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

Here, μ = 5000, σ = 19.3649 and x = 4950. We need to compute P(X >= 4950). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (4950 - 5000)/19.3649 = -2.58

Therefore,
P(X >= 4950) = P(z <= (4950 - 5000)/19.3649)
= P(z >= -2.58)
= 1 - 0.0049 = 0.9951