



 
Hi Lorie, Let X be the amount of annual snowfall in inches then X is distributed as the normal random variable with μ = 109 and σ = 10. Let Xbar be the mean annual snowfall of n = 40 years then Xbar is distributed as the normal random variable with mean μ_{Xbar} = μ = 109 and variance σ^{2}_{Xbar} = σ^{2}/n = 100/40. You want to find Pr(Xbar > 111.8). You first need to convert Xbar to the standard normal distribution by Z = (Xbar  μ)/(σ/√n) and then evaluate Pr[(Xbar  μ)/(σ/√n) > (118  μ)/(σ/√n)] = Pr(Z > (111.8  109)/(10/√40)]. Finally evaluate (111.8  109)/(10/√40) and use the standard normal table to evaluate the probability. Harley  


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