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Given a normally distributed population with a mean of 418.26 and a standard deviation of 31.83....

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?

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Answer #1

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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