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Week 4 Lecture Notes Question
Question in Uniform Density Function
There are many cases in which we may be able to apply the uniform distribution.
Suppose that the research department of a steel factory believes that one of the
company's rolling machines is producing sheets of steel of different thickness. The
thickness is a uniform random variable with values between 150 and 200 millimeters.
Any sheets less than 160 millimeters thick must be scrapped because they are
unacceptable to the buyers. What is the mean and the standard deviation of the X (the
thickness of the sheet produced by this machine), and the fraction of steel sheet
produced by this machine that have to be scrapped?
Mean = (150 + 200)/2 millimeters = 175 millimeters
Standard deviation = (200 – 150)/12 millimeters = 14.43 millimeters
P(x < 160) = (160 – 150) / (200-150) = 10 / 50 = .2
Problem in What Is so Important About the Normal Distributions?
The probability of a defective item coming off a certain assembly line is p = 0.25. A
sample of 400 items is selected from a large lot of these items. What is the probability
90 or less, items are defective?
np = 400  0.25 = 100  5
nq = 400  0.75 = 300  5
So we can use the normal distribution to approximate the binomial probability
distribution.
 = np = 100
 = npq) = 75
z value = (90.5 – ) /  = (90.5 – 100) / 75 = -0.1267
P(x  90) = 0.4496