Question

3. There are two machines available for cutting corks for use in wine bottles. The first produces corks with diameters that are normally distributed with an average of 3 cm and a standard deviation of 0.1 cm. The second machine produces corks with diameters that have a normal distribution with an average of 3.04 cm and a standard deviation of 0.02 cm. Acceptable corks have diameters between 2.9 and 3.1 cm.
a.Which machine is most likely to produce an acceptable cork? Justify your answer
b. Of the 1200 corks cut by the second machine in a working day, approximately, how many are not acceptable?

Answer 1

For Machine 1
Here, μ = 3, σ = 0.1, x1 = 2.9 and x2 = 3.1. We need to compute P(2.9<= X <= 3.1). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (2.9 - 3)/0.1 = -1
z2 = (3.1 - 3)/0.1 = 1

Therefore, we get
P(2.9 <= X <= 3.1) = P((3.1 - 3)/0.1) <= z <= (3.1 - 3)/0.1)
= P(-1 <= z <= 1) = P(z <= 1) - P(z <= -1)
= 0.8413 - 0.1587
= 0.6826

For Machine 2
Here, μ = 3.04, σ = 0.02, x1 = 2.9 and x2 = 3.1. We need to compute P(2.9<= X <= 3.1). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (2.9 - 3.04)/0.02 = -7
z2 = (3.1 - 3.04)/0.02 = 3

Therefore, we get
P(2.9 <= X <= 3.1) = P((3.1 - 3.04)/0.02) <= z <= (3.1 - 3.04)/0.02)
= P(-7 <= z <= 3) = P(z <= 3) - P(z <= -7)
= 0.9987 - 0
= 0.9987

As 99.87% of the output for machine 2 is within the acceptable range, machine 2 is to produce most likely an acceptable cork

b)
Number of non acceptable corks = 1200 * (1 - 0.9987) = 1.56

i.e. ~2 corks