Mathematics: 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 )?

HOT Step-by-step explanation:...

Suppose a simple random sample of size nequals81

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23.01.2022

Mathematics

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

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

(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

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

= $P(Z\leq -2.3)$

= .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

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