Respuesta :
Answer:
a) [tex]P(X<67.5) =P(Z< \frac{67.5-65}{5}) = P(Z<0.5)[/tex]
And we can find this probability using the z table or excel:
[tex]P(z<0.5)=0.691[/tex]
b) [tex]P(64.3<\bar X<66.4)=P(\frac{64.3-\mu}{\sigma_{\bar x}}<\frac{\bar X-\mu}{\sigma_{\bar x}}<\frac{64.3-\mu}{\sigma_{\bar x}})=P(\frac{64.3-65}{1.066}<Z<\frac{66.4-65}{1.066})=P(-0.657<z<1.313)[/tex]
[tex]P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)[/tex]
[tex]P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)=0.905-0.256=0.650[/tex]
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:
[tex]X \sim N(65,5)[/tex]
Where [tex]\mu=65[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(X<67.5)[/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}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<67.5) =P(Z< \frac{67.5-65}{5}) = P(Z<0.5)[/tex]
And we can find this probability using the z table or excel:
[tex]P(z<0.5)=0.691[/tex]
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:
[tex] \bar X \sim N(\mu, \sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}= \frac{5}{\sqrt{22}}=1.066)[/tex]
And for this case we want this probability:
[tex]P(64.3<\bar X<66.4)=P(\frac{64.3-\mu}{\sigma_{\bar x}}<\frac{\bar X-\mu}{\sigma_{\bar x}}<\frac{64.3-\mu}{\sigma_{\bar x}})=P(\frac{64.3-65}{1.066}<Z<\frac{66.4-65}{1.066})=P(-0.657<z<1.313)[/tex]
And we can find this probability on this way:
[tex]P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.657<z<1.313)=P(z<1.313)-P(z<-0.657)=0.905-0.256=0.650[/tex]
Part c
For this case since the sample mean is defined as:
[tex] \bar x = \frac{\sum_{i=1}^n X_i}{n}[/tex]
If we find the expected value for this variable we got:
[tex] E(\bar X)= \frac{1}{n} \sum_{i=1}^n X_i =\frac{n\mu}{n}=\mu[/tex]
And for the variance we have:
[tex] Var(\bar X) =\frac{1}{n^2} \sum_{i=1}^n Var(Xi) =\frac{n \sigma^2}{n^2}= \frac{\sigma^2}{n}[/tex]
And for this reason the deviation is [tex] \frac{\sigma}{\sqrt{n}}[/tex]
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.
