking on the 'thumbs up' button.


Related Solutions

1. A hospital determined that the average duration of a stay at the hospital is 7.1...
1. A hospital determined that the average duration of a stay at the hospital is 7.1 days. Suppose that the population standard deviation for the duration of stays at this hospital is 1.9 days. Researchers randomly select 26 former patients to include in further research on hospital stays. a) Use the central limit theorem to find the approximate probability that total amount of time these 26 patients spent at the hospital exceeds 156 days? b) What assumptions must you make...
The average length of a maternity stay in a U.S. hospital is 2.2 days with a...
The average length of a maternity stay in a U.S. hospital is 2.2 days with a standard deviation of 0.6 days. A sample of 25 women who recently gave birth is taken. Find the probability that the sample mean is greater than 2.4 days. Round your answer to the nearest hundredth.
The average length of a maternity stay in a U.S. hospital is said to be 2.2...
The average length of a maternity stay in a U.S. hospital is said to be 2.2 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital. B)Give the distribution of X. (Round your standard deviation to two decimal places.) C)Give the distribution of ΣX. (Round your standard deviation to two decimal places.)
1. Suppose the equilibrium price for an average hospital stay in the absence of insurance is...
1. Suppose the equilibrium price for an average hospital stay in the absence of insurance is $10,000. At that price, 1000 people are hospitalized each year. Now suppose an insurer offers a policy to lower the out of pocket price of a stay to $100, and at that price, 1200 people are hospitalized. a) How much TOTAL premium revenue must be collected to finance this arrangement? b) How much premium revenue per hospitalized person must be collected? Would the average...
The average length of stay in a chronic disease hospital for a certain type of patient...
The average length of stay in a chronic disease hospital for a certain type of patient is 60 days with a standard deviation of 15 days. If 36 of these patients are randomly selected, find the probability that the average length of stay of these 36 patients is less than 58 days. (a) .1519 (b) .3176 (c) .4483 (d) .2119 (e) .6124
A medical researcher wants to determine if the average hospital stay after a certain procedure is...
A medical researcher wants to determine if the average hospital stay after a certain procedure is less than 11.87 days. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 11.87, Alternative Hypothesis: μ < 11.87. If the researcher randomly samples 25 patients that underwent the procedure and determines their average hospital stay was 10.6 days with a standard deviation of 6.724 days, what is the test statistic and p-value of this test? Question 10 options: 1) Test...
1. A bank has an average asset duration of 4 years and an average liability duration...
1. A bank has an average asset duration of 4 years and an average liability duration of 1.5 years. This bank has total assets of $500 million and total liabilities of $450 million. Currently, market interest rates are 10 percent. If interest rates rise by 1 percent (to 11 percent), what is this bank's change in net worth? a. net worth will decrease by $12.05 million b. net worth will decrease by $15.45 million c. net worth will increase by...
A bank has an average asset duration of 4.2 years and an average liability duration of 2.4 years.
A bank has an average asset duration of 4.2 years and an average liability duration of 2.4 years. This bank has total assets of $650 million and total liabilities of $420 million. Currently, market interest rates are 7 percent. If interest rates fall by 1 percent (to 6 percent), what is this bank's change in net worth?
1. Reimbursement Methods A. A hospital admits 525 patients per month who stay an average of...
1. Reimbursement Methods A. A hospital admits 525 patients per month who stay an average of five (5.0) days per admission. This hospital charges $12,000 per admission and incurs $7,100 in costs ($4,000 fixed and $3,100 variable) per patient. Assume all patients are reimbursed under the same reimbursement method, complete the table and calculate net income for the hospital under each reimbursement method.             A.         80% of Charges:             B.         Cost:             C.         $1,550 per diem:             D.         $7,250 per...
A hospital conducted a survey to estimate the mean length of stay at the hospital. a...
A hospital conducted a survey to estimate the mean length of stay at the hospital. a simple random sample of 100 patients resulted in a sample mean of 4.5 days. the population standard deviation is 4 days. a) Calculate a 90% confidence interval. b)Interpret the results of your confidence interval.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT