## 7.5.1 normal approximation of binomial

to the Binomial Probability Distribution
The goal of this activity is to understand how the normal probability distribution may be used to approximate the binomial probability distribution. State the four criteria for a binomial probability experiment. Clinical trials of the popular allergy medicine Allegra-D found that 13% of the trial subjects reported headache as an adverse reaction to the drug. A follow-up study involves a random sample of 2150 allergy sufferers. What is the probability that less than 250 report headache? To answer this question by hand would involve 250 calculations!

*P*(

*X* < 200) =

*P*(0) +

* P*(1) +

* P*(2) + … +

* P*(249)
Statisticians have found a simpler way. The figure below presents a binomial distribution overlaid with the normal probability distribution with the same mean and standard deviation:
µ =

*n *⋅

*p * and σ =

*n *⋅

*p *⋅ 1
As suggested by the graph and under certain conditions, the normal probability distribution may be used to approximate the binomial probability distribution.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

**The Normal Approximation to the Binomial Probability Distribution **

If

*n*·

*p*·

*q* ≥ 10, the binomial random variable

*X* is approximately normally distributed with

mean µ =

*n *⋅

*p * and standard deviation σ

Consider the binomial random variable

*X* with

*n* = 45 and

*p* = 0.40. The figure below shows a histogram of the probability distribution with a normal curve with µ

We know that the area of the rectangle corresponding to

*X* = 16 represents

*P*(16) of the binomial

probability distribution. The width of the rectangle is 1, so it extends from

*X* = 15.5 to

*X* = 16.5.

Notice that the area under the normal curve from

*X* = 15.5 to

*X* = 16.5 is approximately equal to

the area of the rectangle corresponding to the value of

*X* = 16. Therefore, the area under the

normal curve between

*X* = 15.5 to

*X* = 16.5 is approximately equal to

*P*(16).

The process of adding and subtracting 0.5 from the value of the binomial random variable when

using the normal probability distribution is called the

**correction for continuity**.

Continue to consider the binomial random variable

*X* with

*n* = 45 and

*p* = 0.40. The mean and
1. Verify that it is appropriate to use the normal probability distribution to approximate the
2. Find the value of

*P*(16) using the rules of the binomial random variable.
, determine the probability that

*X*
4. Compare your answers to items 2 and 3. 5. Clinical trials of the popular allergy medicine Allegra-D found that 13% of the trial subjects
reported headache as an adverse reaction to the drug. A follow-up study involves a random sample of 2150 allergy sufferers. What is the probability that less than 250 report headache?
a. Find the mean and standard deviation of the binomial random variable.
b. Determine the area on the normal distribution that approximates the variable. (Don’t
c. Find the z-score(s) for the area on the normal distribution. d. Determine the probability of the area.

Source: http://www.unco.edu/nhs/mathsci/facstaff/powers/m550/ContinuousDistributions/pdf/7.5.1%20Normal%20Approximation%20of%20Binomial.pdf

Rationalising rationing: economic and otherconsiderations in the debate about funding ofElly A. Stolk *, Werner B.F. Brouwer, Jan J.V. Busschbach Institute for Medical Technology Assessment , Erasmus Uni 6 ersity Rotterdam , PO Box 1738,3000 DR Rotterdam , The Netherlands Received 30 May 2000; accepted 10 May 2001 Abstract Although the cost-effectiveness of Viagra for the treatmen

Copyright © 2001 by AMERICAN UROLOGICAL ASSOCIATION, INC.®MEDICAL VERSUS SURGICAL ANDROGEN SUPRESSION THERAPY FORPROSTATE CANCER: A 10-YEAR LONGITUDINAL COST STUDYALBERT J. MARIANI, MONTY GLOVER AND SUZETTE ARITA From the Department of Urology, John A. Burns School of Medicine, University of Hawaii and Kaiser Medical Center, Honolulu, Hawaii Purpose: We provide a relative cost comparison of