Respuesta :
Answer:
a) [tex]X \sim N(\mu=75,\sigma=45)[/tex]
[tex]\bar X \sim N(75,\frac{45}{\sqrt{81}})[/tex]
b) [tex]z=\frac{90-75}{\frac{45}{\sqrt{81}}}=3[/tex]
c) [tex]P(\bar X <90) = P(Z<3)=0.99865[/tex]
d) No. it would not be unusual because more than 5% of all such samples hav means less than 90.
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".
(a) Describe the x distribution and compute the mean and standard deviation of the distribution.
Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=75,\sigma=45)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(75,\frac{45}{\sqrt{81}})[/tex]
(b) Find the z value corresponding to [tex]\bar X = 90[/tex].
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace we got:
[tex]z=\frac{90-75}{\frac{45}{\sqrt{81}}}=3[/tex]
(c) Find [tex]P(\bar X < 90)[/tex].
For this case we can use a table or excel to find the probability required:
[tex]P(\bar X <90) = P(Z<3)=0.99865[/tex]
(d) Would it be unusual for a random sample of size 81 from the x distribution to have a sample mean less than 90? Explain.
For this case the best conclusion is:
No. it would not be unusual because more than 5% of all such samples hav means less than 90.
