Question

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and a standard deviation of 0.1 cm. The second produces corks with diameters that have a normal distribution with mean 3.04 cm and variance 0.0025 cm². The specifications for the diameter of an acceptable cork are between 2.9 cm and 3.1 cm.

Therefore, what should be the mean of the second one to increase the proportion of acceptable corks?

Answer 1

Answer:

Given,

For 1st machine,

mean = 3 , standard deviation = 0.1

P(2.9 < X < 3.1) = P((2.9 - 3)/0.1 < (x-u)/s < (3.1 - 3)/0.1)

= P(- 1 < z < 1)

= P(z < 1) - P(z < - 1)

= 0.8413447 - 0.1586553 [since from z table]

= 0.6827

For second machine,

mean = 3.04 , standard deviation = 0.05

P(2.9 < X < 3.1) = P((2.9 - 3.04)/0.05 < (x-u)/s < (3.1 - 3.04)/0.05)

= P(-2.8 < z < 1.2)

= P(z < 1.2) - P(z < - 2.8)

= 0.8849303 - 0.0025551 [since from z table]

= 0.8824

Here we can say that the second machine is more likely to produce acceptable corks due to the probability to produce acceptable corks for second machine is greater than the first machine.