Chapter 6 – 8A:
Using a Normal Distribution to Approximate
a Binomial Probability Distribution
Graphing a Binomial Probability Distribution Histogram
Lower and Upper Class Boundaries are used to graph the rectangles that represent
the discrete x values on an x axis that is continuous
The Binomial distribution shown below has 5 discrete values for the discrete variable x. The variable
x in the table is represented by discrete whole numbers that are 1 unit apart. In chapter two we used
lower and upper class boundaries on the x axis to represent each rectangle on the x axis. The
discrete x values of 1, 2, 3 , 4 and 5 are not shown on the continuous x axis.
Class
Boundaries
x:
P(x)
.4
0.5 – 1.5
1
.1
1.5 – 2.5
2
.2
2.5 – 3.5
3
.4
3.5 – 4.5
4
.2
.3
P(x)
.2
.1
1
4.5 – 5.5
5
.1
.5
2
1.5
3
2.5
4
3.5
5
4.5
5.5
x
The rectangle that represents 1 in the table is labeled on the graph staring at .5 and ending at 1.5.
The rectangle that represents 2 in the table is labeled on the graph staring at 1.5 and ending at 2.5.
The rectangle that represents 3 in the table is labeled on the graph staring at 2.5 and ending at 3.5.
The rectangle that represents 4 in the table is labeled on the graph staring at 3.5 and ending at 4.5.
The rectangle that represents 5 in the table is labeled on the graph staring at 4.5 and ending at 5.5.
Stat 300 6 – 8A Lecture
Page 1 of 8
© 2012 Eitel
A Binomial Probability Distribution with 5 values for x
x:
P(x)
1
.1
2
.2
3
.4
4
.2
5
.1
In chapter 5 we found the value of P( x) ≥ 3 in the Binomial Probability Distribution above by adding
the values P(3) + P(4) + P(5) . That process worked if the number of rectangles was very small.
What if the Binomial table had x values from 0 to 100
and we needed to find
P( x) ≥ 30
x:
P(x)
:
:
:
:
30
31
:
:
:
:
99
100
P(30)
P(31)
:
:
:
:
P(99)
P(100)
Finding P( x) ≥ 30 would require the calculation of the sum of the all the values for P(x) from P(30) to
P(100) or P(30) + P( 31) + P(32) + ......+ P(100). It would not be practical to find all the values for P(x)
and then find their total. If a Binomial Probability Distribution has a large number of values it becomes
desirable to look for a way to approximate the answer without computing each P(x) and adding all
the values for P(x).
Stat 300 6 – 8A Lecture
Page 2 of 8
© 2012 Eitel
Can we use the the area under a Normal Curve of continuous x values
to approximate the sum of the values of P(x) P(30) + P( 31) + P(32) + ......+ P(100) in the table ?
P(x) ≥ 30
P(x) ≥ 30
= P(30) + P( 31) + P(32) + .....+ P(99) + P(100)
the area under the curve to the right
for a discrete number of P(x) values
of 30 for a continuous normal curve
x:
P(x)
:
:
:
:
30
31
:
:
:
:
99
100
P(30)
P(31)
:
:
:
:
P(99)
P(100)
the area in the
right tail
xrepresents the
P(x > 30 )
x = 30
x
Can we use a Normal Distribution of continuous x values to
approximate a Binomial Probability Distribution ?
Yes we can, under some conditions
and with an adjustment to the value of x.
Stat 300 6 – 8A Lecture
Page 3 of 8
© 2012 Eitel
Continuity Corrections
There is one step that increases that accuracy of the approximation. It involves the way in
which we use the lower and upper class boundaries to label the rectangles on the x axis. To find the
value of P(x) ≥ 3 in a binomial distribution we add the values for P(3) + P(4) + P(5) . It would
seem that we would then find P(x) ≥ 3 in a normal distribution by finding the area under the curve to
the right of 3 but that is not true. We must adjust the value of 3 to reflect the class boundaries we use
when we graph the rectangles on the x axis. We call this the Correction for Continuity. It can be
a challenge to understand this correction so read the next pages carefully.
