Question

A sample of n = 36 observations is drawn from a normal population, with  = 500 and  = 42. Find P(500 < X̅ < 510).

	a.	0.0764
	b.	0.8756
	c.	0.9236
	d.	0.7556
	e.	0.4236

Answer 1

Solution :

Given that ,

mean =   = 500

standard deviation = = 42

n = 36

= 500

=  / n= 42/ 36=7

P(500<     <510 ) = P[(500-500) /7 < ( - ) /   < (510-500) /7 )]

= P(0 < Z <1.43 )

= P(Z < 1.43) - P(Z < 0)

Using z table

=0.9236-0.5

=0.4236