Final answer:
1. The sample size used in this survey was 196. 2. The probability that the point estimate was within ±30 of the population mean is 1.
Step-by-step explanation:
1. To find the sample size used in the survey, we can use the formula for the standard error of the mean:
SE = σ/√n
Where SE is the standard error of the mean, σ is the population standard deviation, and n is the sample size.
Given that the standard error of the mean is 25 and the population standard deviation is 700, we can rearrange the formula to solve for n:
n = (σ/SE)^2
Substituting the values:
n = (700/25)^2
n = 196
Therefore, the sample size used in this survey was 196.
2. To find the probability that the point estimate was within ±30 of the population mean, we can use the normal distribution and the formula for the standard error of the mean:
SE = σ/√n
Where SE is the standard error of the mean, σ is the population standard deviation, and n is the sample size.
In this case, the standard error of the mean is 25 and the population standard deviation is 700. We can use these values to calculate the standard deviation of the sample mean:
SD = SE * √n
SD = 25 * √196
SD = 25 * 14
SD = 350
Next, we can calculate the z-score for ±30:
z = (x - μ) / SD
Positive z-score:
z = (30 - 0) / 350 = 0.0857
Negative z-score:
z = (-30 - 0) / 350 = -0.0857
Using a standard normal distribution table or calculator, we can find the area under the curve for these z-scores:
Area for positive z-score = 0.5359
Area for negative z-score = 0.4641
The total probability of the point estimate being within ±30 of the population mean is the sum of these two areas:
Total probability = Area for positive z-score + Area for negative z-score
Total probability = 0.5359 + 0.4641 = 1
Therefore, the probability that the point estimate was within ±30 of the population mean is 1.