The width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micron and a standard devi

Question

The width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micron and a standard deviation of 0.05 micron. What is the probability that a randomly selected tool will have a width between 0.47 and 0.63 microns? What is a specific width value so that 90% of the tools are below that value?

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Hailey 3 weeks 2021-11-19T19:09:16+00:00 1 Answer 0 views 0

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    2021-11-19T19:11:04+00:00

    Answer:

    Step-by-step explanation:

    Since the width of a tool used for semiconductor manufacturing is assumed to be normally distributed,

    we would apply the formula for normal distribution which is expressed as

    z = (x – µ)/σ

    Where

    x = width of selected tools.

    µ = mean width

    σ = standard deviation

    From the information given,

    µ = 0.5 micron

    σ = 0.05 micron

    We want to find the probability that a randomly selected tool will have a width between 0.47 and 0.63 microns. It is expressed as

    P(0.47 ≤ x ≤ 0.63)

    For x = 0.47,

    z = (0.47 – 0.5)/0.05 = – 0.6

    Looking at the normal distribution table, the probability corresponding to the z score is 0.27

    For x = 0.63,

    z = (0.63 – 0.5)/0.05 = 2.6

    Looking at the normal distribution table, the probability corresponding to the z score is 0.9953

    P(0.47 ≤ x ≤ 0.63) = 0.9953 – 0.27 = 0.7253

    Second question

    90% = 90/100 = 0.9

    Looking at the normal distribution table, the z score corresponding to 90% is 1.285

    Therefore

    1.285 = (x – 0.5)/0.05

    1.285 × 0.05 = x – 0.5

    0.06425 = x – 0.5

    x = 0.5 + 0.06425

    x = 0.56425

    The width value is approximately 0.56

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