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 [tex]\mu=0.13[/tex] ohm and standard deviation [tex]s=0.005[/tex] 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 , [tex]z=\dfrac{x-\mu}{s}[/tex]
The z-value at x= 0.12 will be
[tex]z=\dfrac{0.12-0.13}{0.005}=-2[/tex]
The z-value at x= 0.14 will be
[tex]z=\dfrac{0.14-0.13}{0.005}=2[/tex]
The p-value : [tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
[tex]=0.9772498-(1-P(z<2))\\\\=0.9772498-1+0.9772498=0.9544996[/tex]
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 , [tex]z=\dfrac{x-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
The z-value at x= 0.12 will be
[tex]z=\dfrac{0.12-0.13}{\dfrac{.005}{\sqrt{4}}}=-4[/tex]
The z-value at x= 0.14 will be
[tex]z=\dfrac{0.14-0.13}{\dfrac{.005}{\sqrt{4}}}=4[/tex]
The p-value : [tex]P(-4<z<4)=P(z<4)-P(z<-4)[/tex]
[tex]=0.9999683-(1-P(z<4))\\\\=0.9999683-1+0.9999683=0.9999366[/tex]
The probability that all four in a randomly selected system will meet the specifications = 0.9999366
