Question

Suppose that the mean score on a creativity test is 16 and the standard deviation is 4. You are told that the distribution is normal. Using the approximations for normal curves, how many people would get a score between 12 and 20?

Answer 1

`P(12<X<20)=P((12-\mu)/(\sigma)<(X-\mu)/(\sigma)<(20-\mu)/(\sigma))=P((12-16)/(4)<Z<(20-16)/(4))=P(-1<Z<1)`

And we can find this probability on this way:

`P(-1<Z<1)=P(Z<1)-P(Z<-1) =0.84113-0.15865 =0.6827`

We expect around 68.27% between the two scores provided.

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(16,4)`

Where
`\mu=16` and
`\sigma=4`

We are interested on this probability

`P(12<X<20)`

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(12<X<20)=P((12-\mu)/(\sigma)<(X-\mu)/(\sigma)<(20-\mu)/(\sigma))=P((12-16)/(4)<Z<(20-16)/(4))=P(-1<Z<1)`

And we can find this probability on this way:

`P(-1<Z<1)=P(Z<1)-P(Z<-1) =0.84113-0.15865 =0.6827`

We expect around 68.27% between the two scores provided.