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Chapter 7: Problem 58

Johnson Electronics Corporation makes electric tubes. It is known that the standard deviation of the lives of these tubes is 150 hours. The company's research department takes a sample of 100 such tubes and finds that the mean life of these tubes is 2250 hours. What is the probability that this sample mean is within 25 hours of the mean life of all tubes produced by this company?

Short Answer

Expert verified
The probability that this sample mean is within 25 hours of the mean life of all tubes produced by this company is 0.905 or 90.5%.

Step by step solution

01

Calculate Standard Error

Standard Error (SE) can be calculated using the formula SE = \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation and 'n' is the sample size. In this case, \(\sigma = 150\) hours and 'n' = 100 tubes. Substituting these values into the formula, we get: SE = \(\frac{150}{\sqrt{100}}\) = 15.
02

Calculate Z-Scores

The Z-score, which indicates how many standard errors away from the mean a data point is, can be calculated by the formula Z = \(\frac{x - 碌}{SE}\), where 'x' is the mean of the sample and '碌' is the mean of the population. We are looking for the probability that the sample mean is within 25 hours of the mean life of all tubes. That gives us two z-scores to calculate: one for (x-25) and one for (x+25). For the first z-score, substituting 'x-25' for the population mean (碌), we get Z1 = \(\frac{2250 - 25 - 2250}{15}\) = -1.67. For the second z-score, substituting 'x+25' for the population mean (碌), we get Z2 = \(\frac{2250 + 25 - 2250}{15}\) = 1.67.
03

Calculate Probability

The two z-scores calculated in the previous step determine the range of values within which the sample mean is likely to fall. The probability that the sample mean is within 25 hours of the population mean corresponds to the area under the standard normal curve between z=-1.67 and z=1.67. Using z-table or software that calculates the area under standard normal curve, we fill find that the area associated with z=-1.67 is 0.0475 and the area associated with z=1.67 is 0.9525. So, the probability that the sample mean is within 25 hours of the population mean is 0.9525 - 0.0475 = 0.905.

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