Statistical Quality Control Questions and Answers – Modeling Process Quality – Discrete Distributions – 2

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This set of Statistical Quality Control Interview Questions and Answers for Experienced people focuses on “Modeling Process Quality – Discrete Distributions – 2”.

1. Which of these is not a discrete probability distribution?
a) Hyper geometric Distribution
b) Binomial Distribution
c) Normal Distribution
d) Poisson Distribution
View Answer

Answer: c
Explanation: Hyper geometric distribution, Binomial distribution, and Poisson distribution are all part of discrete probability distribution family. But, Normal distribution is a Continuous distribution.
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2. The mean of hyper geometric probability distribution is expressed as _____
a) μ=nD/N
b) μ=np
c) μ=r/p
d) μ=nN/D
View Answer

Answer: a
Explanation: The mean of hyper geometric distribution is given by,
μ=nD/N.

3. Probability function for hyper geometric probability distribution is _____
a) \(p(x) = \frac{\left(D \atop x\right)\left(N-D \atop n-x\right)}{\left(N \atop n\right)} \)
b) \(p(x) = \frac{\left(D \atop x\right)\left(N-D \atop n-x\right)}{\left(n \atop N\right)} \)
c) \(p(x) = \frac{\left(x \atop D\right)\left(N-D \atop n-x\right)}{\left(N \atop n\right)} \)
d) \(p(x) = \frac{\left(D \atop x\right)\left(n-x \atop N-D\right)}{\left(N \atop n\right)} \)
View Answer

Answer: a
Explanation: The probability function for the hyper geometric probability distribution is evaluated by following equation,
\(p(x) = \frac{\left(D \atop x\right)\left(N-D \atop n-x\right)}{\left(N \atop n\right)} \)
Where D is a number of interest, N is total no. of items in population, x is the no. of sample falling in the class of interest and n is no. of items in a sample.

4. For N=100, D=5, n=10, and x = 2; probability according to hypergeometric distribution will be _____
a) 0.58375
b) 0.92314
c) 0.99336
d) 0.99975
View Answer

Answer: c
Explanation: According to hyper geometric probability distribution,
\(p(x) = \frac{\left(D \atop x\right)\left(N-D \atop n-x\right)}{\left(N \atop n\right)} \)
This gives P(X≤x)=0.99336 for the mentioned values.

5. The independent trials which have either a “success” or a “failure” as an outcome, are called ____
a) Normal trials
b) Bernoulli trials
c) Poisson’s trials
d) Hyper geometric trials
View Answer

Answer: b
Explanation: When a process contain a sequence of independent trials which don’t depend on each other’s result, and which have outcome either a “success” or a “failure”, the trials are called Bernoulli trials.
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6. The binomial distribution is given by _____
a) \(p(x) = \left(n \atop x\right) p^x (1-p)^{n-x}\)
b) \(p(x) = \left(x \atop n\right) p^{n-x} (1-p)^x\)
c) \(p(x) = \left(x \atop n\right) p^x (1-p)^{n-x}\)
d) \(p(x) = \left(n \atop x\right) p^{n-x} (1-p)^x\)
View Answer

Answer: a
Explanation: The binomial distribution is based on Bernoulli trials, which gives the probability distribution as,
\(p(x) = \left(n \atop x\right) p^x (1-p)^{n-x}\).

7. Let us take Bernoulli trial of tossing a coin. If there are 70 independent trials, and finding a head is a success, what will be the variance?
a) 17.5
b) 18.5
c) 4.8
d) 4.18
View Answer

Answer: a
Explanation: Variance of the binomial distribution,
σ2=np(1-p)
For tossing a coin and finding a head, p = 0.5, which gives variance = 17.5.

8. Poisson distribution is given by ____
a) \(p(x) = \frac{e^{-λ} λ^x}{x!}\)
b) \(p(x) = \frac{e^{-x} λ^x}{x}\)
c) \(p(x) = \frac{e^{-λ} λ^x}{x}\)
d) \(p(x) = \frac{e^{-λ} λ^x}{λ!}\)
View Answer

Answer: a
Explanation: The Poisson distribution probability is given by the following equation,
\(p(x) = \frac{e^{-λ} λ^x}{x!}\)
Where λ is a parameter and λ > 0.

9. The mean of a Poisson distribution is _____
a) Greater than λ
b) Lesser than λ
c) Equal to λ
d) Having no relation with λ
View Answer

Answer: c
Explanation: According to Poisson distribution, the mean and the variance, both are always equal to the parameter used in the Poisson probability distribution, i.e. λ.
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10. For a company, which uses Poisson distributions to describe all the defects in its product, which has the parameter λ=4, what will be the probability of finding the selected product from a sample that will have 2 or less than 2 defects?
a) 0.2381
b) 0.2561
c) 0.2104
d) 0.3310
View Answer

Answer: a
Explanation: The Poisson probability distribution which has λ=5, we can determine the above stated probability using,
P{x≤2} = \(\sum_{x=2}^2 \frac{e^{-4} 4^x}{x!}\) = 0.018316 + 0.73263 + 0.146525 = 0.238104.

11. For a experiment following binomial distribution, the mean is 54.3, and the number of independent trials is 121. What will be the probability of failure?
a) 0.55123
b) 0.51000
c) 0.44877
d) 0.45700
View Answer

Answer: a
Explanation: μ=54.3, n=121. We know that, p=μ/n
p=0.44876.
So probability of failure = 1-p = 0.55123.

12. A lot follows hyper geometric distribution for defects found in its items. It contains 100 items out of which, 5 are defective. If 10 items are selected without replacement in a random order, what is the probability of finding 0 defective items?
a) 0.5837521
b) 0.4162479
c) 0.5708992
d) 0.4522222
View Answer

Answer: a
Explanation: According to hyper geometric probability distribution,
\(p(x) = \frac{\left(D \atop x\right)\left(N-D \atop n-x\right)}{\left(N \atop n\right)} \)
Putting values from the question, we get, p(x) = 0.5837521.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn