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

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=(x-\mu)/(s)`

The z-value at x= 0.12 will be

`z=(0.12-0.13)/(0.005)=-2`

The z-value at x= 0.14 will be

`z=(0.14-0.13)/(0.005)=2`

The p-value :
`P(-2<z<2)=P(z<2)-P(z<-2)`

`=0.9772498-(1-P(z<2))\\\\=0.9772498-1+0.9772498=0.9544996`

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=(x-\mu)/((s)/(√(n)))`

The z-value at x= 0.12 will be

`z=(0.12-0.13)/((.005)/(√(4)))=-4`

The z-value at x= 0.14 will be

`z=(0.14-0.13)/((.005)/(√(4)))=4`

The p-value :
`P(-4<z<4)=P(z<4)-P(z<-4)`

`=0.9999683-(1-P(z<4))\\\\=0.9999683-1+0.9999683=0.9999366`

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