To find the probability that the mean lifespan of 18 randomly selected items is greater than 6 years, we can use the Central Limit Theorem. The theorem tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large (which 18 is, in this case). Here's how to solve the problem step by step:

Step 1: Identify the given values

• Population mean (μ) = 5.9 years
• Population standard deviation (σ) = 1.1 years
• Sample size (n) = 18
• Sample mean (X̄) we are interested in = 6 years

Step 2: Standardize the sample mean

To find the probability that the sample mean is greater than 6 years, we first calculate the standard error (SE) of the mean, which accounts for the sample size.

$SE = \frac{\sigma}{\sqrt{n}} = \frac{1.1}{\sqrt{18}} \approx \frac{1.1}{4.242} \approx 0.259$

Step 3: Compute the z-score

The z-score formula is:

$z = \frac{X̄ - \mu}{SE} = \frac{6 - 5.9}{0.259} \approx \frac{0.1}{0.259} \approx 0.386$

Step 4: Find the probability

Now, we look up the z-score in the standard normal distribution table (or use a calculator). The z-score of 0.386 corresponds to a cumulative probability of about 0.650.

Since we are interested in the probability that the mean lifespan is greater than 6 years, we need the area to the right of the z-score. So, the probability is:

$P(X̄ > 6) = 1 - 0.650 = 0.350$

Final Answer:

The probability that the mean lifespan of the 18 randomly selected items is greater than 6 years is approximately 0.35 (or 35%).

Would you like more details on any part of the solution?

Related Questions:

1. What is the probability that the mean lifespan is less than 5.5 years?
2. How would the probability change if the sample size was 30 instead of 18?
3. What is the z-score for a sample mean of 5.7 years?
4. How does the Central Limit Theorem help in this problem?
5. What if the standard deviation was 1.5 years instead of 1.1 years? How would the result change?

Tip:

Always remember that as the sample size increases, the standard error decreases, making the sample mean more likely to cluster around the population mean.