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?

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 ohm and standard deviation 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 ,

The z-value at x= 0.12 will be

The z-value at x= 0.14 will be

The p-value :

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 ,

The z-value at x= 0.12 will be

The z-value at x= 0.14 will be

The p-value :

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

10500 dollars

Step-by-step explanation:

350,000 * .03 = 10500