Answer:
a) [tex]P(X<2.4)=P(\frac{X-\mu}{\sigma}<\frac{2.4-\mu}{\sigma})=P(Z<\frac{2.4-2.5}{0.03})=P(Z<-3.333)=0.00043[/tex]
b) [tex]\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)[/tex]
c) [tex]P(\bar X <2.4)=P(Z<\frac{2.4-2.5}{\frac{0.03}{\sqrt{10}}}=-10.54)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-10.54)\approx 0[/tex]
d) Figure attached
e) If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we can't estimate the probabilities on a easy way.
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]
We are interested on this probability
[tex]P(X<2.4)[/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<2.4)=P(\frac{X-\mu}{\sigma}<\frac{2.4-\mu}{\sigma})=P(Z<\frac{2.4-2.5}{0.03})=P(Z<-3.333)=0.00043[/tex]
And we can find this probability on this way:
[tex]P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)[/tex]
Part b
Since the distribution for X is normal then the distribution for the sample mean is:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)[/tex]
Part c
[tex]P(\bar X <2.4)=P(Z<\frac{2.4-2.5}{\frac{0.03}{\sqrt{10}}}=-10.54)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-10.54)\approx 0[/tex]
Part d
See the figure attached the deviation for the sample mean is lower for this reason we have the pattern in the graph attached.
Part e
If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we can't estimate the probabilities on a easy way.
