Lesson 6
The Normal Distribution as an Approximation to the Binomial Distribution
Suppose a Survey I received requested my email address, was sent to 60,000 people, and the
Percentage of surveys returned with an email address was 2.5%. Suppose the true goal of the
Survey was to get at least 1500 email addresses to be used for marketing purposes. What is
the probability of getting at least 1500 email addresses?
np= (60,000)(0.025) = 1,500 >5
u= 1,500
nq= (60,000) (1 - .025)= 58,500 >5
sigma= [(60,000)(0.025)(.975)]^.5 = 38.2426
z= (x - u)/sigma =
(1500 - 1500) / 38.2426 = 0 = z
This is for an area of 0.50, so the probability of finding “at least 1500 successes” among
60,000 trials is 1 - 0.50 = 0.50
With Acceptance Sampling, a sample of items is randomly selected, and the entire batch is
either accepted or rejected, depending on the results. Company XYZ has manufactured a
large batch of backup power supply units for PC’s, and 7.9 % of them are defective. If the
acceptance sampling plan is to randomly select 82 units and accept the whole batch if at
most 4 units are defective, what is the probability that the entire batch will be accepted?
np = (82)(.079)= 6.478
nq = (82)(1-0.079)= 75.522
np and nq  5 A
  np  6.478
Sigma= (npq)^.5 = [(82)(.079)(1-.079)]^.5 = 2.4426
z =(4.5 – 6.478)/ 2.4426 = -0.8098
Z = -0.8098 is for an area of 0.2147, so the probability that the entire batch will be
accepted if at most 4 units are defective is P (z -0.8098) = 0.2147 . Company XYZ should
change its quality control procedures, since its manufacturing is not producing the desired
results.
The Genetics Institute developed a method to increase the probability of conceiving a girl.
Among 574 women using that method, 525 had baby girls. Assuming that the method
produces boys and girls equally likely, find the probability of getting at least 525 girls among
574 babies. Does the result suggest that the method is effective?
n= 574, p= 0.5, np = 287
= 287
nq= (574)(1-.5) = 287
sigma= (npq)^.5 = [(287)(0.5)]^.5 = 11.9791
z = (524.5 - 287)/ 11.9791 = 19.8262
Z=19.8262 is for an area of 0.9999, so the probability of getting at least 525 girls
among 574 babies is 1 - 0.9999 = 0.0001. The method appears to be effective, since
the probability of getting 525 girls by chance is so small.