Question

natural gas bills in a certain city have mean $77.47 assume the bills are normally distributed with standard deviation $12.96 find the value that seperates the lower 59% of the bills from the rest

Answer 1

Given that,

mean = $\mu$ = $77.47

standard deviation = $\sigma$ = $12.96

Using standard normal table,

P(Z < z) =59 %

=(Z < z) = 0.59

= P(Z < 0.23) = 0.59

z = 0.23

Using z-score formula  

x = z $\sigma$ + $\mu$

x = 0.23 *12.96+77.47

x = 80.4508