To calculate the probability that the average length of life of 100 randomly selected instruments is more than 12.2 months, we need to use the properties of the sampling distribution of the sample mean.
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Understanding the Population Distribution:
- The length of life of the instruments follows a normal distribution with:
- Mean (\(\mu\)) = 12 months
- Standard deviation (\(\sigma\)) = 2 months
- The length of life of the instruments follows a normal distribution with:
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Sampling Distribution of the Sample Mean:
- When considering the sample mean (\(\bar{X}\)) of a random sample of size \(n=100\), the sampling distribution of the sample mean will also be normally distributed with:
- Mean of the sample mean (\(\mu_{\bar{X}}\)) = \(\mu\) = 12 months
- Standard deviation of the sample mean (\(\sigma_{\bar{X}}\)) = \(\frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{100}} = \frac{2}{10} = 0.2\) months
- When considering the sample mean (\(\bar{X}\)) of a random sample of size \(n=100\), the sampling distribution of the sample mean will also be normally distributed with:
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Z-Score Calculation: We want to find the probability that the average is more than 12.2 months, which can be expressed as: \[ P(\bar{X} > 12.2) \] We first calculate the Z-score for \(\bar{X} = 12.2\): \[ Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{12.2 - 12}{0.2} = \frac{0.2}{0.2} = 1 \]
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Finding the Probability: We need to find \(P(Z > 1)\). This can be looked up in a standard normal distribution table, or you can use a calculator or software to find: \[ P(Z > 1) = 1 - P(Z \leq 1) \] From standard normal distribution tables: \[ P(Z \leq 1) \approx 0.8413 \] Therefore: \[ P(Z > 1) \approx 1 - 0.8413 = 0.1587 \]
Thus, the probability that the average length of life of 100 randomly selected instruments will be more than 12.2 months is approximately 0.1587 or 15.87%.