In statistics, we like to study variation. Calculators became relatively inexpensive because its solid-state circuitry is stamped by machine, thus making mass production feasible. One problem with anything that is mass-produced is quality control. The process must somehow be monitored or audited to be sure the output of the process conforms to requirements.
One method of dealing with this problem is lot acceptance sampling, in which items being produced are sampled at various stages of the production process and are carefully inspected. The lot of items from which the sample is drawn is then accepted or rejected, based on the number of defectives in the sample. Lots that are accepted may be sent forward for further processing or may be shipped to customers; lots that are rejected may be reworked or scrapped. For example, suppose a manufacturer of calculators chooses 200 stamped circuits from the day’s production and determines x, the number of defective circuits in the sample. Suppose that up to a 6% rate of defectives is considered acceptable for the process.
a. Find the mean and standard deviation of x, assuming the defective rate is 6%.
n = 200
p = .06
q = .94
µ = np = 200 × .06 = 12
σ2 = npq = 200 × .06 × .94 = 11.28
σ = 3.36
b. Use the normal approximation to determine the probability that 20 or more defectives are observed in the sample of 200 circuits
Test 1: Is np ≥ 5? np = 200 × .06 = 12 yes
Test 2: Is nq ≥ 5? nq = 200 × .94 = 188 yes
P(r ≥ 20) = P(x > 19.5) Correction for continuity
P(x > 19.5) = P(z > 2.23) = 1 – .9871 = .0129
Or normalcdf(2.23,10000) = .0128736783
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Over the years, it has been observed that of all the lawyers who take the state bar exam, only 57% pass.
a. Suppose that this year 850 lawyers are going to take the Tennessee bar exam, how many are expected to pass?
E = µ = np = 850 × .57 = 484.5
b. How many are expected to fail?
850 × .43 = 365.5
What is the probability that
c. 540 or more pass?
First the tests: np ≥ 5 (YES, 484.5); nq ≥ 5 (YES, 365.5)
µ = np = 484.5
σ2 = npq = 308.335
σ = 14.43
P(r ≥ 540) = P(x > 539.5) = P(z > 3.81) = .0000695
d. 500 or fewer pass?
P(r ≤ 500) = P(x < 500.5) = P(z < 1.11) = .8665
e. between 485 and 525, inclusive, pass?
P(485 ≤ r ≤ 525) = P(484.5 < x < 525.5) = P(0 < z < 2.84) = .4977
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A web service reported that 30.71% of all internet users use Firefox as their internet browser. You randomly select 250 internet users and ask each whether he or she uses Firefox.
n = 250
p = .3071
q = .6929
the tests: np ≥ 5 (YES, 76.775); nq ≥ 5 (YES, 173.225)
µ = np = 76.775
σ2 = npq = 53.1973975
σ = 7.29
a. What is the probability that more than 82 will say yes?
P(r > 82) = P(x > 82.5) = P(z > .72) = .2358
b. What is the probability that less than 82 will say yes?
P(r < 82) = P(x < 81.5) = P(z < .65) = .7422
c. What is the probability that exactly 82 will say yes?
P(r = 82) = P(81.5 < x < 82.5) = P(.65 < z < .72) = .0221
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Guidelines:
1. Verify that the binomial distribution applies.
2. Determine if you can use the normal distribution to approximate the binomial variable.
3. Find the mean and the standard deviation for the distribution.
4. Apply the appropriate continuity correction. Shad the corresponding area under the normal curve.
5. Find the corresponding z-scores.
6. Find the probability.
