A random sample of n=100 observations is selected from a population with µ=31 and o=21. Approximate the probabilities shown below.
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The standard deviation has to be adjusted for the given sample size so as to better apply to the sample s=21/√100=2.1. μ stands for the population mean=31, while x̄ is the sample mean. The Z-score shows by how much the sample and population means differ in terms of the standard deviation, Z=(x̄-μ)/s is the test statistic to use in an assumed normal distribution.

a) x̄≥28: Z=(28-31)/2.1=-3/2.1=-1.429. The corresponding probability is p=0.0766. This is the cumulative probability for less than 28, but we need it for greater than 28, so p=1-0.0766=0.9234, or about 92.3%, making it very likely that the sample mean is at least 28.

b) 22.1≤x̄≤26.8: We need two Z-scores: Zhigh=(26.8-31)/2.1=-2; Zlow=(22.1-31)/2.1=-4.238. These correspond to 0.02275 and 0.00001, so the difference is 0.02274. The probability of the mean lying between these two values is therefore about 2.3%.

c) x̄≤28.2: Z=(28.2-31)/2.1=-4/3; p=0.0912, about 9.1%.

d) x̄≥27: Z=(27-31)/2.1=-1.905; p=1-0.0284=0.9716, about 9.7%. See (a) for similar example.

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