Question

Suppose the number of hours to complete this exam for my students is denoted by the random variable X and is normally distributed with the mean of µx hours and the standard deviation of σX = 2.3 hours. Find the probability that a randomly selected student has finished the exam spending 3 to 6 hours, P(3 < X < 6) =? (Use the attached table.)

Answer 1

`P(3<X<6)=P((3-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((3-4.5)/(2.3)<Z<(6-4.5)/(2.3))=P(-0.65<z<0.65)`

And we can find this probability with this difference:

`P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)`

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

`P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)=0.742-0.258=0.484`

Explanation:

Asuming this complete question: Suppose the number of hours to complete this exam for my students is denoted by the random variable X and is normally distributed with the mean of µx hours and the standard deviation of σX = 2.3 hours. Find the probability that a randomly selected student has finished the exam spending 3 to 6 hours, P(3 < X < 6) =? (Use the attached table.)

Let =4.5

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".

Solution to the problem

Let X the random variable that represent the number of hours to complete an examn of a population, and for this case we know the distribution for X is given by:

`X \sim N(4.5,2.3)`

Where
`\mu=4.5` and
`\sigma=2.3`

We are interested on this probability

`P(3<X<6)`

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(3<X<6)=P((3-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((3-4.5)/(2.3)<Z<(6-4.5)/(2.3))=P(-0.65<z<0.65)`

And we can find this probability with this difference:

`P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)`

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

`P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)=0.742-0.258=0.484`