Question

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P (x overbar greater than 81.2 )​? ​(c) What is Upper P (x overbar less than or equals 74.4 )​? ​(d) What is Upper P (77.6 less than x overbar less than 83.2 )​?

Answer 1

Answer:

(a) The sampling distribution of$\overline{X}$ = Population mean = 79

(b)  P ( $\overline{X}$ greater than 81.2 ) =  0.1357

(c) P ($\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

​(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of$\overline{X}$ = Population mean = 79

​(b) What is Upper P ( $\overline{X}$ greater than 81.2 )​ =

$P(\overline{X}> 81.2)$  = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

                    =  $P(Z> 1.1)$

                    = $1 - P(Z< 1.1)$

                    = 1 - .8643 =

                    = 0.1357

(c) What is Upper P ($\overline{X}$ less than or equals 74.4 ) =

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

                    = $P(Z\leq -2.3)$

                    = .0107

​(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

                                = $P(- 0.7< Z< 2.1)$

                                 = $(Z< 2.1) - (Z< -0.7)$

                                  = 0.9821 - .2420

                                   = 0.7401