Question

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)

Answer 1

The capability index of the device is 1.67.

Step-by-step explanation:

It is given that,

The specifications for an electronic device are
`(3\pm0.1)\ A`

Upper specification limit,
`USL=3+0.1=3.1\ A`

Lower specification limit,
`LSL=3-0.1=2.9\ A`

Standard deviation,
`\sigma=0.02`

The capability index of an electronic device is given by :

`C_p=(USL-LSL)/(6\sigma)`

`C_p=(3.1-2.9)/(6* 0.02)`

`C_p=1.666`

or

`C_p=1.67`

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