Question

Consider again the company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 23 mm and an upper specification limit of 23.4 mm. The standard deviation is 0.15 mm and the mean is 23.2 mm. (Round your answer to 4 decimal places.) a. What is the probability that a tire will be too narrow? (Round your answer to 4 decimal places.) b. What is the probability that a tire will be too wide? (Round your answer to 3 decimal places.) c. What is the probability that a tire will be defective? 11

Answer 1

The probability that 23 < X < 23.4 is equal to the blue area under the curve.

3.5 3.0 2.5 2.0 y 15 1.0 0.5 0.0 23 23.4 -0.5 22.7 22.8 22.9 23.0 23.1 23.2 23.3 23.4 23.5 23.6 23.7 X

Since μ=23.2 and σ=0.15 we have:P ( 23 < X < 23.4 )=P ( 23−23.2 < X−μ < 23.4−23.2 )=P ( 23−23.2/0.15 < X−μ/σ < 23.4−23.2/0.15)

Since Z=x−μ/σ , 23−23.2/0.15=−1.33 and 23.4−23.2/0.15=1.33 we have:

P ( 23<X<23.4 )=P ( −1.33<Z<1.33 )

Use the standard normal table to conclude that:

P ( −1.33 < Z < 1.33 )=0.816

The probability outside of 23 and 23.4 is 1 - 0.816 = 0.184

The probability that X>23.4 is equal to the blue area under the curve (Tyres will be too wide)

3.5 3.0 2.5 2.0 y 1.5 1.0 0.5 10.0 23.4 -0.5 22.7 22.8 22.9. 23.0 23.1 23.2 23.3 23.4 23.5 23.6 23.7 X

Since μ=23.2 and σ=0.15 we have:P ( X>23.4 )=P ( X−μ>23.4−23.2 )=P ( X−μ/σ>23.4−23.2/0.15)

Since Z=x−μ/σ and 23.4−23.2/0.15=1.33 we have:

P ( X>23.4 )=P ( Z>1.33 )

Use the standard normal table to conclude that:

P (Z>1.33)=0.092

The probability that X<23 is equal to the blue area under the curve (Tyres will be too narrow)

3.5 3.0 .5 2.0 y 1.5 1.0 0. .5 0.0 23 -0.5 22.7 22.8 22.9 23.0 23.1 23.2 23.3 23.4 23.5 23.6 23.7 х

Since μ=23.2 and σ=0.15 we have:P ( X<23 )=P ( X−μ<23−23.2 )=P (X−μ/σ<23−23.2/0.15)

Since x−μ/σ=Z and 23−23.2/0.15=−1.33 we have:

P (X<23)=P (Z<−1.33)

Use the standard normal table to conclude that:

P (Z<−1.33)=0.0916

The probability that X>23 is equal to the blue area under the curve.

3.5 3.0 2.5 2.0 y 15 1.0 0.5 0.0 23 -0.5 22.7 22.8 22.9 23.0 23.1 23.2. 23.3 23.4 23.5 23.6 23.7 X

Since μ=23.2 and σ=0.15 we have:P ( X>23 )=P ( X−μ>23−23.2 )=P ( X−μ/σ>23−23.2/0.15)

Since Z=x−μ/σ and 23−23.2/0.15=−1.33 we have:

P ( X>23 )=P ( Z>−1.33 )

Step 3: Use the standard normal table to conclude that:

P (Z>−1.33)=0.908

The probability that X<23.4 is equal to the blue area under the curve.

3.5 3.0 2.5 2.0 y 1.5 1.0 0.5 0.0 23.4 -0.5 22.7 22.8 22.9 23.0 23.1 23.2 23.3 23.4 23.5 23.6 23.7 х

Since μ=23.2 and σ=0.15 we have:P ( X<23.4 )=P ( X−μ<23.4−23.2 )=P (X−μ/σ<23.4−23.2/0.15)

Since x−μ/σ=Z and 23.4−23.2/0.15=1.33 we have:

P (X<23.4)=P (Z<1.33)

Step 3: Use the standard normal table to conclude that:

P (Z<1.33)=0.908