Final answer:
The margin of error with a 0.95 probability for the sample mean, given a population variance of 3,600 and a sample size of 144, is approximately 10, which corresponds to option d) 10.
Step-by-step explanation:
The question asked is related to the concept of confidence intervals in statistics, specifically to find the margin of error in a scenario where the variance of a population equals 3,600, and the sample size selected from the population is 144. Given a confidence level of 95%, we want to determine the margin of error for the sample mean.
To calculate the margin of error for the sample mean, the formula we use is Z * (σ/√n), where Z is the Z-score associated with the desired confidence level (which corresponds to 1.96 for a 95% confidence level), σ is the standard deviation of the population, and n is the sample size.
Firstly, we find the standard deviation of the population (σ) by taking the square root of the variance. Since the variance is given as 3,600, the standard deviation is:
σ = √3600 = 60
Next, we calculate the margin of error as follows:
Margin of Error = Z * (σ/√n) = 1.96 * (60/√144) = 1.96 * (60/12) = 1.96 * 5 = 9.8
Thus, the margin of error is approximately 10. Hence, the correct option is d) 10.