Q. 27 - Elementary Statistics

Question

27. Lower Limb Surgery. The study "Intrathecal Sufentanil versus Fentanyl for Lower Limb Surgeries - A Randomized Controlled Trial" (Journal of Anesthesiology Clinical Pharmacology, Vol. 27 . Issue 1. pp. 67-73) by P. Motiani et al. compares two different agents, intrathecal Sufentanil and fentanyl, used in enhancing the anesthesiology of patients receiving major lower limb surgery. One variable compared between the two agents was the amount of blood loss during the surgery. Based on the study, we will assume that using fentanyl, the amount of blood loss during major lower limb surgery is normally distributed with mean $283.3\mathrm{ml}$ and standard deviation $83.3\mathrm{ml}$. Find the percentage of patients whose amount of blood loss during major lower limb surgery using fentanyl is

a. Less than $304\mathrm{ml}$.

b. Between $221$ and $429ml$

c. More than $450ml.$

Step-by-Step Solution

Answer

a. Following major lower limb surgery using fentanyl, $9.87$ percent of patients lose less than $304\mathrm{ml}$ of blood.

b. There are $73.3\%$ of patients whose blood loss during major lower limb surgery using fentanyl is less than $221\mathrm{ml}$ and $429\mathrm{ml}$.

c. More than $450\mathrm{ml}$ of blood was lost during major lower limb surgery using fentanyl in $2.28$ percent of patients.

Part (a) Step 1: Given Information

Calculate the percent of patients whose blood loss during major lower limb surgery involving fentanyl was less than $=304ml$

Part (a) Step 2: Explanation

When the lower limb surgery was performed using fentanyl, the blood loss was normally distributed with a mean of $283.3ml$ and a standard deviation of $83.3ml.$

Formula used:

A normal curve using $X$ has the following equation if $X$ is a continuous random variable with a mean mu and standard deviation sigma:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\mathrm{\infty }<x<\mathrm{\infty },-\mathrm{\infty }<\mu <\mathrm{\infty },\sigma >0$

Furthermore, a normal curve with random variable $Z$ has the following equation:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Mean and standard deviation of $Z$ are $0$ and $1$ , respectively.

Normal curves usually contain two population parameters, one being the mean $\mu$ of the population and the other being the standard deviation $\sigma$of the population.

Calculation:

$\mu =283.3\mathrm{ml} ; \sigma =83.3\mathrm{ml}$

First, determine the probability of the variable $X$ being smaller than $304ml$, i.e., $P(X<304ml)$

Find the value:

$\begin{array}{c}P(X<304ml)=P(X-\mu <304-\mu )\\ =P(X-\mu <304-283.3)\\ =P(\frac{X-\mu }{\sigma }<\frac{304-283.3}{83.3})\\ =P(Z<0.25)\\ =P(0<Z<0.25)\\ =0.0987\end{array}$

The percentage of blood loss will be $0.0987\times 100\%=9.87\%.$

Part (b) Step 1: Given Information

Find the percentage of patients whose amount of blood loss during major lower limb surgery using fentanyl less than $304ml.$

Part (b) Step 2: Explanation

The amount of blood lost during lower limb surgery when using fentanyl is regularly distributed, with a mean of $283.3ml$ and a standard deviation of $83.3ml.$

The following equation describes a normal curve for a continuous random variable $X$with mean $\mu$ and standard deviation $\sigma$:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\mathrm{\infty }<x<\mathrm{\infty },-\mathrm{\infty }<\mu <\mathrm{\infty },\sigma >0$

Furthermore, a normal curve with random variable $Z$ has the following equation:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Normal curves usually contain two population parameters, one being the mean $\mu$ of the population and the other being the standard deviation $\sigma$ of the population.

Calculation:

$\mu =283.3\mathrm{ml};\sigma =83.3\mathrm{ml}$

First, calculate the probability of the variable $X$ being less than $304 ml.$ i.e. $P(221<X<429 ml)$. 

Now calculating its value:

$\begin{array}{c}P(221<X<429ml)=P(221-\mu <X-\mu <429-\mu )\\ =P(221-283.3X-\mu <429-283.3)\\ =P(\frac{221-283.3}{83.3}<\frac{X-\mu }{\sigma }<\frac{429-283.3}{83.3})\\ =P(-0.75<Z<1.75)\\ =P(-0.75<Z<0)+P(0<Z<1.75)\\ =0.2734+0.4599=0.7333\end{array}$

The percentage of blood loss will be $0.7333\times 100\%=73.33\%.$

Part (c) Step 1: Given Information

Determine the percentage of patients who had less than $304ml$ blood loss during major lower limb surgery with fentanyl.

Part (c) Step 2: Explanation

The volume of blood lost during lower limb surgery when using fentanyl is regularly distributed, with a mean of $283.3ml$and a standard deviation of $83.3 ml.$

The following equation describes a normal curve for a continuous random variable $X$with mean $\mu$ and standard deviation $\sigma$:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\mathrm{\infty }<x<\mathrm{\infty },-\mathrm{\infty }<\mu <\mathrm{\infty },\sigma >0$

Furthermore, a normal curve with random variable $Z$ has the following equation:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Normal curves usually contain two population parameters, one being the mean $\mu$ of the population and the other being the standard deviation $\sigma$ of the population.

Calculation:

$\mu =283.3\mathrm{ml} ; \sigma =83.3\mathrm{ml}$

Determine the probability of the variable $X$ such that it is less than $304\mathrm{ml}$ that is, $P(X>450ml)$

Find its value:

$\begin{array}{c}P(X>304ml)=P(X-\mu >450-\mu )\\ =P(X-\mu >450-283.3)\\ =P(\frac{X-\mu }{\sigma }>\frac{450-283.3}{83.3})\\ =P(Z>2)\\ =0.5-P(0<Z<2)\\ =0.5-0.4772=0.0228\end{array}$

The percentage of blood loss will be $0.0228\times 100\%=2.28\%.$