Question

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?

Answer 1

a)
`P(1<X<3)=0.7036`

b)
`P(X<4)=0.9991`

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:

`X \sim N(1.5,0.8)`

Where
`\mu=1.5` and
`\sigma=0.8`

We are interested on this probability

`P(1<X<3)`

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

`z=(x-\mu)/(\sigma)`

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

`P(1<X<3)=P((1-\mu)/(\sigma)<(X-\mu)/(\sigma)<(3-\mu)/(\sigma))=P((1-1.5)/(0.8)<Z<(3-1.5)/(0.8))=P(-0.625<z<1.875)`

And we can find this probability on this way:

`P(-0.625<z<1.875)=P(z<1.875)-P(z<-0.625)`

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

`P(-0.625<z<1.875)=P(z<1.875)-P(z<-0.625)=0.9696-0.2660=0.7036`

3) Part b

We are interested on this probability

`P(X<4)`

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

`z=(x-\mu)/(\sigma)`

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

`P(X<4)=P((X-\mu)/(\sigma)<(4-\mu)/(\sigma))=P(Z<(4-1.5)/(0.8))=P(z<3.125)`

And we can find this probability on this way:

`P(z<3.125)=0.9991`