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N ormal Approximation to the Binomial Distribution Prepared by E.G. Gascon.

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Presentation on theme: "N ormal Approximation to the Binomial Distribution Prepared by E.G. Gascon."— Presentation transcript:

1 N ormal Approximation to the Binomial Distribution Prepared by E.G. Gascon

2 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

3 Binomial Problem solved using the Standard Normal Curve Use the same Z- score as with the Normal Distribution, except for x use the point where the rectangle touches the curve What does that mean? Say the Data point is here… 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

4 Modification to Mean and Standard Deviation n is the number of trials, and p is the probability of success on a single trial

5 Problem 9.4# 17 (a) p =.05, n = 75 a.) Exactly 5 defectives X = 5, therefore the lower limit is 4.5 and the upper limit is 5.5 of the rectangle 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 Probability of exactly 5 defectives is 16.84% Notice the value for x being used To find Z even though There are 5 defects

6 Problem 9.4# 17 (b) p =.05, n = 75 b.) No Defects X = 0, therefore the lower limit is -.5 and the upper limit is.5of the rectangle 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 Probability of exactly 0 defectives is 3.05%

7 Problem 9.4# 17 - c p =.05, n = 75 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. 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 Probability of 1 or more defectives is 95.73%

8 Questions / Comments / Suggestions Please post questions, comments, or suggestions in the main forum regarding this presentation.


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