The amounts of electricity bills for all households in a cit

The amounts of electricity bills for all households in a city have a skewed probability distribution with a mean of $138 and a standard deviation of $30. Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be between $132 and $136.

Solution

Normal Distribution
Mean ( u ) =138
Standard Deviation ( sd )=30
Number ( n ) = 75
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 132) = (132-138)/30/ Sqrt ( 75 )
= -6/3.4641
= -1.7321
= P ( Z <-1.7321) From Standard Normal Table
= 0.04163
P(X < 136) = (136-138)/30/ Sqrt ( 75 )
= -2/3.4641 = -0.5774
= P ( Z <-0.5774) From Standard Normal Table
= 0.28185
P(132 < X < 136) = 0.28185-0.04163 = 0.2402