The mean weight of an adult is 70 kilograms with a standard deviation of 8 kilograms. If 87 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 1.2 kilograms? Round your answer to four decimal places.
Population mean, = 70 kg
Standard deviation, = 8 kg
Sample size, n = 87
For sampling distribution of mean, P( < A) = P(Z < (A - )/)
= = 70 kg
=
=
= 0.8577 kg
P(sample mean would differ from the population mean by more than 1.2 kilograms) = P(Z < -1.2/) + P(Z > 1.2/)
= P(Z < -1.2/0.8577) + P(Z > 1.2/0.8577)
= P(Z < -1.40) + 1 - P(Z < 1.40)
= 0.0808 + 1 - 0.9192
= 0.1616
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