Final answer:
To find the 10th percentile of the sample mean for TV sets in U.S. households, calculate the standard error, find the Z-score corresponding to the 10th percentile, and use these to determine the sample mean at this percentile. The 10th percentile is approximately 2.052.
Step-by-step explanation:
Finding the 10th Percentile of the Sample Mean
To find the 10th percentile of the sample mean for TV sets in U.S. households, given the population mean (μ) is 2.24 and the standard deviation (σ) is 1.4, with a sample size (n) of 90 households, you'll use the Central Limit Theorem (CLT). The CLT allows us to assume that the sampling distribution of the sample mean is approximately normally distributed since n ≥ 30. First, we calculate the standard error (SE) of the sample mean, which is σ divided by the square root of n (SE = σ/√n). Then, we use the standard normal distribution to find the Z-score that corresponds to the 10th percentile. Finally, we convert this Z-score back to the original scale using the population mean and the standard error.
Step-by-Step Solution:
- Calculate standard error (SE): SE = 1.4 / √90.
- Look up the Z-score for the 10th percentile in standard normal distribution tables (Z0.10).
- Calculate the 10th percentile of the sample mean: X10th = μ + Z0.10 × SE.
The exact calculation would require a statistics table or software to find the precise Z-score for the 10th percentile. In this case, Z0.10 is approximately -1.28. The calculation proceeds as follows:
SE = 1.4 / √90 ≈ 0.147
X10th = 2.24 + (-1.28 × 0.147) ≈ 2.24 - 0.188 ≈ 2.052
The 10th percentile of the sample mean for TV sets in U.S. households is approximately 2.052.