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Example: Uniform Distribution
Chapter 6. Continuous
Random Variables
A continuous random variable has a
uniform distribution if its values are
spread evenly over a certain range.
Reminder:
Continuous random variable
takes infinitely many values
Those values can be associated with
measurements on a continuous scale
(without gaps or interruptions)
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Density Curve
Example: Uniform Distribution
A density curve is the graph of a
continuous probability distribution. It
must satisfy the following properties:
A continuous random variable has a
uniform distribution if its values are
spread evenly over a certain range.
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Example: voltage output of an
electric generator is between 123 V
and 125 V. The actual voltage level
may be anywhere in this range.
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1. The total area under the curve must
equal 1.
2. Every point on the curve must have a
vertical height that is 0 or greater.
(That is, the curve cannot fall below
the x-axis.)
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Using Area to Find Probability
Area and Probability
Given the uniform distribution illustrated, find
the probability that a randomly selected
voltage level is greater than 124.5 volts.
Because the total area under the
density curve is equal to 1,
there is a correspondence
between area and probability.
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Shaded area
represents
voltage levels
greater than
124.5 volts.
The area
corresponds
to probability:
P = 0.25.
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6
Standard Normal Distribution
Standard Normal Distribution
The standard normal distribution is a bellshaped probability distribution with  = 0
and  = 1. The total area under its density
curve is equal to 1.
Standard normal distribution
has the following properties:
1. Its graph is bell-shaped
2. It is symmetric about its center
3. Its mean is equal to 0
( = 0)
4. Its standard deviation is 1 ( = 1)
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Examples
Standard Normal Distribution:
Areas and Probabilities
Thermometers are supposed to give readings
of 0ºC at the freezing point of water.
Probability that the
standard normal random
variable takes values
less than z is given by
the area under the curve
from the left up to z
(blue area in the figure)
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Since these instruments are not perfect, some
of them give readings below 0ºC and others
above 0ºC.
Assume the following for the readings:
• The mean is 0ºC
• The standard deviation is 1ºC
• They are normally distributed
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Example 1
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Look at Table A-2
If one thermometer is randomly selected, find
the probability that, at the freezing point of
water, the reading is less than 1.27º.
µ=0
σ=1
P (z < 1.27) = ???
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Using StatCrunch
Using StatCrunch
1.27
P(z<1.27): Enter ‘1.27’ in left box with ‘<=‘
then hit Compute
Select Stat, Calculators, Normal
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Using StatCrunch
Example 1
P (z < 1.27) = 0.8980
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P(z<1.27) = 0.8979577
(In the right box)
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Example 1
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Example 2
If thermometers have an average (mean) reading of
0 degrees and a standard deviation of 1 degree for
freezing water, and if one thermometer is randomly
selected, find the probability that it reads (at the
freezing point of water) above –1.23 degrees.
P (z < 1.27) = 0.8980
µ=0
σ=1
P (z > –1.23) = ???
The probability of randomly selecting a
thermometer with a reading less than 1.27º
is 0.8980.
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Example 2
µ=0
Example 2
σ=1
µ=0
σ=1
Method 1
Method 2
P = P(z > –1.23)
P = 1 – P(z < 1.23)
P = 1 – 0.10934855
P = 0.89065145
P ≈ 0.8907
P = P(z > –1.23)
P = 0.89065146
P ≈ 0.8907
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Example 2
Example 3
A thermometer is randomly selected. Find the
probability that it reads (at the freezing point of water)
between –2.00 and 1.50 degrees.
P (z > 1.23) = 0.8907
σ=1
µ=0
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P ( –2.00 < z < 1.50) = ???
Thus, there is a 89.80% change the picked
thermometer will read above -1.23
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Notation
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P(a < z < b) = P(z < b) – P(z < a)
P(z < a)
denotes the probability that the z score is less than a.
P(z > a)
P(a < z < b)
P(z < b)
P(z < a)
denotes the probability that the z score is greater than a.
Example:
P(a < z < b)
denotes the probability that the z score is between a and b.
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P(-1.5 < z < 0.7) = P(z < 0.7) – P(z < -1.5)
P(-1.5 < z < 0.7) = 0.7580364 – 0.0668072
P(-1.5 < z < 0.7) = 0.6912
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Finding z Scores
When Given Probabilities
Finding z Scores
When Given
Probabilities
5% or 0.05
(z score will be positive)
1.645
Finding the z-score separating 95% bottom
values from 5% top values.
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Finding z Scores
When Given Probabilities
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Finding z Scores
When Given Probabilities - cont
Enter the probability
on the right box then
hit Compute
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The z-score will be in
the left box.
(One z score will be negative and the other positive)
Example:
For top 5% (i.e. 0.05)
z-score= 1.6449
Finding the Bottom 2.5% and Upper 2.5%
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Normal distributions
that are not standard
Notation
We use za to represent the z-score
separating the top a from the bottom 1-a.
All normal distributions have bell-shaped
density curves.
Examples: z0.025 = 1.96, z0.05 = 1.645
A normal distribution is standard if its mean
 is 0 and its standard deviation  is 1.
A normal distribution is not standard if its
mean  is not 0, or its standard deviation 
is not 1, or both.
Area = a
We can use a simple conversion that allows
us to standardize any normal distribution so
that Table A-2 can be used.
