Question

On Yelp the average review of a restaurant is 3 stars with standard deviation 1.5. If a random sample of 36 review are selected, what is the probability that the sample average, , is between 2 and 3.5?

Answer 1

Solution :

Given that ,

mean =   = 3

standard deviation = = 1.5

n = 36

= 3

=  / n= 1.5/ 36=0.25

P(2<     < 3.5) = P[(2-3) /0.25 < ( - ) /   < (3.5-3) /0.25 )]

= P(-4 < Z <2 )

= P(Z <2 ) - P(Z < -4)

Using z table

=0.9772-0

probability= 0.9772