Monday, February 8, 2016

Unizor - Bernoulli Statistics - Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on

Bernoulli Statistics - Problems

Problem 1

A quality control at some parts manufacturer has determined that out of 10,000 sampled parts made by this manufacturer 300 were defective.
Determine the probability of manufacturing the defective part and a margin of error with the level of certainty equal to 0.9545.


Consider Bernoulli random variable ξ that takes the value 1 with an unknown probability P if a part is defective and takes the value 0 otherwise.
As we know, mathematical expectation and variance of this random variable are:
E(ξ) = P
Var(ξ) = P·(1−P)

Then the empirical frequency of defective parts is expressed as
η = (ξ1+ξ2+...+ξN) / N
Here N=10,000 and a single value of random variable η is 300/10000=0.03.

We assume that the distribution of η is close to Normal with mathematical expectation and variance, expressed in terms of unknown probability P as
E(η) = N·E(ξ) / N = P
Var(η) = σ² = N·Var(ξ) / N² =
= P(1−P) / N which is not greater than 1 / (4N)

Since the unknown probability P equals to mathematical expectation of η and a single value of η is an unbiased approximation to this expectation, we can say that the probability P is, approximately, equal to 0.03.

To determine the margin of error, recall that for a Normal random variable with certainty level of 0.9545 its values are within an interval of 2σ from its mathematical expectation, where σ is a standard deviation.
In our case, though we don't know σ exactly, we know that it is bounded from above:
σ is not greater than 1/(2√N) = 0.005

Therefore, with certainty level of 0.9545 we can state that an unknown probability P is on a distance (that is, with a margin of error) of no more then
2σ = 1/√N = 0.01
from empirical expectation 0.03, which can be expressed as:
Prob{P∈[0.02;0.04]} is not less than 0.9545

Problem 2

A quality control at some parts manufacturer has determined that out of 10,000 sampled parts made by this manufacturer 300 were defective.
What certainty level can we attribute to the following (narrower than in a previous problem) evaluation of probability P of manufacturing a defective part:


Notice that in this case we are talking about a margin of error 0.005 around the empirical sample average 0.03, that we can use as an unbiased evaluation of probability P. This margin of error equals to an upper bound of a standard deviation of our random variable σ. Therefore, as we know, the probability of a normal random variable to be in the vicinity of σ from its mathematical expectation equals to 0.6825.
Therefore, the level of certainty for this evaluation is:
Prob{P∈[0.025;0.035]} is not less than 0.6825
As you see, more precise evaluation can be made with less certainty.

Problem 3

Now our purpose is to determine the volume N of the sample set of parts required to evaluate the probability of manufacturing a defective part within a margin of error Δ=0.005 with certainty level p=0.9545.

The required certainty level can be assured with evaluation of probability P by its empirical value (presumed, normally distributed) if margin of error Δ is equal to 2σ.
Since σ is not greater than 1/(2√N) and margin of error Δ=2σ=0.005, we can find N:
N = 200² = 40000
So, it takes 40,000 parts to examine to achieve a precision of our evaluation to be within margin of error of 0.005 .

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