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

Answer:

a) $P(X$

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

$P(z$

b) $P(64.3$

$P(-0.657$

$P(-0.657$

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.

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".  

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

$P(X$

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:

$P(X$

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

$P(z$

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:

$\bar X \sim N(\mu, \sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}= \frac{5}{\sqrt{22}}=1.066)$

And for this case we want this probability:

$P(64.3$

And we can find this probability on this way:

$P(-0.657$

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

$P(-0.657$

Part c

For this case since the sample mean is defined as:

$\bar x = \frac{\sum_{i=1}^n X_i}{n}$

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

$E(\bar X)= \frac{1}{n} \sum_{i=1}^n X_i =\frac{n\mu}{n}=\mu$

And for the variance we have:

$Var(\bar X) =\frac{1}{n^2} \sum_{i=1}^n Var(Xi) =\frac{n \sigma^2}{n^2}= \frac{\sigma^2}{n}$

And for this reason the deviation is $\frac{\sigma}{\sqrt{n}}$

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.