Question

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 0.9973, what standard deviation is needed? Round your answer to 4 decimal places.

Answer 1

Answer:

The standard deviation is 0.0002 inch

Step-by-step explanation:

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

$\mu = 0.002$

The probability that a dot meets specifications is between 0.0014 and 0.0026 inch is 0.9973

So,P(0.0014<x<0.0026)=0.9973

Formula: $Z=\frac{x-\mu}{\sigma}$

So,$P(\frac{0.0014-0.002}{\sigma}<Z<\frac{0.0026-0.002}{\sigma})=0.9973$

$P(\frac{-0.0006}{\sigma}<Z<\frac{0.0006}{\sigma})=0.9973$

$2P(\frac{0.0006}{\sigma})=1-0.9973$

$P(\frac{0.0006}{\sigma})=0.0027$

Using Z table

$\frac{0.0006}{\sigma}=2.99998$

$\sigma =0.0002$

Hence The standard deviation is 0.0002 inch