Question

According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours.

a)Find the probability of randomly selecting a household that has at least 1950 kilwatt-hours.

b)What percentage of the households in this area have a monthly electric consumption of 1970.00 to 2370.00 kilowatt-hours?

c) The Company sent a notice to bill Johnson informing him that 90 % of the households use less electricity than her does. What is Bills monthly electricity consumption?

Answer 1

Mean= 1650 KWH and standard deviation = 320 kwh

a) The probability of randomly selecting a household that has at least 1950 kilwatt-hours is 0.1736

Z= 1950-1650/320 = 300/320 = 0.9375

P(Z>0.9375)= 1-NORMDIST(Z)= 1-0.825749288= 0.1746

b) P(1970<X<2370)=

Z= 1970-1650/320= 1

Z= 2370-1650/320= 2.25

P(1<Z<2.25)= NORMDIST(2.25) - NORMDIST (1)= 0.987775527 - 0.841344746= 0.146430781

c) Z corresponding to 0.9 is 1.28

Z= X-M/SIGMA

Z*SIGMA= X-M

Z*SIGMA+M= X

1.28*320+1650=X

409.6+1650=X

2059.6 KWH =X