Download Normal Approximation of a Binomial Probability

Normal Approximation to the Binomial
Distribution
Prepared by E.G. Gascon
Properties of
Normal Approximation of a Binomial
Distribution
• The area under the is made up of rectangles.
• Each rectangle touches the normal curve at two points .
• A value of a binomial experiment is the midpoint of a given
rectangle.
• The total area under the curve is always = 1. (This agrees with the
fact that the sum of the probabilities in any distribution is 1.)
Drawing a Standard Normal Curve: http://www.tushar-mehta.com/excel/c
Binomial Problem solved
using the Standard Normal Curve
Use the same Zscore as with
the Normal
Distribution,
except for x use
the point
where the
rectangle
touches the
curve
Say the Data point is
here…
What does that mean?
Then the Lower
value is …
The
Upper
value
is…
IF the data point is 250, then the lower value is 249.5 and the upper value is 250.5
Modification to Mean and Standard Deviation
  np
  np(1  p)
n is the number of trials, and p is the probability of success on a single
trial
Problem 9.4# 17 (a)
 p = .05, n = 75
  75  .05  3.75
  75  .05(1  .05)  75  .05(1  .05
Notice the value
for x being used
To find Z even though
There are 5 defects
3.5625  1.89
a.) Exactly 5 defectives
X = 5, therefore the lower limit is 4.5 and the upper limit is 5.5 of the rectangle
  3.75,   1.89,
4.5  3.75
1.89
zl  .40
zl 
  3.75,   1.89,
5.5  3.75
zu 
1.89
zu  .93
Probability of exactly 5 defectives is 16.84%
•Looking up the values in the normal
table
•Area to left of upper is.8238
and area to left of lower is .6554.
•.8238 - .6554 = .1684 is the area of
the rectangle
Problem 9.4# 17 (b)
 p = .05, n = 75
  75  .05  3.75
  75  .05(1  .05)  75  .05(1  .05
3.5625  1.89
b.) No Defects
X = 0, therefore the lower limit is -.5 and the upper limit is .5of the rectangle
  3.75,   1.89,
  3.75,   1.89,
.5  3.75
1.89
zl  2.25
.5  3.75
zu 
1.89
zu  1.72
zl 
Probability of exactly 0 defectives is 3.05%
•Looking up the values in the normal
table
•Area to left of upper is..0427
and area to left of lower is .0122.
•.0427 - .0122 = .0305 is the area of
the rectangle
Problem 9.4# 17 - c
 p = .05, n = 75
  75  .05  3.75
  75  .05(1  .05)  75  .05(1  .05
3.5625  1.89
c.) At least 1 defect , meaning that there are 1 or more defects
X = 1, therefore lower limit is .5 the upper limit is 1.5, use that lower limit so that
the “1” will be included in the calculations of greater than.
  3.75,   1.89,
.5  3.75
1.89
zl  1.72
zl 
Probability of 1 or more defectives is 95.73%
•Looking up the values in the normal
table
•Area to left of the limit is..0427
which is less than .5
•.1 - .0427, is the area to the right
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Suggestions
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