Question

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.

Answer #1

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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Let x be a random variable that represents the weights in
kilograms (kg) of healthy adult female deer (does) in December in a
national park. Then x has a distribution that is approximately
normal with mean μ = 70.0 kg and standard deviation σ = 8.1 kg.
Suppose a doe that weighs less than 61 kg is considered
undernourished.
(a) What is the probability that a single doe captured (weighed
and released) at random in December is undernourished? (Round your...

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