Suppose that the mean value of interpupillary 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 interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x 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.)

Answer:[tex]P(64 <X<66)=0.6826[/tex]Step-by-step explanation:1) Previous conceptsNormal 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".  Let X the random variable that represent the interpupillary distance (the distance between the pupils of the left and right eyes) 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]And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]On this case  [tex]\bar X \sim N(65,\frac{5}{\sqrt{25}})[/tex]We are interested on this probability[tex]P(64<\bar X<66)[/tex]And the best way to solve this problem is using the normal standard distribution and the z score given by:[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]If we apply this formula to our probability we got this:[tex]P(64<\bar X<66)=P(\frac{64-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{66-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex][tex]=P(\frac{64-65}{\frac{5}{\sqrt{25}}}<Z<\frac{66-65}{\frac{5}{\sqrt{25}}})=P(-1<z<1)[/tex]And we can find this probability on this way:[tex]P(-1<z<1)=P(z<1)-P(-1)[/tex]And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  [tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.8413-0.1587=0.6826[/tex]The other way to solve this problem is using the empirical rule.The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).And since the standard deviation for the mean is 1, then on the interval from 64 to 66 we are within one deviation from the mean 65. so Using this rule we have that the % of the data between 64 and 66 would be approximately 68%.