Chapter 7, Problem 4RE

Take your medicine: Medication used to treat a certain condition is administered by syringe. The target dose in a particular application is 10 milligrams. Because of the variations in the syringe, in reading the scale: and in mixing the fluid suspension, the actual dose administered is normally distributed with mean $\mu =10$ milligrams and standard deviation $\sigma =1.6$ milligrams.

1. What is the probability that the dose administered is between 9 and 11.5 milligrams?
2. Find the 98th percentile of the administered dose.
3. If a clinical overdose is defined as a dose larger than 15 milligrams, what is the probability that a patient will receive an overdose?

To determine

To find: the probability that the dose administered is between $\text{9 and 11}\text{.5}$ milligrams.

Answer to Problem 4RE

The required answer is $0.56$ .

Explanation of Solution

Given Information:

A normal population has mean $\mu =10$ milligrams

standard deviation $\sigma =1.6$ milligrams.

Required Calculations:

It is asked in the question to find $P(9<X<11.5)$

Now,

  $\begin{array}{l}P(9<X<11.5)=P\left(\frac{9-10}{1.6}<\frac{X-10}{1.6}<\frac{11.5-10}{1.6}\right)\\ P(9<X<11.5)=P(-0.625<Z<0.9375)\end{array}$

Using normal table

  $P(9<X<11.5)=0.56$

Calculation:

To compute the proportion of the population that is between $\text{9 and 11}\text{.5}$ milligrams using Minitab follow the steps given below.

1. Click CaIc, then select Probability distributions and then go to Normal.

2. Select the Cumulative option.

3. Enter $10$ in the Mean field and enter $1.6$ in the standard deviation fields

4. Enter $9$ in the input constant field.

5. Click Ok. The final output is given below.

Cumulative Distribution function

Normal with mean $=10$ and standard deviation $=1.6$

  $\begin{array}{l}xP\left(X\le x\right)\\ 90.265986\end{array}$

Similarly, compute the probability when $x\text{ is 11}\text{.5}$ . The final output is given below.

Cumulative distribution function.

Normal with mean $=10$ and standard deviation $=1.6$ .

  $\begin{array}{l}x\text{P}\left(X\le x\right)\\ 11.50.825749\end{array}$

The required probability that $X$ is between $9\text{ and 11}\text{.5}$ is

  $\begin{array}{l}P\left(9<x<11.5\right)\\ =P\left(x<11.5\right)-P\left(x<9\right)\\ =0.83-0.27\\ =0.56\end{array}$

The required answer is $0.56$ .

To determine

To find: the $98\text{th}$ percentile of the administered dose.

Answer to Problem 4RE

The $98\text{th}$ percentile of the population is $13.3$ .

Explanation of Solution

Given Information:

The actual dose administered is normally distributed with mean $\mu =10$ milligrams

standard deviation $\sigma =1.6$ milligrams.

Calculation:

To compute the $98th$ percentile using Excel follow the steps given below.

1. Click on Insert function.

2. Select NORMINV

3. Enter $0.98$ in the probability field, enter $10$ in the mean field and enter $1.6$ in the standard deviation field.

The above output shows that the ${98}^{th}$ percentile of the population is $13.3$ .

The required answer is $13.3$ .

To determine

To compute: the probability that the dose administered is greater than $15$ milligrams.

Answer to Problem 4RE

The probability that the dose administered is greater than $15\text{mg}$ is $0.001$ .

Explanation of Solution

Given Information:

A normal population has mean $\mu =10$ milligrams

standard deviation $\sigma =1.6$ milligrams.

Required calculations:

it is asked in the question to find $P\left(X>15\right)$

  $\begin{array}{l}P\left(X>15\right)=P\left(\frac{X-10}{1.6}>\frac{15-10}{1.6}\right)\\ P\left(X>15\right)=P\left(Z>3.125\right)\end{array}$

Using normal table

  $P\left(X>15\right)=0.001$

Calculation steps:

Compute the probability that the dose administered is greater than $15$ milligrams. The final screenshot is given below.

Cumulative Distribution Function

Normal with mean $=10$ and standard deviation $=1.6$

  $\begin{array}{l}x\text{ P}\left(X\le x\right)\\ 150.999111\end{array}$

The required probability is,

  $\begin{array}{l}P\left(x>15\right)\\ =1-P\left(x<15\right)\\ =1-0.999\\ =0.001\end{array}$

The probability that the dose administered is greater than $15$ milligrams is very low. Hence, the patient will receive an overdose.