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The mean and standard deviation are given below. Find the required probability and determine whether the given sample mean wo
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Population mean μ = 21 and a standard deviation σ = 1.22

We want to find the probability of a sample mean being less than 21.2

P(xbar < 21.2) = P((xbar - µ)/ (σ/√n)) < (21.2– 21)/ (1.22/sqrt(64))

             = P( Z < 1.31)

               = 0.9049 (Using Statistical table)

The probability is 0.9049

The given sample mean being less than 21.2 would be considered unusual because it has less than 90% probability.

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