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asambeis [7]
3 years ago
9

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

ally, the failure rate for LED light bulbs that the company manufactures is 3%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that
Mathematics
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

The probability that none of the LED light bulbs are​ defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are​ defective?

Solution:

Let the random variable <em>X</em> represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.

A random sample of <em>n</em> = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.

The probability mass function of <em>X</em> is:

P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...

Compute the probability that none of the LED light bulbs are​ defective as follows:

P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}

                =1\times 1\times 0.737424\\=0.737424\\\approx 0.7374

Thus, the probability that none of the LED light bulbs are​ defective is 0.7374.

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Step-by-step explanation:

Given equations are:

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Subtracting equation 2 from eqn 1

(6x-4y)-(4x - 4y) = -36+64\\6x-4y-4x+4y=28\\2x=28\\Dividing\ both\ sides\ by\ 2\\\frac{2x}{2}=\frac{28}{2}\\x=14\\Putting\ x=14\ in\ equation\ 1\\-4y=-64-6(14)\\-4y=-64-84\\-4y=-148\\Dividing\ both\ sides\ by\ -4\\\frac{-4y}{-4}=\frac{-148}{-4}\\y=37

The solution of system is x=14, y=37

Keywords: Linear equation, Elimination method

Learn more about linear equations at:

  • brainly.com/question/4706270
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