A production process manufactures alternators for tanks. On the average, 1.5% of the alternators will not perform up to specifications. When a shipment of 100 alternators is received at the plant, they are tested, and if more than 2 are defective, the shipment is returned to the manufacturer. What is the probability of returning a shipment

Answers

Answer 1

Answer:

0.1902 = 19.02% probability of returning a shipment

Step-by-step explanation:

For each alternator, there are only two possible outcomes. Either it is defective, or it is not. The probability of an alternator being defective is independent of any other alternator, which means that the binomial probability distribution is used 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.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

1.5% of the alternators will not perform up to specifications.

This means that [tex]p = 0.015[/tex]

Shipment of 100 alternators

This means that [tex]n = 100[/tex]

What is the probability of returning a shipment?

Probability of more than 2 defective, which is:

[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]. So

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

[tex]P(X = 0) = C_{100,0}.(0.015)^{0}.(0.985)^{100} = 0.2206[tex]

[tex]P(X = 1) = C_{100,1}.(0.015)^{1}.(0.985)^{99} = 0.3360[/tex]

[tex]P(X = 2) = C_{100,2}.(0.015)^{2}.(0.985)^{98} = 0.2532[/tex]

Then

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2206 + 0.3360 + 0.2532 = 0.8098[/tex]

Then

[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.8098 = 0.1902[/tex]

0.1902 = 19.02% probability of returning a shipment


Related Questions

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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I hope this is helpful for you

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Answer:

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Answer:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

Step-by-step explanation:

First we find out the area of circle, and then subtract the area of triangle from it to find the area of the shaded region

Area of Circle

Here, r=radius

area of circle= π×r²

                     = π × 5²     (radius=half of diameter)

                     = 25π

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Area of Triangle

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Answers

Answer:

[tex]Shaded = 212.5cm^2[/tex]

Step-by-step explanation:

Given

See attachment

Required

The shaded region

First, calculate the area of the complete rectangle

[tex]Area =Length * Width[/tex]

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Next, calculate the area of the triangle

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Answers

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Answers

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Answers

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When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 400 cm^3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?

Answers

Answer:

The volume increases at 35.71cm^3/min

Step-by-step explanation:

Given

[tex]PV^{1.4} = C[/tex]

[tex]V = 400cm^3[/tex]

[tex]P =80kPa[/tex]

[tex]\frac{dP}{dt} =-10kPa/min[/tex]

Required

Rate at which volume increases

[tex]PV^{1.4} = C[/tex]       [tex]V = 400cm^3[/tex]       [tex]P =80kPa[/tex]

Differentiate: [tex]PV^{1.4} = C[/tex]  

[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = \frac{d}{dt}C[/tex]

By differentiating C, we have:

[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = 0[/tex]

Rewrite as:

[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} + V^{1.4}*\frac{dP}{dt} = 0[/tex]

Solve for [tex]\frac{dV}{dt}[/tex]

[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} =- V^{1.4}*\frac{dP}{dt}[/tex]

[tex]\frac{dV}{dt} =- \frac{V^{1.4}*\frac{dP}{dt} }{P*(1.4)*V^{0.4}}[/tex]

Substitute values

[tex]\frac{dV}{dt} =- \frac{400^{1.4}*-10 }{80*(1.4)*400^{0.4}}[/tex]

[tex]\frac{dV}{dt} =\frac{400*10 }{80*1.4}[/tex]

[tex]\frac{dV}{dt} =\frac{4000 }{112}[/tex]

[tex]\frac{dV}{dt} =35.71cm^3/min[/tex]

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