Respuesta :
Answer:
[tex]P(16.6 < \bar X < 22.6) = P(\frac{16.6-19}{3} <Z< \frac{22.6-19}{3})= P(-0.8 < Z < 1.2)[/tex]
[tex]P(16.6 < \bar X < 22.6) =P(-0.8<Z<1.2) = P(Z<1.2)-P(Z<-0.8) = 0.88493- 0.211855= 0.673[/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 weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(19,6)[/tex]
Where [tex]\mu=19[/tex] and [tex]\sigma=6[/tex]
And we select n =4 fish. For this case we want to find this probability:
[tex] P(16.6 < \bar x < 22.6) [/tex]
And since the distribution for X is normal then the distribution for the sample mean is also normal and given by:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}}=3)[/tex]
And the z score is given by:
[tex]z = \frac{\bar x -\mu}{\sigma_{\bar x}}[/tex]
And if we apply this formula we got:
[tex]P(16.6 < \bar X < 22.6) = P(\frac{16.6-19}{3} <Z< \frac{22.6-19}{3})= P(-0.8 < Z < 1.2)[/tex]
And we can find this probability with this operation using the normal standard table or excel:
[tex] =P(-0.8<Z<1.2) = P(Z<1.2)-P(Z<-0.8) = 0.88493- 0.211855= 0.673[/tex]
