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A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2000 hours is at least 0.95?

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Answer:

Step-by-step explanation:

From the given information:

Assuming we have an integer c to represent the components in the stocks.

Thus, the needed probability can be expressed as:

[tex]P(X_1+X_2+ ........ X_c \ge 2000)[/tex]

To break this down, we have:

[tex]P(\sum X_i \ge 2000) = P \Bigg(\dfrac{\sum X_i - 100n}{\sqrt{n \ Var (X)}} \ge \dfrac{2000-100 n }{\sqrt{n \times 900}} \Bigg )[/tex]

[tex]P \Bigg(\dfrac{\sum X_i - 100n}{\sqrt{n \ Var (X)}} \ge \dfrac{2000-100 n }{\sqrt{n \times 900}} \Bigg )= P (Z \ge 0.95 ) \ \ \ because \ Z_{0.05} = -1.65 \\ \\ \\ \dfrac{2000-100 n }{\sqrt{n \times 900}} = -1.65 \\ \\ \\ \dfrac{100\ n-2000 }{\sqrt{900n}} = 1.65 \\ \\ 100 n -1.65 \sqrt{900 n }-2000 = 0[/tex]

By solving the equation:

n = 23

Thus, relating to the needed condition; n ≥ 23

The needed number of the components that should be in stock should be  at least 23.

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