Question

Suppose the local Best Buy store averages 480 customers per day entering the facility with a standard deviation of 110 customers. A random sample of 48 business days was selected. What is the probability that the average number of customers in the sample is more than 500? (round the final answer to 4 decimals)

Answer 1

Solution :

Given that ,

mean = = 480

standard deviation = = 110

n = 48

= 480

= / n = 110/ 48 = 15.88

P( > 500) = 1 - P( < 500)

= 1 - P[( - ) / < (500-480) /15.88 ]

= 1 - P(z <1.26 )

Using z table

= 1 - 0.8962

= 0.1038

probability=0.1038