Download Normal Random Variables

Normal Random Variables
Let the random variable X represent the weight of any randomly selected pumpkin in our patch, then X is a
continuous random variable, since it represents an amount to be measured (not counted). Suppose that further
analysis indicated that the weight of our pumpkins has a normal distribution, with a mean weight of 15 pounds
and a standard deviation of 2 lb. Now our variable is a Normal Random Variable. Draw the probability density
curve here:
The probability that the weight of any randomly selected pumpkin would fall in any interval is equal to the area
under the curve above that interval. If everyone knew some calculus we could find the area that way, but since
most of us don’t, we will turn to our calculator.
Find the probability that X falls in an interval (a,b): Press 2nd and VARS (to choose DISTR). Then select
2:normalcdf(. Once “normalcdf(“ is pasted on your home screen you need to enter the information as
follows: normalcdf (min interval value, max interval value, mean, standard deviation) then press ENTER to
see your results.
Example: Find the probability that a randomly selected pumpkin has a weight between 13.5 and 15.5 pounds.
Enter: normalcdf(13.5, 15.5, 15, 2) = .372
Example: Find the percentage of pumpkins that have a weight between 14 and 16.5 pounds.
Enter: normalcdf(14, 16.5, 15, 2) = .465 = 46.5%
Practice: Find the probability that a randomly selected pumpkin has a weight between 15.8 and 17.9 pounds.
Enter:
(Answer should be .271)
Practice: Find the percentage of pumpkins that have a weight between 15 and 19.2 pounds.
Enter:
(Answer should be .482 = 48.2%)
Open-Ended Intervals: Sometimes one end of your interval will be open-ended. In this situation you will need
to make-up your own endpoint for the interval. To be sure you have included enough area under the curve, my
rule is that you must choose a number at least four standard deviations from the mean. In this case, since the
mean is 15 and the standard deviation is 2, four standard deviations is 4(2) = 8. So your lower limit must be
smaller than 15 – 8 or 7, and your upper limit must be larger than 15 + 8 or 23. But remember, you can never
pick an endpoint too far from center, only too close, so why not just pick one FAR FAR away.
Example: Find the probability that a randomly selected pumpkin has a weight greater than 12 pounds. Since
there is no upper endpoint I will choose 100 (anything bigger than 23 would have been fine, but no sense
cutting it close)
Enter: normalcdf(12, 100, 15, 2) = .933
Example: What percentage of pumpkins have a weight less than 14.5 pounds? Since there is no lower endpoint
I will choose 0 (anything less than 7 would have been fine – quite often 0 is a good choice for a lower endpoint,
but not always)
Enter: normalcdf(0, 14.5, 15, 2 ) = .401 = 40.1%
Practice: Find the probability that a randomly selected pumpkin has a weight less than 18.5 pounds.
Enter:
(Answer = .960)
Practice: What percentage of pumpkins have a weight greater than 19 pounds?
Enter:
(Answer = .0228 = 2.28%)
Going Backwards: Sometimes the question will be asked in reverse. In other words you will be given a
probability or a percentage and asked to find the range of X values. For these questions we will use the
“invnorm” function, found by pressing 2nd and DISTR, on our calculator which finds the x value that has the
area under the curve to it’s left that we input. The values are entered as follows: invnorm (area to the left, mean,
standard deviation).
Example: 74% of all pumpkins have a weight less than what?
Enter: invnorm(.74, 15, 2) =16.3 lbs
Example: There is a .1 probability that a randomly selected pumpkin will be heavier than what weight?
Enter: invnorm(.9, 15, 2) = 17.6 lbs
Practice: 32% of all our pumpkins are lighter than what?
Enter:
(Answer = 14.1 lbs)
Practice: The middle weight range for 80% of all the pumpkins will fall in what weight range?
Note that the invnorm function must be used twice here, once for each endpoint
Enter:
(Answer: between 12.4 and 17.6 lbs)