In one region, the September energy consumption levels for single-family homes a

Question

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. a. For a randomly selected home, find the probability that the September energy consumption level is less than 1000 kWh. b. For a randomly selected home, find the probability that the September energy consumption level is greater than 1178 kWh. c. For a randomly selected home, find the probability that the September energy consumption level is between 900 kWh and 1200 kWh. d. For a simple random sample of 100 homes, find the probability that the mean September energy consumption level is less than 1000 kWh.

Explanation / Answer

Here we wre given that,mean=1050 and standard deviation=218

a.

we have to find P(X<1000)=P(Z<-0.229) =0.4093

Using Excel function NORMSDIST(-0.229) we will get probability 0.4093

b.

we have to find P(X>1178)=P(Z>0.587) =1-P(Z<0.587)=1-0.7215=0.2786

Using Excel function NORMSDIST(0.587) we will get probability .7215

c.

P(900<X<1200)=P(X<1200)-P(Z<900)

=P(Z< 0.755) -P(Z<-0.688)

=0.7543-0.2452

=0.5085

P(Xbar<1000)=P(Z<1000-1050/210/sqrt(100))

=P(Z<-2.293)=0.01091

Hope this will helpful. Thanks and God Bless You

Thumbs up if it helps you :-)

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