A company generally purchases large lots of a certain kind of electronic device. A method is used that rejects a lot if 4 or more defective units are found in a random sample of 100 units. ​(a) What is the probability of rejecting a lot that is ​3% ​defective? ​(b) What is the probability of accepting a lot that is ​4% ​defective?

Answers

Answer 1

Using the binomial distribution, it is found that there is a:

a) 0.3526 = 35.26% probability of rejecting a lot that is 3% defective.

b) 0.4295 = 42.95% probability of accepting a lot that is 4% defective.

For each device, there are only two possible outcomes, either it is defective, or it is not. The probability of a device being defective is independent of any other device, hence the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

x is the number of successes. n is the number of trials. p is the probability of a success on a single trial.

In this problem, the sample has 100 units, hence [tex]n = 100[/tex].

Item a:

3% of the pieces are defective, hence [tex]p = 0.03[/tex].

The probability is:

[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]

In which:

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

Hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{100,0}.(0.03)^{0}.(0.97)^{100} = 0.0476[/tex]

[tex]P(X = 1) = C_{100,1}.(0.03)^{1}.(0.97)^{99} = 0.1471[/tex]

[tex]P(X = 2) = C_{100,2}.(0.03)^{2}.(0.97)^{98} = 0.2252[/tex]

[tex]P(X = 3) = C_{100,3}.(0.03)^{3}.(0.97)^{97} = 0.2275[/tex]

Then:

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0476 + 0.1471 + 0.2252 + 0.2275 = 0.6474[/tex]

[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.6474 = 0.3526[/tex]

0.3526 = 35.26% probability of rejecting a lot that is 3% defective.

Item b:

4% of the pieces are defective, hence [tex]p = 0.04[/tex].

Lot is accepted if less than 4 units are defective, hence:

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = 0) = C_{100,0}.(0.04)^{0}.(0.96)^{100} = 0.0169[/tex]

[tex]P(X = 1) = C_{100,1}.(0.04)^{1}.(0.96)^{99} = 0.0703[/tex]

[tex]P(X = 2) = C_{100,2}.(0.04)^{2}.(0.96)^{98} = 0.1450[/tex]

[tex]P(X = 3) = C_{100,3}.(0.04)^{3}.(0.96)^{97} = 0.1973[/tex]

Then:

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0169 + 0.0703 + 0.1450 + 0.1973 = 0.4295[/tex]

0.4295 = 42.95% probability of accepting a lot that is 4% defective.

A similar problem is given at https://brainly.com/question/24863377


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3 Answers: Choice AChoice CChoice E

=============================================================

Explanation:

Check out the diagram below. I've plotted the regression line A(t) = -1.4t+96 on the same xy grid as the given scatterplot. This line tries to get as close as possible to all of the points.

From here, we plug in each of the temperatures mentioned in the five answer choices (10, 15, 68.5, 55, 0). The outputs form the second column of the table in that same diagram.

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The other temperature values are handled the same way. Choice B is false because t = 15 does not lead to A(t) = 85. Instead, it's A(15) = 75 meaning it's 10 degrees off the mark.

Choice C is true though. Plugging t = 68.5 into A(t) leads to A(t) = 0.1 which rounds to 0 assuming we're only worried about whole numbers. If your teacher is considering decimal values, then choice C is unfortunately false. For now I'll assume your teacher wants you to round.

If we tried t = 55, then we'd get A(15) = -1.4*55+96 = 19 which is not 5 as we'd want. This rules out choice D.

Lastly, plug in t = 0 to find that A(t) = 96. This leads to choice E being one of the answers.

For the graph attached statements A, C and E are true.

     Function that fits the given data in the graph,

A(t) = -1.4t + 96

Here, t = temperature in Fahrenheit

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By substituting the value of 't' in the equation we can get the amount of oil used A(t) to heat the house.

Option A

By substituting t = 10°F,

A(10) = -1.4(10) + 96

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Therefore, statement is true.

Option B

For t = 15°F,

A(15) = -2.4(15) + 96

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Statement is False.

Option C

For t = 68.5°F

A(5) = -1.4(68.5) + 96

       = 0.1 gallons

       ≈ 0 gallons

Statement is True.

Option D

For t = 55°F,

A(55) = -1.4(55) + 96

         = 19 gallons

Statement is False.

Option E

For t = 0°F

A(0)  = -1.4(0) + 96

        = 96 gallons

Statement is true.

     Therefore, statements given in options A, C and E are True.

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