Answer to Question #149426 in Statistics and Probability for John David Manabat

Question

A process was designed to cut metal rods for use in an assembly of mechanical devices. The lengths of the metal rods cut are approximately normally distributed with a mean of 2.20 cm and a standard deviation of 0.04 cm. If a particular type of mechanical device requires metal rods with lengths between 2.22 and 2.25 cm, what percentage of the metal rods would be expected to be rejected?

Answer 1

to find  percentage of the metal rods would be expected to be rejected, we should find percentage of rods with lengths less than 2.22 and with lengths more than 2.25 and add them.

P=P(x<2.22)+P(x>2.25)=1-P(x<2.25)+P(x<2.22)

P(x<2.22)

z=$\frac{x-\mu}{\sigma}$ =(2.22-2.2)/0.04=0.5

using table, P(z=0.5)=0.69146

P(x<2.25)

z=(2.25-2.2)/0.04=1.25

using table, P(z=1.25)=0.89435

P=1-0.89435+0.69146=0.7971=79.71% would be rejected