Having used people-counting devices at the entry to shopping centre, it is known that the average...

Having used people-counting devices at the entry to shopping centre, it is known that the average number of shoppers visiting this centre during any one-hour period is 448 shoppers, with a standard deviation of 21 shoppers. What is the probability that a random sample of 49 different one-hour shopping periods will yield a sample mean between 441 and 446 shoppers?

please type it do not handwritten

Expert Solution

Solution :

Given that ,

mean = = 448

standard deviation = = 21

n = 49

= 448

= / $\sqrt$ n= 21 / $\sqrt$ 49=3

P(441< <446 ) = P[(441-448) /3 < ( - ) / < (446-448) / 3)]

= P(-2.33 < Z <-0.67 )

= P(Z < -0.67) - P(Z <-2.33 )

Using z table

=0.2514-0.0099

=0.2415

probability=0.2415

orchestra answered 2 years ago

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