Z(.674) = 0.25
So, (52-μ)/5 = .674
μ = 52-5*0.674 = 48.63
The length of Paulo’s lunch break follows a normal distribution with mean μ minutes and standard
deviation 5 minutes. On one day in four, on average, his lunch break lasts for more than 52 minutes.
Find the value of μ.
1 answer
1 answer