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olasank [31]
3 years ago
15

The lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour. there

are nine batteries in a package. what lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (round your answer to two decimal places.)
Mathematics
1 answer:
Montano1993 [528]3 years ago
6 0
The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by

P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \  \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)

Given that t<span>he lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour and that the probability that </span><span>nine batteries in a package exceeds a certain value, x, is </span>5% or 0.05. Then the value x is given by

P(X\ \textgreater \ x)=0.05 \\  \\ \Rightarrow1-P\left(z\ \textless \ \frac{x-12}{\frac{1}{\sqrt{9}}} \right)=0.05 \\  \\ \Rightarrow P\left(z\ \textless \ \frac{x-12}{\frac{1}{3}} \right)=1-0.05=0.95 \\  \\ \Rightarrow P(z\ \textless \ 3(x-12))=P(z\ \textless \ 1.645) \\  \\ \Rightarrow3(x-12)=1.645 \\  \\ \Rightarrow x-12= \frac{1.645}{3} =0.5483 \\  \\ \Rightarrow x=0.5483+12\approx12.55
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snow_lady [41]

Answer:

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Step-by-step explanation:

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(\sqrt[n]{a})^n=a\\\\(a+b)=a^2+2ab+b^2

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You need to square both sides of the equation:

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17-x=(x+3)^2

Simplifying, you get:

17-x=x^2+2(x)(3)+3^2\\\\17-x=x^2+6x+9\\\\x^2+6x+9+x-17=0\\\\x^2+7x-8=0

Factor the quadratic equation. Find two numbers whose sum be 7 and whose product be -8. These are: -1 and 8:

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4=4 (It checks)

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Therefore, the solution is:

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