The mean weight of an adult is 67 kilograms with a variance of 81.
If 57 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 1.5 kilograms? Round your answer to four decimal places.
Given that, mean (μ) = 67 kilograms
variance = 81
=> standard deviation = √(81) = 9
sample size (n) = 57
We want to find,
Therefore, required probability is 0.2076
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