Respuesta :
Answer:
[tex] P(\bar X <63) [/tex]
And we can solve this using the following z score formula:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we use this formula we got:
[tex] z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4[/tex]
So we can find this probability equivalently like this:
[tex] P( Z<-2.4) = 0.0082[/tex]
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".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(63.6,2.5)[/tex]
Where [tex]\mu=63.6[/tex] and [tex]\sigma=2.5[/tex]
We select n =100. 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 this probability:
[tex] P(\bar X <63) [/tex]
And we can solve this using the following z score formula:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we use this formula we got:
[tex] z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4[/tex]
So we can find this probability equivalently like this:
[tex] P( Z<-2.4) = 0.0082[/tex]
