A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. ​(a) Describe the sampling distribution of x overbarx. ​(b) What is Upper P (x overbar greater than 81.5 )P x>81.5​? ​(c) What is Upper P (x overbar less than or equals 69.5 )P x≤69.5​? ​(d) What is Upper P (73.4 less than x overbar less than 84.05 )P 73.4

Answer:a) [tex]P(\bar X>81.5)=1-0.933=0.067[/tex]b) [tex]P(\bar X<69.5)=0.0062[/tex]c) [tex]P(73.4<\bar X<84.05)=0.8755[/tex]  Step-by-step explanation:1) Previous conceptsNormal 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".  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=77,\sigma=27)[/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(77,\frac{27}{\sqrt{81}})[/tex]Part aWe want this probability:[tex]P(\bar X>81.5)=1-P(\bar X<81.5)[/tex]The best way to solve this problem is using the normal standard distribution and the z score given by:[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]If we apply this formula to our probability we got this:[tex]P(\bar X >81.5)=1-P(Z<\frac{81.5-77}{\frac{27}{\sqrt{81}}})=1-P(Z<1.5)[/tex][tex]P(\bar X>81.5)=1-0.933=0.067[/tex]Part bWe want this probability:[tex]P(\bar X\leq 69.5)[/tex]If we apply the formula for the z score to our probability we got this:[tex]P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z<-2.5)[/tex][tex]P(\bar X\leq 69.5)=0.0062[/tex]Part cWe are interested on this probability[tex]P(73.4<\bar X<84.05)[/tex]  If we apply the Z score formula to our probability we got this:[tex]P(73.4<\bar X<84.05)=P(\frac{73.4-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{84.05-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex][tex]=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}<Z<\frac{84.05-77}{\frac{27}{\sqrt{81}}})=P(-1.2<z<2.35)[/tex]And we can find this probability on this way:[tex]P(-1.2<z<2.35)=P(z<2.35)-P(z<-1.2)[/tex]And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  [tex]P(-1.2<z<2.35)=P(z<2.35)-P(z<-1.2)=0.9906-0.1151=0.8755[/tex]