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Consider again that the company making tires for bikes is concerned about the exact width of its cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. What is the probability that a tire will either be too wide or too narrow?

Answer :

JeanaShupp

Answer: 0.1835

Step-by-step explanation:

Let x be a random variable that represents the width of the tires .

As per given , we have

[tex]\mu=23\ mm[/tex]

[tex]\sigma=0.15\ mm[/tex]

The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm.

∵ [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

Then for x=22.8, [tex]z=\dfrac{22.8-23}{0.15}=-1.33[/tex]  

For x=23.2, [tex]z=\dfrac{23.2-23}{0.15}=1.33[/tex]  

The probability that a tire will either be too wide or too narrow :-

[tex]P(x<22.8)+P(x>23.2)\\\\=P(z<-1.33)+P(z>1.33)\\\\=(1-P(z\leq1.33)+(1-P(z<1.33)) \ [\because P(Z<-z)=1-P(z<z) \ ,\ P(Z>z)=1-P(Z<z)]\\\\=2-2(P(z<1.33))\ \ [\because P(Z\leq z)=P(Z<z)]\\\\=2-2(0.9082408)\ \ \text{[Using z-value table]}\\\\=0.1835184\approx0.1835[/tex]

Hence, the probability that a tire will either be too wide or too narrow= 0.1835

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