Question

The true mean of the cost of a certain type of surgeries is ? = 5000 riyals. Also, it is known that the population standard deviation of the costs is ? = 150. Find the probability that a sample of size ? = 60 surgeries yields a sample mean (?̅) more than 4950 Riyals.

Answer 1

Solution :

= / n = 150 / 60

P( > 4950) = 1 - P( < 4950)

= 1 - P[( - ) / < (4950 - 5000) / 150 / 60 ]

= 1 - P(z < -2.582)

= 0.9951