Errors in filling prescriptions A large number of preventable errors (e.g., overdoses, botched operations, misdiagnoses) are being made by doctors and nurses in U.S. hospitals. A study of a major metropolitan hospital revealed that of every 100 medications prescribed or dispensed, 1 was in error,

but only 1 in 500 resulted in an error that caused significant problems for the patient. It is known that the hospital prescribes and dispenses 60,000 medications per year.

  1. What is the expected proportion of errors per year at this hospital? The expected proportion of significant errors per year?
  2. Within what limits would you expect the proportion significant errors per year to fall?

Short Answer

Expert verified
  1. The expected proportion of errors per year at this hospital is 600 and the expected proportion of significant errors per year is 120.
  2. The proportion of significant errors per year falls between 0.00164 and 0.00236.

Step by step solution

01

Given information

There is a study of a major metropolitan hospital. The study revealed that in every 100 medications prescribed or dispensed, 1 is an error. But there is 1 error from a significant problem in every 500 medications. The hospital prescribed or dispensed 60000 medications per year.

02

Calculate the expected proportion

Consider the probability of having an error p=1100=0.01.

And the probability of having a significant error ps=1500=0.002.

a.

The expected proportion of error per year of the hospital is,

μ=np=60000×0.01=600

The expected proportion of significant error per year of the hospital is,

μs=nps=60000×0.002=120

03

Determine the fall limit

The limit of the expected proportion significant error per year fall is, ps±2ps1psn.

So,

ps±2ps1psn=0.002±2×0.002×10.00260000=0.0022×0.002×10.00260000,0.002+2×0.002×10.00260000=0.0020.000364,0.002+0.000364=0.00164,0.00236

Thus, the required limit is 0.00164,0.002360.00164,0.00236.

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Most popular questions from this chapter

Question: Refer to Exercise 5.3.

a. Find the sampling distribution of s2.

b. Find the population variance σ2.

c. Show that s2is an unbiased estimator of σ2.

d. Find the sampling distribution of the sample standard deviation s.

e. Show that s is a biased estimator of σ.

Question: Refer to Exercise 5.5, in which we found the sampling distribution of the sample median. Is the median an unbiased estimator of the population mean m?

Producing machine bearings. To determine whether a metal lathe that produces machine bearings is properly adjusted, a random sample of 25 bearings is collected and the diameter of each is measured.

  1. If the standard deviation of the diameters of the bearings measured over a long period of time is .001 inch, what is the approximate probability that the mean diameter xof the sample of 25 bearings will lie within.0001 inch of the population mean diameter of the bearings?
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Plastic fill process. University of Louisville operators examined the process of filling plastic pouches of dry blended biscuit mix (Quality Engineering, Vol. 91, 1996). The current fill mean of the process is set at μ= 406 grams, and the process fills standard deviation is σ= 10.1 grams. (According to the operators, “The high level of variation is since the product has poor flow properties and is, therefore, difficult to fill consistently from pouch to pouch.”) Operators monitor the process by randomly sampling 36 pouches each day and measuring the amount of biscuit mix in each. Considerx the mean fill amount of the sample of 36 products. Suppose that on one particular day, the operators observe x= 400.8. One of the operators believes that this indicates that the true process fill mean for that day is less than 406 grams. Another operator argues thatμ = 406, and the small observed value is due to random variation in the fill process. Which operator do you agree with? Why?

Refer to Exercise 5.3.

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