Question

Women have head circumferences that are normally distributed with a mean given by μ = 22.51 in​., and a standard deviation given by σ = 0.9 in. a. If a hat company produces​ women's hats so that they fit head circumferences between 22.1 in. and 23.1 ​in., what is the probability that a randomly selected woman will be able to fit into one of these​ hats?

Answer 1

`P(22.1<X<23.1)=P(-0.456<Z<0.656)=P(Z<0.656)-P(Z<-0.456)=0.744-0.324=0.420`

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

`X \sim N(22.51,0.9)`

Where
`\mu=22.51` and
`\sigma=0.9`

We are interested on this probability

`P(22.1<X<23.1)`

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(22.1<X<23.1)=P((22.1-\mu)/(\sigma)<(X-\mu)/(\sigma)<(23.1-\mu)/(\sigma))=P((22.1-22.51)/(0.9)<Z<(23.1-22.51)/(0.9))=P(-0.456<Z<0.656)`

And we can find this probability on this way:

`P(-0.456<Z<0.656)=P(Z<0.656)-P(Z<-0.456)`

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

`P(-0.456<Z<0.656)=P(Z<0.656)-P(Z<-0.456)=0.744-0.324=0.420`