If P( x ≥ a) change to P( x ≥ a − .5)
Binomial Probability P( x ≥ 3)
Normal probability
P( x > 2.5)
P(x)
x > 2.5
µx = n • p
σ x = n • p• q
x
For the Binomial Probability P( x ≥ 3) means to start at the rectangle for x = 3 and add all the
rectangles to the right. The rectangle for x = 3 starts at 2.5 and we want to add it and the ones to
the right of it. For the normal curve we need to start at 2.5 on x axis and go to the right x > 2.5
Stat 300 6 – 8A Lecture
Page 4 of 8
© 2012 Eitel
If P( x > a) change to P( x > a + .5)
Binomial Probability P( x > 3)
Normal probability P( x > 3.5)
P(x)
x > 3.5
µx = n • p
σ x = n • p• q
x
For the Binomial Probability P( x > 3) means to start at the rectangle for x = 4 and add all the rectangles
to the right. The rectangle for x = 4 starts at 3.5 and we want to add it and the ones to the right of it.
For the normal curve we need to start at 3.5 on x axis and go to the right x > 3.5
If P( x ≤ a) change to P( x ≤ a + .5)
Binomial Probability P( x ≤ 3)
Normal probability P( x < 3.5)
P(x)
x < 3.5
µx = n • p
σ x = n • p• q
x
For the Binomial Probability P( x ≤ 3) means to start at the rectangle for x = 3 and add all the rectangles
to the left. The rectangle for x = 3 starts at 3.5 and we want to add it and the ones to the left of it.
For the normal curve we need to start at 3.5 on x axis and go to the left x < 3.5
Stat 300 6 – 8A Lecture
Page 5 of 8
© 2012 Eitel
If P( x < a) change to P( x < a − .5)
Binomial Probability P( x < 3)
Normal probability P( x < 2.5)
P(x)
x < 2.5
µx = n • p
σ x = n • p• q
x
For the Binomial Probability P( x < 3) means to start at the rectangle for x = 2 and add all the rectangles
to the left. The rectangle for x = 2 starts at 2.5 and we want to add it and the ones to the left of it. For
the normal curve we need to start at 2.5 on x axis and go to the left x < 2.5
Stat 300 6 – 8A Lecture
Page 6 of 8
© 2012 Eitel
Continuity Corrections
Example 1 P( x > 40 )
Example 2 P( x > 40 )
P( x ≥ a) is adjusted to P( x ≥ a − .5)
P( x > a) is adjusted to P( x ≤ a + .5)
P( x ≥ 40) is adjusted to P(x > 39.5)
P( x > 40) is adjusted to P(x > 40.5)
x > 40
includes 40
and all values to the right
x > 40
Does not include 40
It includes all values to the
right of 40
40
40
x
40.5
39.5
39.5
40.5
x
x > 40.5
A) at least 40
A) greater than 40
B) no less than 40
B) more than 40
C) greater then or equal to 40
Example 3
P( x < 40 )
Example 4
P( x < 40 )
If P( x ≤ a) is adjusted to P( x ≤ a + .5)
P( x < a) is adjusted to P( x < a − .5)
P( x ≤ 40) is adjusted to P( x < 40.5)
P( x < 40) is adjusted to P(x < 39.5)
x < 40
includes 40
and all values to the left
x < 40
Does not include 40
It includes all values to the
left of 40
40
39.5
40
40.5
x < 40.5
x
39.5
x < 39.5
A) at most 40
A) fewer than 40
B) no more than 40
B) less than 40
40.5 x
C) less than or equal to 40
Stat 300 6 – 8A Lecture
Page 7 of 8
© 2012 Eitel
Using a Normal Distribution to Approximate a Binomial Distribution
1. Check to be sure the binomial distribution is approximately normal.
If n • p ≥ 5 and n • q ≥ 5
Then the Binomial Distribution can be considered approximatley normal
2. Find the mean and standard deviation for the normal distribution.
Use the population mean µ = n • p
and Standard Deviation σ = n • p• q
3. Find the continuity correction for x
Continuity Corrections
If P( x ≥ a) is adjusted to P( x ≥ a − .5)
If P( x ≤ a) is adjusted to P( x ≤ a + .5)
If P( x > a) is adjusted to P( x > a + .5)
If P( x < a) is adjusted to P( x ≤ a − .5)
Write the probability question in algebraic terms using the corrected value for a.
x > x ± .5
or
x < x ± .5
µx = n • p
σ x = n • p• q
4. Show the set up to convert the adjusted x value to z and then shade the area on the z graph that
corresponds to that specified probability.
z=
(adjusted x − µ)
n • p• q
z > z1
or
z < z2
0
1
Z
5. State the answer to the probability question. Round P(x) to 2 decimal places
Stat 300 6 – 8A Lecture
Page 8 of 8
© 2012 Eitel