Area = 1-a
za
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Converting to a Standard
Normal Distribution
Conversion Formula
Let x be a score for a normal distribution
with mean  and standard deviation 
We convert it to a z score by this formula:
z=
z=
x–

x–µ

(round z scores to 2 decimal places)
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Example – Weights of Passengers
Finding x Scores
When Given Probabilities
Weights of taxi passengers have a normal
distribution with mean 172 lb and standard deviation
29 lb. If one passenger is randomly selected, what is
the probability he/she weighs less than 174 pounds?
P(x < 174) = ???
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µ = 172 and σ = 29
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x − µ 174 – 172
z= σ =
= 0.069
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P(z < 0.069) = 0.5279
1. Use StatCrunch to find the z score
corresponding to the given probability
(the area to the left).
2. Use the values for µ, , and the z score found
in step 1, to find x:
x = µ + (z • )
P(x < 174) = 0.5279
(If z is located to the left of the mean, be sure
that it is a negative number.)
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Example – Lightest and Heaviest
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Central Limit Theorem
The weights of taxi passengers have a normal
distribution with mean 172 lb and standard deviation 29
lb. Determine what weight separates the lightest 99.5%
from the heaviest 0.5%?
The Central Limit Theorem tells us that the
distribution of the sample mean x for a
sample of size n approaches a normal
distribution, as the sample size n increases.
µ = 172 and σ = 29
P(x < ???) = 0.995
P(z < 0.995) = 2.575
x = µ + σ*z
= 172 + 2.575*29
= 246.7
Separating weight: 247 ib
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Central Limit Theorem – cont.
Central Limit Theorem
Conclusions:
Given:
1. The distribution of the sample mean x will,
as the sample size increases, approach a
normal distribution.
1. The random variable x has a distribution
(which may or may not be normal) with mean
µ and standard deviation .
2. The mean of that normal distribution is the
same as the population mean µ.
2. A random sample of size n is selected from
the population.
3. The standard deviation of that normal
distribution is  n. (So it is smaller than
the standard deviation of the population.)
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Practical Rules:
Formulas
1. For samples of size n larger than 30, the
distribution of the sample mean can be
approximated by a normal distribution.
the mean
µx = µ
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the standard deviation
x = n
2. If the original population is normally
distributed, then for any sample size n, the
sample means will be normally distributed.
3. We can apply Central Limit Theorem if either
n>30 or the original population is normal.
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Example: Water Taxi Passengers
Example – cont
Assume the population of taxi passengers is
normally distributed with a mean of 172 lb and
a standard deviation of 29 lb
a) Find the probability that if an individual man
is randomly selected, his weight is greater
than 175 lb.
a) Find the probability that
if an individual
passenger is randomly
selected, his weight is
greater than 175 lb.
µ = 172
σ = 29
P(x > 175) = ???
b) Find the probability that
20 randomly selected
passengers will have a
mean weight that is
greater than 175 lb.
n = 20
µx = µ
σx = σ/ 𝒏
P(x > 175) = ???
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µ = 172
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σ = 29
x − µ 𝟏𝟕𝟓 −𝟏𝟕𝟐
Z = σ = 𝟐𝟗 = 0.10
P(z > 0.10) = 0.4602
P(x > 175) = 0.4602
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Example – cont
Example - cont
b) Find the probability that 20 randomly
selected men will have a mean weight that is
greater than 175 lb (so that their total weight
exceeds the safe capacity of 3500 pounds).
µ = 172 σ = 29 n = 20
Z=
a) Find the probability that if an individual passenger is
randomly selected, his weight is greater than 175 lb.
P(x > 175) = 0.4602
x − µx x −µ 𝟏𝟕𝟓 −𝟏𝟕𝟐
= σ =
= 0.4626
𝟐𝟗
σx
𝒏
b) Find the probability that 20 randomly selected
passengers will have a mean weight that is greater
than 175 lb.
𝟐𝟎
P(z > 0.4626) = 0.3228
P(x > 175) = 0.3228
It is much easier for an individual to deviate from
the mean than it is for a group of 20 to deviate
from the mean.
P(x > 175) = 0.3228
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Review
Approximation of a Binomial Distribution
with a Normal Distribution
Binomial Probability Distribution
If np  5 and nq  5
1. The procedure must have a fixed number of trials, n.
2. The trials must be independent.
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3. Each trial must have all outcomes classified into two
categories (commonly, success and failure).
4. The probability of success p remains the same in all trials
(the probability of failure is q=1-p)
Then µ = np and  =
npq
and the random variable has
Solve by binomial probability formula or calculator
a
distribution.
(normal)
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Procedure for Using a Normal Distribution
to Approximate a Binomial Distribution
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Continuity Correction
When we use the normal distribution (which is
a continuous probability distribution) as an
approximation to the binomial distribution
(which is discrete), a continuity correction is
made to a whole number x in the binomial
distribution by representing the number x by
the interval from
x – 0.5 to x + 0.5
(that is, adding and subtracting 0.5).
1. Verify that both np  5 and nq  5. If not, you cannot
use normal approximation to binomial.
2. Find the values of the parameters µ and  by
calculating µ = np and  = npq.
3. Identify the discrete whole number x that is relevant to
the binomial probability problem. Use the continuity
correction (see next).
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at least 8
(includes 8 and above)
Example:
Finding the Probability of
“At Least 122 Men” Among 213 Passengers
more than 8
(doesn’t include 8)
at most 8
(includes 8 and below)
fewer than 8
(doesn’t include 8)
The value 122 is represented by the interval (121.5,122.5)
exactly 8
The values “at least 122 men” are represented by
the interval starting at 121.5.
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