The specifications for an electronic device are 3.00 ± 0.10 amps. If the standard deviation is 0.02, what is the capability index? Write your answer to 2 decimal places (X.XX)

Respuesta :

Answer:

The capability index of the device is 1.67.

Explanation:

It is given that,

The specifications for an electronic device are [tex](3\pm0.1)\ A[/tex]

Upper specification limit, [tex]USL=3+0.1=3.1\ A[/tex]

Lower specification limit, [tex]LSL=3-0.1=2.9\ A[/tex]

Standard deviation, [tex]\sigma=0.02[/tex]

The capability index of an electronic device is given by :

[tex]C_p=\dfrac{USL-LSL}{6\sigma}[/tex]

[tex]C_p=\dfrac{3.1-2.9}{6\times 0.02}[/tex]

[tex]C_p=1.666[/tex]

or

[tex]C_p=1.67[/tex]

So, the capability index of the device is 1.67. Hence, this is the required solution.

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