Solution Found!
Medication used to treat & certain condition is
Chapter 4, Problem 7SE(choose chapter or problem)
Medication used to treat a certain condition is administered by syringe. The target dose in a particular application is μ. 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 μ and variance \(\sigma^{2}\).
a. What is the probability that the dose administered differs from the mean μ by less than σ?
b. If X represents the dose administered, find the value of z so that \(P(X<\mu+z \sigma)=0.90\).
c. If the mean dose is 10 mg, the variance is \(2.6 \mathrm{mg}^{2}\), and a clinical overdose is defined as a dose larger than 15 mg, what is the probability that a patient will receive an overdose?
Equation transcription:
Text transcription:
\sigma^{2}
P(X<\mu+z \sigma)=0.90
2.6 {mg}^{2}
Questions & Answers
QUESTION:
Medication used to treat a certain condition is administered by syringe. The target dose in a particular application is μ. 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 μ and variance \(\sigma^{2}\).
a. What is the probability that the dose administered differs from the mean μ by less than σ?
b. If X represents the dose administered, find the value of z so that \(P(X<\mu+z \sigma)=0.90\).
c. If the mean dose is 10 mg, the variance is \(2.6 \mathrm{mg}^{2}\), and a clinical overdose is defined as a dose larger than 15 mg, what is the probability that a patient will receive an overdose?
Equation transcription:
Text transcription:
\sigma^{2}
P(X<\mu+z \sigma)=0.90
2.6 {mg}^{2}
ANSWER:
Answer:
Step 1 of 3:
(a)
In this question we have asked to find the probability that the dose administered differs from the mean by less than
.
Actual dose administered is normally distributed with mean and variance
.
We know that if X is a random variable whose probability density function is normal with mean and variance
, then we can write as X
Where N is called the normal distribution ( also called the Gaussian distribution) whose probability density function can be written as
Figure 1 represent a plot of the normal distribution (probability distribution function plot) with mean and standard deviation
Figure 1 Probability density function of a normal random variable with mean
and standard deviation
Sometimes we sample any population from a normal population with value of mean 0 and standard deviation 1 and this normal population is called the standard normal population and we represent it by the “z-score” of x.
In the question we are asking the probability that the dose administered differs from the mean by less than
.
So from the figure we can find the probability that a normal random variable is within one standard deviation of its mean
is the area under the normal curve between
and
. (using table)
Area to the left of z = - 1 is 0.1587
Area to the left of z = 1 is 0.8413
Hence the area between this 0.8413 - 0.1587 = 0.6826 and it is also visible from the figure 1.
So the probability that the dose administered differs from the mean by less
is equal to 0.6826.