Question

A survey among students at a certain university revealed that the number of hours spent studying the week before final exams was approximately normally distributed with mean 25 and standard deviation 6. What proportion of students studied between 25 and 34 hours

Answer 1

`P(25<X<34)=P((25-\mu)/(\sigma)<(X-\mu)/(\sigma)<(34-\mu)/(\sigma))=P((25-25)/(6)<Z<(34-25)/(6))=P(0<z<1.5)`

And we can find this probability with this difference:

`P(0<z<1.5)=P(z<1.5)-P(z<0)`

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

`P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433`

Explanation:

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 hous spent studying the week before final exams of a population, and for this case we know the distribution for X is given by:

`X \sim N(25,6)`

Where
`\mu=25` and
`\sigma=6`

We are interested on this probability

`P(25<X<34)`

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(25<X<34)=P((25-\mu)/(\sigma)<(X-\mu)/(\sigma)<(34-\mu)/(\sigma))=P((25-25)/(6)<Z<(34-25)/(6))=P(0<z<1.5)`

And we can find this probability with this difference:

`P(0<z<1.5)=P(z<1.5)-P(z<0)`

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

`P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433`