At the end of section 8.3 it was noted that the samples of serum cholesterol lev

Question

At the end of section 8.3 it was noted that the samples of serum cholesterol levels of size 2.5- drawn from a population with mean =211 mg/100 ml and standard deviation = 46 mg/100 ml- the probability that a sample mean X lies within the interval (193.0, 229.0) is .95. Furthermore, the probability that the mean lies below 22.6 mg/100ml is .95, and the probability that is it above 195.9 mg/100ml is .95. For all three of these events to happen simultaneously, the sample mean X would have to lie in the interval (195.9, and 226.1) What is the probability that this occurs?

Explanation / Answer

we calculate z=(x-mean)/sd

for 195.9 z-value z1=(195.9-211)/46=-.83889

for 226.1 z-value z2=(226.1-211)/46=0.328261

p(z1<z<z2)=0.628643-0.200766=0.427877

required probability of P(195.9,226.1)=0.427877

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