A company making tires for bikes is concerned about the exact width of their 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. (Round your answer to 4 decimal places.) What is the process capability index for the process

For this process, the capability index is 0.4444

Step-by-step explanation:

Given the data in the question;

Lower specification limit = 22.8 mm

Upper specification limit = 23.2 mm

The standard deviation σ = 0.15 mm

Mean μ = 23

capability index Cpk = ?

Now, Using the formula;

Cpk = min{ [(USL - μ)/3σ], [(μ - LSL)/3σ]

where USL is the Upper specification limit, LSL is the Lower specification limit, σ is the standard deviation, μ is the mean

so we substitute

Cpk = min{ [(23.2 - 23 )/(3×0.15)], [(23 - 22.8)/(3×0.15)] }

Cpk = min{ [0.2 / 0.45 ], [0.2 / 0.45] }

Cpk = min{ [ 0.4444 ], [0.4444] }

Therefore, for this process, the capability index is 0.4444

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9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841