Question

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 1

Answer:

a) $P(\bar X>81.5)=1-0.933=0.067$

b) $P(\bar X$

c) $P(73.4$  

Step-by-step explanation:

1) 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".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=77,\sigma=27)$  

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case  $\bar X \sim N(77,\frac{27}{\sqrt{81}})$

Part a

We want this probability:

$P(\bar X>81.5)=1-P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >81.5)=1-P(Z$

$P(\bar X>81.5)=1-0.933=0.067$

Part b

We want this probability:

$P(\bar X\leq 69.5)$

If we apply the formula for the z score to our probability we got this:

$P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z$

$P(\bar X\leq 69.5)=0.0062$

Part c

We are interested on this probability

$P(73.4$  

If we apply the Z score formula to our probability we got this:

$P(73.4$

$=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}$

And we can find this probability on this way:

$P(-1.2$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1.2$