Question

A bookstore chain claims that the annual gross revenue for all of their stores has a mean revenue of 350000 with a standard deviation of 82000, A random sample of 121 stores is selected.

A. Find the mean and the standard deviation for this sampling distribution

B. Find the probability that the mean annual gross revenue is between 350,000 and 360,000

Answer 1

Solution :

Given that ,

mean =   = 350000

standard deviation = =82000     

n = 121

mean of sampling distribution = = 350000

standard deviationof sampling distribution = =  / n= 82000 / 121=7454.5455

P(350000<     <360000 ) = P[(350000-350000) / 7454.5455< ( - ) /   < (360000-350000) /7454.5455 )]

= P( 0< Z <1.34)

= P(Z < 1.34) - P(Z <0 )

Using z table

=0.9099-0.5

=0.4099

probability= 0.4099