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MATH Resources for Academic Success
Continuous Probability Distributions
Before we begin, you need to understand the difference between a discrete variable and a continuous
variable. Recall, if a variable can take on any value between two specified
values, it is continuous (like weight, height, time.) Otherwise it is discrete
(like the possible numbers of dots on a die.)
Just as variables can be either continuous or discrete, so can distributions. A
distribution that is discrete has a finite number of possibilities so we can
calculate the probability of each one and put them in a table (which we
have done before now.) However, a distribution that is continuous has a sample space with infinite
possibilities so we can’t put them in a table AND the probability of any one particular value will be zero—not
very useful. Therefore, for continuous distributions we will look at the probability of a value being in an
interval BETWEEN, say, a and b.
Temperature is a continuous variable since the temperature is not limited to just being 73 degrees, 74
degrees, 75 degrees, etc. It can be anywhere in between, say 73.84604 degrees. Therefore, it makes no sense
to find the probability of a temperature EXACTLY EQUAL to 74 degrees. Instead, we will talk about the
probability that our variable is BETWEEN two values. For example, we can look at the probability of the
temperature being between 73.5 and 74.5 degrees.
To compute these probabilities, we look at a probability density function (PDF). Since the possibilities are
infinite, we cannot model these probabilities with a table. Instead, we use an equation or a graph. In the
graph models, the probability that a random variable assumes a value between a and b is equal to the area
under the density function bounded by a and b. Keep in mind that similar to a discrete probability
distribution, a PDF must satisfy two things:
1. The total area under the graph is 1 (100%). This
is the same as the sum of all probabilities being 1
for a discrete variable.
2. The graph is nonnegative everywhere it's
defined. This is the same as each probability of a
discrete variable having to be nonnegative.
If the distribution is a bell-shaped curve (like the one in the
picture), we call it normal. Since we need to figure out the area under the curve for our interval, what we'd
like to be able to do is extend the concept of the Empirical Rule and z-scores because this would tell us how
much of the data (i.e. the proportion, which is the same as probability) lies in our given interval.
For example we already know from the Empirical Rule that for a normal distribution, 68% of the data is
between μ - σ and μ + σ. But how much of the data is between (say) -3.1 and 1.4 standard deviations away
from the mean? In turns out we can compute this using the normalcdf() function on the TI-84. The idea,
remember, is area = proportion =
To find the probability that the data is
To find the z-score for a
probability. If we can find areas (with
within the interval:
given probability:
normcdf() for example) we can use that
2nd vars2
2nd vars3
normcdf(lower bound, upper bound, mean, st dev) invnorm(area, mean, st dev.)
to talk about proportion and probability.
Developed by Glenn Bryce for NFCC, ed. 2015
MATH Resources for Academic Success
Example) The waiting times in Wendy's drive-thrus are normally distributed with mean 138.5 second and
standard deviation 29 seconds. I highly suggest you draw a picture of this distribution with mean and three
std devs up and down labeled. I have done that below but you should be able to do it as well.
a. What's the probability that a randomly selected
driver will wait less than 100 seconds?
From our sketch we can see we are looking for the
area under the bit of the curve that is to the left of
100 seconds: the interval from –∞ to 100 given a
mean of 138.5 and sd of 29.
We can use a z-score table or on the TI84 normcdf(-1E99, 100, 138.5, 29) = 0.0922. So the probability
of a random driver waiting less than 100 seconds is 0.0922 (about 9%.) This is also the proportion of all
drivers who wait less than 100 seconds, so for example in a sample of n=1000, you could expect about
(0.0922)*1000 = 922 drivers to wait less than 100 seconds.
b. What proportion of cars spend between 2 and 3 minutes in the drive-thru?
The proportion of cars that wait between 2 and 3 minutes (that is, between 120 and 180 seconds) is
the same as the area under the normal curve between 120 and 180. We could estimate this area since
these times are about 1 standard deviation which is about 68% so we shouldn’t be surprised that
normcdf(120, 180, 138.5, 29) = 0.662. That means that a proportion of 0.622 cars wait between 2 and
3 minutes; another way to say that is that 62.2% of all cars wait between 2 and 3 minutes in the drivethru. (0.622 is also the probability that a randomly selected car waits between 2 and 3 minutes.)
c. Find the time for which 95% of drivers wait less than that time.
This is the 95th percentile, or in other words, the area to the left of this score is 0.95. So, this is an
inverse problem where we know the area but NOT the time. We can use a z-score table or on the TI84
invNorm(0.95, 138.5, 29) = 186.2. Our conclusion is that 95% of drivers wait less than 186.2 seconds.
RESOURCES
Continuous, Discrete, and PDFs:
http://www.khanacademy.org/math/statistics/v/probability-density-functions
http://stattrek.com/probability-distributions/discrete-continuous.aspx
Mr. Bryce’s Calculator Demonstrations:
http://www.screencast.com/t/UhYMxxBO1d9a
http://www.screencast.com/t/vEH4aGbG3FHO
Normally Distributed? You decide!: https://www.khanacademy.org/math/probability/statisticsinferential/normal_distribution/v/ck12-org-normal-distribution-problems-qualitative-sense-of-normaldistributions
Developed by Glenn Bryce for NFCC, ed. 2015