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Wires manufactured for a certain computer system are specified to have a resistance of between 0.12 and 0.14 ohm. The actual measured resistances of the wires produced by Company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm. The computer firm orders 70% of the wires used in its systems from Company A and 30% from Company B.What is the probability that a randomly selected wire from Company B’s production lot will meet the specifications?

Respuesta :

Answer:

0.9544

Step-by-step explanation:

Let the measured resistances of the wire production by Company B = x

Define the standard normal variable [tex]Z=\frac{x-0.13}{0.005}[/tex]

The probability that a randomly selected wire from Company B's production lot will meet the specification is :

P = (0.12 ≤ X ≤ 0.14)

  = [tex]P(\frac{0.12-0.13}{0.005}\leq Z\leq \frac{0.14-0.13}{0.005})[/tex]

 = [tex]P(-2\leq Z\leq 2)[/tex]

 = [tex]P(Z\leq 2)-P(Z\leq -2)[/tex]

 = 0.9772 - (1 - 0.9972)

 = 0.9544

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