A resistor utilized in a critical electronic component is required to have a resistance of 45 ohms /- 0.32 ohms. The current manufacturer of this resistor has found that its manufacturing process produces a mean value of 50 ohms and has a standard deviation of 0.16 ohms. What would be the C P of this process

Respuesta :

Answer:

The value is  [tex]CP = 0.667[/tex]

Step-by-step explanation:

From the question we are told that

    The resistance is  [tex]R = 45 \pm 0.32\ \Omega[/tex]

   The mean is  [tex]\mu = 50 \Omega[/tex]

    The standard deviation is  [tex]\sigma = 0.16 \Omega[/tex]

Generally the upper  limit  of the resistance specified in the question(i.e  upper specification limit (USL )) is

        [tex]USL = 45 + 0.32 = 45.32 \ \Omega[/tex]  

Generally the lower  limit of the resistance specified in the question(i.e  lower specification limit (USL )) is

        [tex]LSL = 45 - 0.32 = 44.68 \ \Omega[/tex]

Generally the CP of the process of manufacturing the resistor is mathematically represented as  

         [tex]C P = \frac{ USL - LSL }{ 6 * \sigma }[/tex]

=>     [tex]CP = \frac{45.32 - 44.68 }{6 * 0.16}[/tex]

=>     [tex]CP = \frac{0.64 }{0.96}[/tex]

=>     [tex]CP = 0.667[/tex]