Suppose the mean blood glucose of a population of adults is 162 with a standard deviation σ=58. From sample sizes of 36 adults find the proportion of sample means having values of at least 170.
P(xˉ⩾170)=P(nxˉ−μσ⩾n170−μσ)==P(Z⩾36170−16258)=1−Φ(0.827586)==1−0.79605=0.20395P\left( \bar{x}\geqslant 170 \right) =P\left( \sqrt{n}\frac{\bar{x}-\mu}{\sigma}\geqslant \sqrt{n}\frac{170-\mu}{\sigma} \right) =\\=P\left( Z\geqslant \sqrt{36}\frac{170-162}{58} \right) =1-\varPhi \left( 0.827586 \right) =\\=1-0.79605=0.20395P(xˉ⩾170)=P(nσxˉ−μ⩾nσ170−μ)==P(Z⩾3658170−162)=1−Φ(0.827586)==1−0.79605=0.20395
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