Answer:
The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is 0.1972
Explanation:
The energy consumption level for a single family in September is normally distributed, therefore to solve this problem, we are going to use the z score. Z score shows the relationship of a group of values to the mean measured in terms of standard deviation from the mean.
From the question, the mean(m) = 1050 kWh
Standard deviation(s) = 218 kWh
The formula for the z score(z) where x is the raw score is given as:
z = [tex]\frac{x-m}{s}[/tex]
Therefore to get the probability that the September energy consumption level is between 1100 kWh and 1225 kWh, we calculate the z score for 1100 kWh and then for 1225 kWh.
For 1100 kWh, [tex]z=\frac{x-m}{s} = \frac{1100-1050}{218} = 0.23[/tex]
For 1225 kWh, [tex]z=\frac{x-m}{s} = \frac{1225-1050}{218} =0.80[/tex]
The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is given by:
P(1100<x<1225) = P(0.23<z<0.8) = 0.78814 - 0.59095 = 0.1972
The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is 0.1972
