Final answer:
Using the Central Limit Theorem and standard normal distribution, we can find the z-score for a mean of 111.8 watts and then calculate the tail probability that the sampled mean of amplifiers exceeds this value.
Step-by-step explanation:
To find the probability that the mean output of a sampled group of amplifiers is greater than 111.8 watts, we use the Central Limit Theorem (CLT). This theorem states that the sampling distribution of the sample mean will be normally distributed around the population mean if the sample size is sufficiently large, which is true for our case with 44 amplifiers being sampled.
The population mean (μ) is given as 110 watts, and the population variance (σ^2) is 100 watts^2. The standard deviation (σ) is the square root of the variance, so σ = 10 watts. To find the standard error (SE), which is the standard deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size (n). Thus, SE = σ / √n = 10 / √44.
Next, we calculate the z-score to see how many standard errors 111.8 watts is away from the mean. The z-score formula is (X - μ) / SE, where X is the value we're interested in (111.8 watts). Substituting the values, we get:
-
- SE = 10 / √44 ≈ 1.51
-
- Z = (111.8 - 110) / 1.51 ≈ 1.19
Using a standard normal distribution table or calculator, we find the probability corresponding to a z-score of 1.19. This probability tells us the likelihood that the sample mean is below 111.8 watts, so we subtract this from 1 to find the probability that the sample mean is above 111.8 watts.
Assuming the z-score corresponds to a probability of P, the answer would be:
-
- Probability (mean > 111.8 watts) = 1 - P