SOLUTION: A sample of 110 adults has a mean body temperature of 98.18. a. Assuming that the population of healthy adults has a mean body temperature of mu = 98.6 with a standard deviatio

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Question 1176007: A sample of 110 adults has a mean body temperature of 98.18.
a. Assuming that the population of healthy adults has a mean body temperature of mu = 98.6 with a standard deviation of sigma = 0.58, approximate the probability that a sample of 110 has a mean less than 98.18. (P(xBar sub 110) <= 98.18)
b. What if anything do the results say about the belief that the average body temperature is 98.6?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Normal Distribution:    μ = 98.6  and    σ = .58
n = 118  x̄ = 98.16
 z = (98.16 - 98.6)/.58/√110 = 1.8083 |  z+=blue+%28x+-+mu%29%2Fblue%28sigma%2Fsqrt%28n%29%29
P(x̄ < 98.16) = P(x̄ ≤ 98.16) = P(z ≤ 1.8083) = .9647

b. results lend support the claim that 98.6 is the mean temperature.
 
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