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.
Solution :
=
/
n = 150 /
60
P(
> 4950) = 1 - P(
< 4950)
= 1 - P[(
-
) /
< (4950 - 5000) / 150 /
60 ]
= 1 - P(z < -2.582)
= 0.9951
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