SOLUTION: In one region, the September energy consumption levels for single family homes are found to be normally distributed with a mean of 2995 kwh and a standard deviation of 1050 kwh. a

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Question 44281This question is from textbook Elementary Statistics
: In one region, the September energy consumption levels for single family homes are found to be normally distributed with a mean of 2995 kwh and a standard deviation of 1050 kwh.
a. for a randomly selected home, find the probability that the September energy consumption level is below 200 kwh.
b. for a randomly selected home, find the probability that the September energy consumption level is between 2100 kwh and 2200 kwh
c. for a randomly selected home, find the probability taht the September energy consumption level is above 1900 kwh.
d. find the energy consumption level that separates the bottom 15% from the top 85% This question is from textbook Elementary Statistics

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
In one region, the September energy consumption 
levels for single family homes are found to be 
normally distributed with a mean of 2995 kwh and 
a standard deviation of 1050 kwh.

a. for a randomly selected home, find the 
probability that the September energy consumption 
level is below 200 kwh.

Note: If you are supposed to use z-scores
and tables instead of the TI-83 or better 
calculator, then post again, as I'm only going
to tell you how to do it with the calculator.

On your TI-83 or better calculator:

Press 2nd VARS  [to get DIST menu]

Press 2         [to get normalcdf(  ]

After " normalcdf(  " type this:

    -10^99, 200, 2995, 1050)

so that you see this on the main
screen

normalcdf(-10^99,200,2995,1050)

Then press ENTER and you read

.0038850368

===================================== 

b. for a randomly selected home, find the 
probability that the September energy 
consumption level is between 2100 kwh
and 2200 kwh

On your TI-83 or better calculator:

Press 2nd VARS  [to get DIST menu]

Press 2         [to get normalcdf(  ]

After " normalcdf(  " type this:

     2100,2200,2995,1050)

so that you see this on the main
screen

normalcdf(2100,2200,2995,1050)

Then press ENTER and you read

.0274807743

==========================================

c.  for a randomly selected home, find the 
probability that the September energy 
consumption level is above 1900 kwh.

On your TI-83 or better calculator:

Press 2nd VARS  [to get DIST menu]

Press 2         [to get normalcdf(  ]

After " normalcdf(  " type this:

1900,10^99,2995,1050)

so that you see this on the main
screen

normalcdf(1900,10^99,2995,1050)

Then press ENTER and you read

.8514927476

====================================

d. find the energy consumption level 
that separates the bottom 15% from the
top 85%

On your TI-83 or better calculator:

Press 2nd VARS  [to get DIST menu]

Press 3         [to get invNorm(  ]

After " invNorm(  " type this:

.15,2995,1050)

so that you see this on the main
screen

invNorm(.15,2995,1050)

Then press ENTER and you read

1906.744951

=============================

Note: to use

normalcdf( 

enter it this way:

normalcdf(LOWER BOUND. UPPER BOUND, MEAN, STANDARD DEVIATION)

If there is no lower bound, that is, if it is 
all the way to the left, use -10^99 as the lower
bound.

If there is no upper bound, that is, if it is 
all the way to the right, use 10^99 as the upper 
bound.

Note: to use

invNorm( 

enter it this way:

invNorm(LOWER PERCENTAGE EXPRESSED AS A DECIMAL, MEAN, STANDARD DEVIATION)


Edwin