There are two machines available for cutting corks intended for use in wine bott

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 0.10 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. What is the probability that the first machine produces an acceptable cork? (Round your answer to four decimal places.) What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places. ) Which machine is more likely to produce an acceptable cork? the first machine the second machine

Explanation / Answer

Let X1 be the first machine produces corks with diameters.

and X2 be the second machine produces corks with diameters.

X1 has normal distribution with mean = 3 cm and sd = 0.10 cm.

X2 has normal distribution with mean = 3.04 cm and sd = 0.02 cm

And also given that acceptable corks have diameters between 2.9 cm and 3.1 cm.

1) What is the probability that the first machine produces an acceptable cork?

That is here we have to find P(2.9 < X1 < 3.1).

First we have to find z-score for 2.9 and 3.1.

z = (x1 - mean) / sd

z-score for x1=2.9 is,

z = (2.9 - 3) / 0.1 = -1

z-score for x1 = 3.1 is,

z = (3.1 - 3) / 0.1 = 1

That is now we have to find P(-1 < Z < 1).

P(-1 < Z < 1) = P(Z <=1) - P(Z <=-1)

These probabilties we can find by using EXCEL.

syntax is,

=NORMSDIST(z)

P(Z <=1) = 0.8413

P(Z <=-1) = 0.1587

P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6827

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What is the probability that the second machine produces an acceptable cork?

That is here we have to find P(2.9 < X2 < 3.1).

First we have to find z-score for 2.9 and 3.1.

z = (x2 - mean) / sd

z-score for x2=2.9 is,

z = (2.9 - 3.04) / 0.02 = -7

z-score for x2 = 3.1 is,

z = (3.1 - 3.04) / 0.02 = 3

That is now we have to find P(-7 < Z < 3).

P(-7 < Z < 3) = P(Z <=1) - P(Z <=-7)

These probabilties we can find by using EXCEL.

syntax is,

=NORMSDIST(z)

P(Z <=3) = 0.9987

P(Z <=-7) = 0.0000

P(-7 < Z < 3) = 0.9987 - 0.0000 = 0.9987

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