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.

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

so what is the question

step-by-step explanation:

answer: r there any options?

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