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Chapter 6: Problem 83

A machine at Kasem Steel Corporation makes iron rods that are supposed to be 50 inches long. However, the machine does not make all rods of exactly the same length. It is known that the probability distribution of the lengths of rods made on this machine is normal with a mean of 50 inches and a standard deviation of \(.06\) inch. The rods that are either shorter than \(49.85\) inches or longer than \(50.15\) inches are discarded. What percentage of the rods made on this machine are discarded?

Short Answer

Expert verified
The percentage of the rods made on this machine that are discarded is the sum of the probabilities calculated in step 2. To find this exact number, you would need to refer to a z-table or use a software that can compute it based on the z-scores calculated in step 1.

Step by step solution

01

Calculate the z-scores for the cutoff points

The formula for the z-score is \( z = \frac{x - \mu}{\sigma} \), where \(x\) is the value from the dataset, \(\mu\) is the mean and \(\sigma\) is the standard deviation. Using this formula, we find the z-scores for 49.85 and 50.15.
02

Find the probabilities corresponding to these z-scores

Using a standard normal (z) table or a suitable software, we find the probability that a rod will be shorter than 49.85 inches (z < z-value for 49.85 inches) and the probability that a rod will be longer than 50.15 inches (z > z-value for 50.15 inches).
03

Calculate the total probability of rods being discarded

To find the total percentage of rods discarded, we add the probabilities found in Step 2. The result is the percentage of rods shorter than 49.85 inches or longer than 50.15 inches, which is the percentage of rods that are discarded.

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Most popular questions from this chapter

Let \(x\) be a continuous random variable that follows a normal distribution with a mean of 550 and \(a\) standard deviation of 75 . a. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is 0250 . b. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is \(.9345\). c. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is approximately \(.0275\). d. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is approximately \(.9600\). e. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4700\) and the value of \(x\) is less than \(\mu\). f. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4100\) and the value of \(x\) is greater than \(\mu\).

Find the \(z\) value for each of the following \(x\) values for a normal distribution with \(\mu=30\) and \(\sigma=5\) a. \(x=39\) b. \(x=19\) c. \(x=24\) d. \(x=44\)

Obtain the following probabilities for the standard normal distribution. a. \(P(z>-.98)\) b. \(P(-2.47 \leq z \leq 1.29)\) c. \(P(0 \leq z \leq 4.25)\) d. \(P(-5.36 \leq z \leq 0)\) e. \(P(z>6.07)\) f. \(P(z<-5.27)\)

Major League Baseball rules require that the balls used in baseball games must have circumferences between 9 and \(9.25\) inches. Suppose the balls produced by the factory that supplies balls to Major League Baseball have circumferences normally distributed with a mean of \(9.125\) inches and a standard deviation of \(.06\) inch. What percentage of these baseballs fail to meet the circumference requirement?

Determine the value of \(z\) so that the area under the standard normal curve a. in the right tail is \(.0500\) b. in the left tail is \(.0250\) c. in the left tail is \(.0100\) d. in the right tail is \(.0050\)
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