Question

The army reports that the distribution of head circumference among sale soldiers is approximately normal with mean 22.8 inches and standard deviation 1.1 inches.


a. A male soldier whose head circumference is 23.9 inches would be at what

percentile?


b. The army's helmet supplier regularly stocks helmets that fit soldiers with head

circumferences between 20 and 26 inches. Anyone with a head circumference

outside that interval requires a customized helmet order. What percent of soldiers

require custom helmets?

Answer 1

Answer:

a) We can find the z score for the value of 23.9:

$z=\frac{23.9-22.8}{1.1}=1$

And we can find the percentile with the following probability:

$P(z<1) =0.841$

So it's approximately the 84 percentile for this case.

b) $P(20<X<26)=P(\frac{20-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{26-\mu}{\sigma})=P(\frac{20-22.8}{1.1}<Z<\frac{26-22.8}{1.1})=P(-2.55<z<2.91)$

And we can find this probability like this:

$P(-2.55<z<2.91)=P(z<2.91)-P(z<-2.55)$

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

$P(-2.55<z<2.91)=P(z<2.91)-P(z<-2.55)=0.998-0.005=0.993$

So for this case we can conclude that approximately 99.3% of the soldiers satisfy the requirements.

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".  The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z<b)$

Let X the random variable that represent the head circumference of a population, and for this case we know the distribution for X is given by:

$X \sim N(22.8,1.1)$  

Where $\mu=22.8$ and $\sigma=1.1$

Solution to the problem

Part a

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

We can find the z score for the value of 23.9:

$z=\frac{23.9-22.8}{1.1}=1$

And we can find the percentile with the following probability:

$P(z<1) =0.841$

So it's approximately the 84 percentile for this case.

Part b

For this case we can find the following probability:

$P(20<X<26)=P(\frac{20-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{26-\mu}{\sigma})=P(\frac{20-22.8}{1.1}<Z<\frac{26-22.8}{1.1})=P(-2.55<z<2.91)$

And we can find this probability like this:

$P(-2.55<z<2.91)=P(z<2.91)-P(z<-2.55)$

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

$P(-2.55<z<2.91)=P(z<2.91)-P(z<-2.55)=0.998-0.005=0.993$

So for this case we can conclude that approximately 99.3% of the soldiers satisfy the requirements.