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tensa zangetsu [6.8K]
3 years ago
8

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. Suppose that the speci

fications 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.
Mathematics
1 answer:
Leni [432]3 years ago
7 0

Answer:

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 <em>μ</em> and the standard deviation is <em>σ</em>,

68% of data lies within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>,

95% of data lies within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>,

99.7% of data lies within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 3<em>σ.</em>

<em />

From the question, <em>μ</em> = 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 <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 3<em>σ.</em>

<em>μ </em>- 3<em>σ</em> = 0.0014

0.002 - 3<em>σ</em> = 0.0014

3<em>σ</em> = 0.0006

<em>σ</em> = 0.0002

Hence, the standard deviation is 0.0002 inch

We can check with the other end of the range:

<em>μ </em>+ 3<em>σ</em> = 0.0026

3<em>σ</em> = 0.0026 - 0.002

3<em>σ</em> = 0.0006

<em>σ</em> = 0.0002

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