Answer:
The probability that 2 or more of the original 5,100 components may fail during the useful life of the product is:
= 0.001
Step-by-step explanation:
Probability of operating without failure = 0.999
The probability of failed component = 0.001 (1 - 0.999)
The number of components of the electronic office product = 5,100
The number of components that may fail, given the above successful operation = 5.1 (0.001 * 5,100).
Therefore, the probability that 2 or more of the original 5,100 components may fail during the useful life of the product = 0.001
Probability of component failure is the likelihood that a component fails during the useful life of the product. It is expressed as the number of likely failed components divided by the total number of components. This result can be left in decimal form or expressed as a percentage.
Following are the solution to the given expression:
[tex]\to \text{p = p (failure)} = 1 -0.999 = 0.001 \\\\[/tex]
We employ the binomial approximation of the normal distribution with mean.
[tex]\to \mu = np = 5100(0.001) = 5.1 \\\\ \to \text{standard deviation} \ (\sigma) = \sqrt{np(1-P)}[/tex]
[tex]= \sqrt{5100(0.001)(1-0.001)} \\\\ = \sqrt{5.1 \times (0.999)} \\\\= \sqrt{5.0949 } \\\\= 2.2572[/tex]
It require [tex]p(x\geq 2)[/tex] but we find [tex]p(x \geq 1.5)[/tex], because of [tex]0.5[/tex] correction.
Here
[tex]\to z = \frac{X-\mu}{\sigma }[/tex]
[tex]=\frac{1.5-5.1}{2.2572} \\\\ =\frac{-3.6}{2.2572} \\\\ = -1.59 \\\\[/tex]
Calculating the required probability:
[tex]\to p(z \geq -1.59)[/tex]
[tex]=1+ p(z< 1.59) \\\\= 1+ 0.9440826[/tex] , by standard normal table
[tex]= 1.9440826 \approx 1.944[/tex]
Learn more about the probability:
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