Question

Suppose that the mean value of interpupillary distance for all adult males is 65 mm, and the population standard deviation is 5 mm.

a) Find the probability that a male selected at random will have an interpupillary distance less that 67.5 mm. Assume that the distribution of interpupillary distances is approximately normal.

b) If a sample of 22 adult males is selected, what is the probability that the sample average distance x is between 64.3 and 66.4 mm?

c) Given that the sample size in part (b) is small, explain why we could still use the z-table to answer the question.

Answer 1

a)
`P(X<67.5) =P(Z< (67.5-65)/(5)) = P(Z<0.5)`

And we can find this probability using the z table or excel:

`P(z<0.5)=0.691`

b)
`P(64.3<\bar X<66.4)=P((64.3-\mu)/(\sigma_(\bar x))<(\bar X-\mu)/(\sigma_(\bar x))<(64.3-\mu)/(\sigma_(\bar x)))=P((64.3-65)/(1.066)<Z<(66.4-65)/(1.066))=P(-0.657<z<1.313)`

`P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)`

`P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)=0.905-0.256=0.650`

c) And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.

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".

Part a

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:

`X \sim N(65,5)`

Where
`\mu=65` and
`\sigma=5`

We are interested on this probability

`P(X<67.5)`

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(X<67.5) =P(Z< (67.5-65)/(5)) = P(Z<0.5)`

And we can find this probability using the z table or excel:

`P(z<0.5)=0.691`

Part b

For this case the distirbution of the sample mean is also normal since the distribution for the random variable X is normal and is given by:

`\bar X \sim N(\mu, \sigma_(\bar x)=(\sigma)/(√(n))= (5)/(√(22))=1.066)`

And for this case we want this probability:

`P(64.3<\bar X<66.4)=P((64.3-\mu)/(\sigma_(\bar x))<(\bar X-\mu)/(\sigma_(\bar x))<(64.3-\mu)/(\sigma_(\bar x)))=P((64.3-65)/(1.066)<Z<(66.4-65)/(1.066))=P(-0.657<z<1.313)`

And we can find this probability on this way:

`P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)`

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

`P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)=0.905-0.256=0.650`

Part c

For this case since the sample mean is defined as:

`\bar x = (\sum_(i=1)^n X_i)/(n)`

If we find the expected value for this variable we got:

`E(\bar X)= (1)/(n) \sum_(i=1)^n X_i =(n\mu)/(n)=\mu`

And for the variance we have:

`Var(\bar X) =(1)/(n^2) \sum_(i=1)^n Var(Xi) =(n \sigma^2)/(n^2)= (\sigma^2)/(n)`

And for this reason the deviation is
`(\sigma)/(√(n))`

And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.