Question

Maryland is trying to build new off-shore windmills to generate power. Estimates of their production assume that the wind in that area will be approximately Normally distributed with a mean speed of 55 knots and a standard deviation of 17 knots. Assuming these are the correct population mean and population standard deviation, If we take 20 measurements, what is the probability that the mean windspeed during those measurements is between 50 and 58 knots?

Answer 1

X be the speed of the wind in the area       
X follows normal distribution mean μ and standard deviation σ       
Given μ = 55  σ = 17    
n = 20 Sample Size      
By the Central Limit Theorem we know that       
X̅ follows normal distribution mean μ and standard deviation σ'       
where μ = 55  σ' = σ/√n = 17/√20 = 3.8013    
        
To find P(50 < X̅ < 58)       
P(50 < X̅ < 58) = P(X̅ < 58) - P(X̅ < 50)       
We find these probability using Excel function NORM.DIST       
P(50 < X̅ < 58) = NORM.DIST(58, 55, 3.8013, TRUE)        
   - NORM.DIST(50, 55, 3.8013, TRUE)     
                = 0.7850 - 0.0942       
                = 0.6908       
Probability that the mean windspeed is between 50 and 58 knots = 0.6908