Suppose that the mean value of inter-pupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm.

(a) If the distribution of inter-pupillary distance is normal and a sample of n =25 adult males is to be selected, what is the probability that the sample average distance for these 25 will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.)

(b) Suppose that a sample of 100 adult males is to be obtained. Without assuming that inter-pupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) ?

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

[tex]X \sim N(65,5)[/tex]

Where [tex]\mu=65[/tex] and [tex]\sigma=5[/tex]

Part a

Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

We want to find this probability:

[tex]P(64< \bar X <66)=P(\frac{64-65}{\frac{5}{\sqrt{25}}}<Z<\frac{66-65}{\frac{5}{\sqrt{25}}})[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(-1<Z<1)=P(Z<1)-P(Z<-1) =0.8413-0.1587= 0.6827[/tex]

Part b

[tex]P(64< \bar X <66)=P(\frac{64-65}{\frac{5}{\sqrt{100}}}<Z<\frac{66-65}{\frac{5}{\sqrt{100}}})[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(-2<Z<2)=P(Z<2)-P(Z<-2) =0.9773-0.0228= 0.9545[/tex]