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 standard deviation .1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation .02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. Which machine is more likely to produce an acceptable cork?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 3 ( first machine )

standard deviation = $\sigma$ = 0.1

P(2.9 < x < 3.1) = P[(2.9 - 3)/0.1 ) < (x - $\mu$ ) /$\sigma$  < (3.1 - 3) / 0.1) ]

= P(-1 < z < 1)

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

Using z table,

= 0.8413 - 0.1587

= 0.6826

mean = $\mu$ = 3.04 ( second machine )

standard deviation = $\sigma$ = 0.02

P(2.9 < x < 3.1) = P[(2.9 - 3.04)/0.02 ) < (x - $\mu$ ) /$\sigma$  < (3.1 - 3.04) / 0.02) ]

= P(-7 < z < 3)

= P(z < 3) - P(z < -7)

Using z table,

= 0.9987 - 0

= 0.9987

second machine is more likely to produce an acceptable cork