math
posted by sanjay on .
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.

Z(.674) = 0.25
So, (52μ)/5 = .674
μ = 525*0.674 = 48.63