Question

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

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

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.

Calculate the expected proportion

Consider the probability of having an error $p=\frac{1}{100}=0.01$.

And the probability of having a significant error $p_s=\frac{1}{500}=0.002$.

a.

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

$\begin{array}{c}\mu =np\\ =60000\times 0.01\\ =600\end{array}$

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

$\begin{array}{c}{\mu }_s=n{p}_s\\ =60000\times 0.002\\ =120\end{array}$

Determine the fall limit

The limit of the expected proportion significant error per year fall is, $\left(p_s\pm 2\sqrt{\frac{p_s\left(1-p_s\right)}{n}}\right)$.

So,

$\begin{array}{c}\left(p_s\pm 2\sqrt{\frac{p_s\left(1-p_s\right)}{n}}\right)=\left(0.002\pm 2\times \sqrt{\frac{0.002\times \left(1-0.002\right)}{60000}}\right)\\ =\left[\left(0.002-2\times \sqrt{\frac{0.002\times \left(1-0.002\right)}{60000}}\right),\left(0.002+2\times \sqrt{\frac{0.002\times \left(1-0.002\right)}{60000}}\right)\right]\\ =\left[\left(0.002-0.000364\right),\left(0.002+0.000364\right)\right]\\ =\left[0.00164,0.00236\right]\end{array}$

Thus, the required limit is $\left[0.00164,0.00236\right]$$\left[0.00164,0.00236\right]$.