Respuesta :
Answer:
a) [tex] P(X<2.4) = P(Z<\frac{2.4-2.5}{0.03}) = P(Z<-3.33) = 0.000434[/tex]
b) Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X}=2.5[/tex]
[tex] \sigma_{\bar X}=\frac{0.03}{\sqrt{10}}=0.00949[/tex]
c) [tex] P(\bar X<2.4)[/tex]
We can use the z score formula given by [tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using the z score formula we got:
[tex] P(\bar X<2.4) = P(Z<\frac{2.4-2.5}{0.00949}) = P(Z<-10.54) \approx 0[/tex]
d) Figure attached.
As we can see we have less deviation for the sample mean and for this reason we have more values near to the mean.
e) Since we just have the mean and the deviation we can't estimate the probability if we don't know the distribution.
The z score can't be applied if we have a normal distribution.
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 weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.5,0.03)[/tex]
Where [tex]\mu=2.5[/tex] and [tex]\sigma=0.03[/tex]
And we want this probability:
[tex] P(X<2.4)[/tex]
We can use the z score formula given by [tex] z = \frac{x -\mu}{\sigma}[/tex]
And using the z score formula we got:
[tex] P(X<2.4) = P(Z<\frac{2.4-2.5}{0.03}) = P(Z<-3.33) = 0.000434[/tex]
Part b
Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X}=2.5[/tex]
[tex] \sigma_{\bar X}=\frac{0.03}{\sqrt{10}}=0.00949[/tex]
Part c
[tex] P(\bar X<2.4)[/tex]
We can use the z score formula given by [tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using the z score formula we got:
[tex] P(\bar X<2.4) = P(Z<\frac{2.4-2.5}{0.00949}) = P(Z<-10.54) \approx 0[/tex]
Part d
Figure attached.
As we can see we have less deviation for the sample mean and for this reason we have more values near to the mean.
Part e
Since we just have the mean and the deviation we can't estimate the probability if we don't know the distribution.
The z score can't be applied if we have a normal distribution.
