Given a normally distributed population with a mean of 418.26 and a standard deviation of 31.83. What is the probability of the average of 3 randomly selected items being greater than 331.2?
Solution :
Given that ,
mean = = 418.26
standard deviation = = 31.83
_{} = / n = 31.83 / 3 = 18.3771
P( > 331.2) = 1 - P( < 331.2)
= 1 - P[( - _{} ) / _{} < (331.2 - 418.26) / 18.3771]
= 1 - P(z < -4.74)
= 1 - 0
= 1
probability = 1
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