Question

Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms. The actual measured resistances of wires produced by company A have a normal probability distribution with mean .13 ohm and standard deviation .005 ohm. a. (15 pts) What is the probability that a randomly selected wire from company A’s production will meet the specifications? z = (X-µ)/σ P(X<.12) = P(z<( (.12- .14)/.005) = b. (10 pts) If four of these wires are used in each computer system and all are selected from company A. What is the probability that all four in a randomly selected system will meet the specifications?

Answer 1

Answer: a) 0.9544996

b) 0.9999366

Step-by-step explanation:

Given : The actual measured resistances of wires produced by company A have a normal probability distribution with mean $\mu=0.13$ ohm and standard deviation $s=0.005$ ohm.

Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms.

Let x be the random variable that represents the value of resistance in wires.

Using formula for z-score , $z=\dfrac{x-\mu}{s}$

The z-value at x= 0.12 will be

$z=\dfrac{0.12-0.13}{0.005}=-2$

The z-value at x= 0.14 will be

$z=\dfrac{0.14-0.13}{0.005}=2$

The p-value : $P(-2$

$=0.9772498-(1-P(z$

Hence, the probability that a randomly selected wire from company A’s production will meet the specifications = 0.9544996

b) Sample size : n= 4

Using formula for z-score , $z=\dfrac{x-\mu}{\dfrac{s}{\sqrt{n}}}$

The z-value at x= 0.12 will be

$z=\dfrac{0.12-0.13}{\dfrac{.005}{\sqrt{4}}}=-4$

The z-value at x= 0.14 will be

$z=\dfrac{0.14-0.13}{\dfrac{.005}{\sqrt{4}}}=4$

The p-value : $P(-4$

$=0.9999683-(1-P(z$

The probability that all four in a randomly selected system will meet the specifications = 0.9999366