Researchers at a pharmaceutical company have found that the effective time duration of a safe dosage of pain relief drug is normally distributed with the mean 1.5 hours and standard deviation 0.8 hours For a patient selected at random
a) what is the probability that the drug will be effective between 1 and 3 hours after taken
?
b) what is the probability the drug will be effective for 4 hours or less?

Respuesta :

Answer:

a) [tex]P(1<X<3)=0.7036[/tex]

b) [tex]P(X<4)=0.9991[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

Let X the random variable that represent the duration of a safe dosage of pain relief drug, and for this case we know the distribution for X is given by:

[tex]X \sim N(1.5,0.8)[/tex]  

Where [tex]\mu=1.5[/tex] and [tex]\sigma=0.8[/tex]

We are interested on this probability

[tex]P(1<X<3)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(1<X<3)=P(\frac{1-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(\frac{1-1.5}{0.8}<Z<\frac{3-1.5}{0.8})=P(-0.625<z<1.875)[/tex]

And we can find this probability on this way:

[tex]P(-0.625<z<1.875)=P(z<1.875)-P(z<-0.625)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-0.625<z<1.875)=P(z<1.875)-P(z<-0.625)=0.9696-0.2660=0.7036[/tex]

3) Part b

We are interested on this probability

[tex]P(X<4)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=P(Z<\frac{4-1.5}{0.8})=P(z<3.125)[/tex]

And we can find this probability on this way:

[tex]P(z<3.125)=0.9991[/tex]