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

Answer:

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(22.1<X<23.1)=P(\frac{22.1-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{23.1-\mu}{\sigma})=P(\frac{22.1-22.51}{0.9}<Z<\frac{23.1-22.51}{0.9})=P(-0.456<Z<0.656)$

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.  

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