Design specifications that require a key dimension on a product measure 100 ± 10 units. A centered process is being considered for producing

Question

Design specifications that require a key dimension on a product measure 100 ± 10 units. A centered process is being considered for producing this product which has a standard deviation of four units.
(a) What can you say (quantitatively) regarding the process capability?
(b) Suppose the process average shifts to 92. Calculate the new process capability.
(c) What can you say about the process after the shift? Approximately what percentage of the items produced will be defective?

in progress 0
Savannah 3 weeks 2021-10-01T11:18:27+00:00 2 Answers 0

Answers ( )

    0
    2021-10-01T11:19:41+00:00

    Answer:

    a. Process Capability = 0.8333

    b. New Process Capability = 0.1667

    c. Probability = 0.30854

    Step-by-step explanation:

    The Process capability of a process is determined using the capability index C_{pk} . This capability index helps in determining whether the output of a process lies within the specification limits.

    a. Given that

    Mean, X = 100

    Lower Specification Limit, LSL = 100 – 10 = 90

    Upper Specification Limit, USL = 100 + 10 = 110

    Standard Deviation, \alpha = 4

    Process Capability can therefore be calculated using the formula :

    C_{pk}= min[\frac{X-LSL}{3\alpha }, \frac{USL-X}{3\alpha },]\\= min[\frac{(100-90)}{3*4 }, \frac{(110-100)}{3*4 },]\\=min[\frac{10}{12}, \frac{10}{12}]\\=min[0.8333, 0.8333] = 0.8333

    Hence, the process capability value for the process is 0.833

    b. When the process average shifts to 92, the new process mean would be 92

    X = 92

    LSL = 90

    USL = 110

    Standard Deviation = 4

    C_{pk} = min[\frac{X - LSL}{3\alpha }, \frac{USL - X}{3\alpha }]\\=min[\frac{(92-90)}{3*4}, \frac{(110-92)}{3*4}]\\=min(0.1667, 1.5)= 0.1667

    c. Probability of the defective unit after the shift :

    We calculate the Z score of LSL like so :

    Z_{LSL}=\frac{LSL - X}{\alpha }\\= \frac{90-92}{4}\\=- \frac {2}{4} = -0.5

    Using the “NORMDIST(Z)” function in Excel, we find the probability associated with Z_{LSL} as 0.30854

    Therefore the probability that the output will be less than 90 units is 0.30854

    0
    2021-10-01T11:19:54+00:00

    Answer:

    Please see explanation below for answers

    Step-by-step explanation:

    Parameters:

    Mean (μ)= 92 & Standard deviation (σ)= 4

    a. Upper specification limit USL = 100+10 = 110 units while

    Lower specification limit LSL = 100-10 = 90 units

    b. Process Capability Index, Cpk = min[USL−μ/3σ,μ−LSL/3σ]

    Cpk = min[110-92/12,92-90/12]

    Cpk = min[1.5,0.17]

    Cpk = 0.17

    c. From (b) above, since Cpk >0, The process is still within the product specification limit.

    From the normal distribution

    z = USL−μ/σ;μ−LSL/σ

    z = 4.5;0.5

    z= 4.5 for the upper specification and 0.5 for the lower specification

    Using the normal distribution table at z= 4.5;

    0.000340% are defective at the upper limit.

    Using the normal distribution table at z= 0.5;

    30.15% are defective at the lower limit.

    Therefore total % defect = 30.15%+0.000340%=30.15%

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )