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A certain industrial process yields a large number of steel cylinders whose lengths are distributed normal with mean \(3.25\) inches and standard deviation \(0.05\) inches. If two such cylinders are chosen at random and placed end to end what is the probability that their combined length is less than \(6.60\) inches?

Short Answer

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The probability that the combined length of two randomly picked cylinders is less than 6.60 inches is approximately 0.9213 or 92.13%.
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Step 1: Understand the Distribution of Cylinder Lengths

The length of each cylinder follows a normal distribution with mean \(3.25\) inches and standard deviation \(0.05\) inches. Therefore, the process of choosing one cylinder and measuring its length can be considered as drawing a random sample from this distribution.

Step 2: Find the Distribution of the Sum of Two Cylinders

We are interested in the combined length of two independently chosen cylinders. The sum of two independent normally distributed random variables is also a normal distribution. The mean of this new distribution is the sum of the means of the original distributions, and the variance is the sum of the original variances. Therefore, the combined length of two cylinders also follows a normal distribution, with mean \(3.25 + 3.25 = 6.5\) inches, and variance \(0.05^2 + 0.05^2 = 0.005\). The standard deviation is the square root of the variance, which is \( \sqrt{0.005} = 0.0707 inches\).

Step 3: Find the Z-Score for 6.60 inches

Next, convert the length of 6.60 inches to a standard score (or Z-score) by subtracting the mean and dividing by the standard deviation. The z-score (z) is calculated as \(z = (X - \mu)/\sigma\), where X is the value to be converted, \(\mu\) the mean, and \(\sigma\) the standard deviation. So, \(z = (6.60 - 6.5)/0.0707 = 1.4142\).

Step 4: Compute the Probability

Finally, find the probability that a normally distributed random variable is less than the Z-Score, which is equivalent to looking up the Z-score in a standard normal distribution table or using a calculator with a normal distribution function. The probability that the combined length of the two cylinders less than 6.60 inches is approximately 0.9213 or 92.13%.

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