Answer:
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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the population mean and standard deviation are given below.
find the required probability and determine
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