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The line width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer.

What is the probability that a line width is greater than 0.62 micrometer?

What is the probability that a line width is between 0.47 and 0.63 micrometer?

The line width of 90% of samples is below what value?

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  1. 29 May, 05:37
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    a. 0.82%

    b. 71.11%

    c. 0.564 micrometer

    Step-by-step explanation:

    What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.

    We have that z is equal to:

    z = (x - m) / (sd)

    x is the value to evaluate, m the mean, sd the standard deviation

    a.

    So for 0.62 copies we have:

    z = (0.62 - 0.5) / (0.05)

    z = 2.4

    and this value represents 0.9918

    p (x > 0.62) = 1 - 0.9918

    p (x > 0.62) = 0.0082 = 0.82%

    b.

    So for 0.47 copies we have:

    z = (0.47 - 0.5) / (0.05)

    z = - 0.6

    and this value represents 0.2742

    So for 0.63 copies we have:

    z = (0.63 - 0.5) / (0.05)

    z = - 2.6

    and this value represents 0.9953

    p (0.47 > x > 0.63) = 0.9953 - 0.2742

    p (0.47 > x > 0.63) = 0.7211 = 72.11 %

    c.

    So for x copies we have:

    p = 0.9, It represents z = 1.28

    1.28 = (x - 0.5) / (0.05)

    x = 1.28*0.05 + 0.5

    x = 0.564 micrometer
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