The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weights less than

Question

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places.

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Eloise 3 weeks 2022-01-01T18:13:44+00:00 1 Answer 0 views 0

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    2022-01-01T18:15:16+00:00

    Answer:

    a)  P(X<9.85)= P(Z < \frac{9.85-10}{0.15}) = P(Z<-1)=0.159

    And for the other case:

    tex] P(X>10.15)[/tex]

     P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z<1)=0.159

    So then the probability of being defective P(D) is given by:

     P(D) = 0.159+0.159 = 0.318

    And the expected number of defective in a sample of 1000 units are:

     X= 0.318*1000= 318

    b)  P(X<9.85)= P(Z < \frac{9.85-10}{0.05}) = P(Z<-3)=0.00135

    And for the other case:

    tex] P(X>10.15)[/tex]

     P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z<3)=0.00135

    So then the probability of being defective P(D) is given by:

     P(D) = 0.00135+0.00135 = 0.0027

    And the expected number of defective in a sample of 1000 units are:

     X= 0.0027*1000= 2.7

    c) For this case the advantage is that we have less items that will be classified as defective

    Step-by-step explanation:

    Assuming this complete question: “Motorola used the normal distribution to determine the probability of defects and the number  of defects expected in a production process. Assume a production process produces  items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected  number of defects for a 1000-unit production run in the following situation.

    Part a

    The process standard deviation is .15, and the process control is set at plus or minus  one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces  will be classified as defects.”

    Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Solution to the problem

    Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

    X \sim N(10,0.15)  

    Where \mu=10 and \sigma=0.15

    We can calculate the probability of being defective like this:

     P(X<9.85)

    And we can use the z score formula given by:

     z=\frac{x-\mu}{\sigma}

    And if we replace we got:

     P(X<9.85)= P(Z < \frac{9.85-10}{0.15}) = P(Z<-1)=0.159

    And for the other case:

    tex] P(X>10.15)[/tex]

     P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z<1)=0.159

    So then the probability of being defective P(D) is given by:

     P(D) = 0.159+0.159 = 0.318

    And the expected number of defective in a sample of 1000 units are:

     X= 0.318*1000= 318

    Part b

    Through process design improvements, the process standard deviation can be reduced
    to .05. Assume the process control remains the same, with weights less than 9.85 or  greater than 10.15 ounces being classified as defects.

     P(X<9.85)= P(Z < \frac{9.85-10}{0.05}) = P(Z<-3)=0.00135

    And for the other case:

    tex] P(X>10.15)[/tex]

     P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z<3)=0.00135

    So then the probability of being defective P(D) is given by:

     P(D) = 0.00135+0.00135 = 0.0027

    And the expected number of defective in a sample of 1000 units are:

     X= 0.0027*1000= 2.7

    Part c What is the advantage of reducing process variation, thereby causing process control  limits to be at a greater number of standard deviations from the mean?

    For this case the advantage is that we have less items that will be classified as defective

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