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21 July, 17:30

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. Suppose that the specifications require the dot diameter to be between 0.0014 and 0.0026 inch. If the probability that a dot meets specifications is to be 0.9963, what standard deviation is needed? Round your answer to 4 decimal places.

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  1. 21 July, 19:59
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    0.0002 inch

    Step-by-step explanation:

    The empirical rule of the normal distribution, the 68-95-99.7 rule, means

    if the mean is μ and the standard deviation is σ,

    68% of data lies within μ - σ and μ + σ,

    95% of data lies within μ - 2σ and μ + 2σ,

    99.7% of data lies within μ - 3σ and μ + 3σ.

    From the question, μ = 0.002.

    The required range is 0.0014 to 0.0026.

    With a probability of 0.9963, then 0.9963 * 100% = 99.63% should lie within the range. This approximately corresponds to μ - 3σ and μ + 3σ.

    μ - 3σ = 0.0014

    0.002 - 3σ = 0.0014

    3σ = 0.0006

    σ = 0.0002

    Hence, the standard deviation is 0.0002 inch

    We can check with the other end of the range:

    μ + 3σ = 0.0026

    3σ = 0.0026 - 0.002

    3σ = 0.0006

    σ = 0.0002
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