Respuesta :
Answer:
a) [tex]\bar X \sim N(\mu=1750, \frac{\sigma}{\sqrt{n}}=\frac{250}{\sqrt{15}}=64.550)[/tex]
b) [tex]P(\bar X >1800)=P(Z>\frac{1800-1750}{\frac{250}{\sqrt{15}}}=0.775)[/tex]
Using the complement rule and the normal standard table or excel we have:
[tex]P(Z>0.775)=1-P(Z<0.775)= 1-0.781=0.219[/tex]
c) [tex]P(1650<\bar X <1850)=P(\frac{1650-1750}{\frac{250}{\sqrt{15}}}<Z<\frac{1850-1750}{\frac{250}{\sqrt{15}}})= P(-1.549<Z<1.549)[/tex]
We can find this probability with the following operation:
[tex] P(-1.549< Z< 1.549) = P(Z<1.549)-P(Z<-1.549) = 0.939-0.061= 0.879[/tex]
d) [tex]1750-1.96\frac{250}{\sqrt{15}}=1623.48[/tex]
[tex]1750+1.96\frac{250}{\sqrt{15}}=1876.517[/tex]
So on this case the 95% confidence interval would be given by (1623.48;1876.517)
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 volume of overnight bags of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1750,250)[/tex]
Where [tex]\mu=1750[/tex] and [tex]\sigma=250[/tex]
Since the distribution for X is normal the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu=1750, \frac{\sigma}{\sqrt{n}}=\frac{250}{\sqrt{15}}=64.550)[/tex]
Part b
For this case we want this probability:
[tex] P(\bar X >1800)[/tex]
We can use the z score given by:
[tex] z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Using this formula we got:
[tex]P(\bar X >1800)=P(Z>\frac{1800-1750}{\frac{250}{\sqrt{15}}}=0.775)[/tex]
Using the complement rule and the normal standard table or excel we have:
[tex]P(Z>0.775)=1-P(Z<0.775)= 1-0.781=0.219[/tex]
Part c
For this case we want this probability:
[tex] P(1650<\bar X <1850)[/tex]
Using the z score formula we got:
[tex]P(1650<\bar X <1850)=P(\frac{1650-1750}{\frac{250}{\sqrt{15}}}<Z<\frac{1850-1750}{\frac{250}{\sqrt{15}}})= P(-1.549<Z<1.549)[/tex]
We can find this probability with the following operation:
[tex] P(-1.549< Z< 1.549) = P(Z<1.549)-P(Z<-1.549) = 0.939-0.061= 0.879[/tex]
Part d
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]1750-1.96\frac{250}{\sqrt{15}}=1623.48[/tex]
[tex]1750+1.96\frac{250}{\sqrt{15}}=1876.517[/tex]
So on this case the 95% confidence interval would be given by (1623.48;1876.517)
