Question

The distribution of the average amount of sleep per night gotten by college students is roughly bell-shaped with mean 412 minutes and standard deviation 68 minutes. The proportion of those who get an average of less than eight hours (480 minutes) of sleep per night is about:

Answer 1

`P(X<480)=P((X-\mu)/(\sigma)<(480-\mu)/(\sigma))=P(Z<(480-412)/(68))=P(z<1)`

`P(z<1)=0.841`

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 scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(412,68)`

Where
`\mu=412` and
`\sigma=68`

We are interested on this probability

`P(X<480)`

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<480)=P((X-\mu)/(\sigma)<(480-\mu)/(\sigma))=P(Z<(480-412)/(68))=P(z<1)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<1)=0.841`