The Hawkins Company randomly samples 10 items from every lar

diferira7c 2021-12-17 Answered
The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected.
If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that meets the contract requirements will end up being 100 percent inspected is approximately...
A) .086
B) .41
C) .68
D) There is insufficient information to answer

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Expert Answer

Jimmy Macias
Answered 2021-12-18 Author has 4834 answers

Step 1
Introduction:
Denote X as the random variable of interest here, which is the number of defective items in the random sample of 10 items.
The batch from which the 10 items are randomly selected, is a large batch, which has 5% defectives. Due to the large size of the batch, it can be assumed that the selection of the 10 items are practically independent of each other, with probability of “success” or probability that each item is defective is \(\displaystyle{p}={0.05}\ {\quad\text{or}\quad}\ {5}\%\).
Thus, it may be assumed that, X has a binomial distribution with parameters, \(\displaystyle{n}={10}\) packages, \(\displaystyle{p}={0.05}\).
Since p denotes the probability of success, the probability of failure is, \(\displaystyle{q}={1}-{p}={0.95}\).
Step 2
Calculation:
If \(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({n},{p}\right)}\), then the probability mass function of X is:
\(f(x)=\begin{cases}(\begin{array}{c}n\\x\end{array})p^{x}q^{n-x};x=0,1,...,n;\& 0\end{cases}\) Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.
Thus, the probability mass function of X here is:
\[f(x)=\begin{cases}(\begin{array}{c}10\\x\end{array})(0.05)^{x}(0.95)^{10-x};& x=0,1,2,...,10\\0; & otherwise\end{cases}\]
The entire batch, that is, 100% will be inspected if 2 or more defectives are found in the random sample of 10 items from that batch, that is, if \(\displaystyle{X}\geq{2}\). For a batch that meets the contract requirements (that is, a batch with 5% defectives), the probability that it will end up being 100% inspected is calculated below:
\(\displaystyle{P}{\left({X}\geq{2}\right)}={1}-{P}{\left({X}{ < }{2}\right)}\)
\(\displaystyle={1}-{\left[{P}{\left({X}\le{1}\right)}\right]}\)
\(\displaystyle={1}-{\left[{P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}\right]}\)
\(\displaystyle={1}-{\left[{f{{\left({0}\right)}}}+{f{{\left({1}\right)}}}\right]}\)
\(=1-\begin{cases}(\begin{array}{c}10\\0\end{array})(0.05)^{0}(0.95)^{10-0}+\begin{cases}(\begin{array}{c}10\\1\end{array})\end{cases}\end{cases}\)(0.05)^{1}(0.95)^{10-1}]\]
\(\displaystyle={1}-{\left[{\left({1}\right)}\cdot{\left({1}\right)}\cdot{\left({0.95}\right)}^{{{10}}}+{\left({10}\right)}\cdot{\left({0.05}\right)}\cdot{\left({0.95}\right)}^{{{9}}}\right]}\)
\(\displaystyle\approx{1}-{\left[{0.599}+{0.315}\right]}\)
\(\displaystyle={1}-{0.914}\)
\(\displaystyle={0.086}\).
Thus, the probability that the batch will end up being 100% inspected is Option A) 0.086.

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Orlando Paz
Answered 2021-12-19 Author has 830 answers
Answer:
\(\displaystyle{0.9138}={91.38}\%\)
Step-by-step explanation:
For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
\(\displaystyle{P}{\left({X}={x}\right)}={C}_{{{n},{x}}}\cdot{p}^{{{x}}}\cdot{\left({1}-{p}\right)}^{{{n}-{x}}}\)
In which \(\displaystyle{C}_{{{n},{x}}}\) is the number of different combinations of x objects from a set of n elements, given by the following formula.
\(\displaystyle{C}_{{{n},{x}}}={\frac{{{n}!}}{{{x}!{\left({n}-{x}\right)}!}}}\)
And p is the probability of X happening.
10 items
This means that \(\displaystyle{n}={10}\)
5 percent of the items shipped can be defective.
This means that \(\displaystyle{p}={0.05}\)
Probability that a batch that meets the contract requirements will be shipped without further inspection
Probability of 1 or fewer defects.
So \(\displaystyle{P}{\left({X}\le{1}\right)}={P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}\)
In which \(\displaystyle{P}{\left({X}={x}\right)}={C}_{{{n},{x}}}\cdot{p}^{{{x}}}\cdot{\left({1}-{p}\right)}^{{{n}-{x}}}\)
\(\displaystyle{P}{\left({X}={0}\right)}={C}_{{{10},{0}}}\cdot{\left({0.05}\right)}^{{{0}}}\cdot{\left({0.95}\right)}^{{{10}}}={0.5987}\)
\(\displaystyle{P}{\left({X}={1}\right)}={C}_{{{10},{1}}}\cdot{\left({0.05}\right)}^{{{1}}}\cdot{\left({0.95}\right)}^{{{9}}}={0.3151}\)
\(\displaystyle{P}{\left({X}\le{1}\right)}={P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}={0.5987}+{0.3151}={0.9138}\)
So the answer is:
\(\displaystyle{0.9138}={91.38}\%\)
0
content_user
Answered 2021-12-27 Author has 11052 answers

Answer:
The answer is 0.9138
Step-by-step explanation:
This is a question of Binomial Probability Combination and the formula that will be used in this question is:
\(P(x)=nCx \cdot p^{x} \cdot q^{n-x}\)
Where,
\(n \Rightarrow\) Finite sample number, here it is equal to 10
\(p \Rightarrow\) Success, here it is equal to 5% defective or 
\(q \Rightarrow\) Failure, here it is equal to (1 - p) i.e.
\(x \Rightarrow\) Number of defective products
Now, in order to to pass inspection and be shipped  needs to be equal to '0' or '1'.
So,
Putting x=0 in the formula along with the values mentioned above we get
\(P(0)=10C0 \cdot 0.05^{0} \cdot 0.95^{10-0}\)
\(P(0)=1 \cdot 1 \cdot 0.5987\)
\(P(0)=0.5987\)
Similarly,
Putting x=1 in the formula along with the values mentioned in the bullets above we get
\(P(1)=10C1 \cdot 0.05^{1} \cdot 0.95^{10-1}\)
\(P(1)=10 \cdot 0.05 \cdot 0.6302\)
\(P(1)=0.3151\)
Now, in order to get the actual probability we need to add \(P(0)\ and\ P(1)\) because there is a chance that either there is no defective product or there is 1 defective in the shipped-batch from which sample was taken. Hence,
\(P(0)+P(1)=0.5987+0.3151\)
\(=0.9138\) (Answer)

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