- The standard error of the sample mean is \(\sigma_{\bar{X}} = \frac{8}{\sqrt{32}} \approx 1.4142\).
- The standardized \(z\)-scores for the bounds \(58\) and \(62\) are \(z = \pm 1.41\).
- Using the standard normal distribution, the probability is \(P(-1.41 < Z < 1.41) = P(Z < 1.41) - P(Z < -1.41) \approx 0.9207 - 0.0793\).
- The final probability is \(\boxed{0.8414}\).
Explanation
- Analyze the data and distribution properties
The population mean lifetime of the smoke detectors is \(\mu = 60\) months, and the population standard deviation is \(\sigma = 8\) months. We are taking a random sample of \(n = 32\) smoke detectors. According to the Central Limit Theorem, since the sample size \(n \ge 30\), the sampling distribution of the sample mean \(\bar{X}\) follows a normal distribution with a mean equal to the population mean \(\mu_{\bar{X}} = \mu = 60\) months. We need to find the probability that the sample mean \(\bar{X}\) falls between \(58\) and \(62\) months, that is, \(P(58 < \bar{X} < 62)\).
- Calculate the standard error of the mean
The standard error of the mean, \(\sigma_{\bar{X}}\), measures the standard deviation of the sampling distribution and is calculated as:
\[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{32}} \approx 1.4142\]
This value will be used to standardize our sample mean bounds into \(z\)-scores.
- Calculate and round the z-scores
To find the probability, we convert the sample mean bounds \(\bar{X}_1 = 58\) and \(\bar{X}_2 = 62\) into standard normal \(z\)-scores using the formula \(z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}\):
For \(\bar{X}_1 = 58\):
\[z_1 = \frac{58 - 60}{1.4142} \approx -1.4142\]
Rounding to two decimal places, we get \(z_1 = -1.41\).
For \(\bar{X}_2 = 62\):
\[z_2 = \frac{62 - 60}{1.4142} \approx 1.4142\]
Rounding to two decimal places, we get \(z_2 = 1.41\).
- Determine the probability using z-scores
Now, we find the probability \(P(-1.41 < Z < 1.41)\) using the standard normal distribution:
\[P(-1.41 < Z < 1.41) = P(Z < 1.41) - P(Z < -1.41)\]
Using standard normal distribution tables:
\(P(Z < 1.41) \approx 0.9207\)
\(P(Z < -1.41) \approx 0.0793\)
Therefore:
\[P(-1.41 < Z < 1.41) = 0.9207 - 0.0793 = 0.8414\]
Examples
In quality control, manufacturing companies often use these types of probability calculations to monitor the reliability of their products. By checking if the average lifespan of a sample of products (like smoke detectors) falls within an expected range, the company can determine if their production process is stable or if it requires adjustment, ensuring consistent quality for customers.