Answer:
the correct answer is option B.
Explanation:
We can use the formula for the margin of error for a confidence interval for a population mean with a known population standard deviation:
Margin of error = z*σ/√n
where z is the critical value from the standard normal distribution for the desired confidence level (95% in this case), σ is the population standard deviation, and n is the sample size.
We are given that the margin of error is $135 and σ is $538. We need to find the sample size, n. To do this, we first need to find the appropriate value of z for a 95% confidence level. Using a standard normal distribution table or calculator, we can find that z = 1.96.
Substituting the values into the margin of error formula and solving for n, we have:
$135 = 1.96*($538)/√n
Squaring both sides and solving for n, we get:
n = [1.96*($538)/$135]^2 ≈ 62
Therefore, a sample size of 62 business students must be randomly selected to estimate the mean monthly earnings of business students at one college with 95% confidence that the sample mean is within $135 of the population mean, assuming the population standard deviation is known to be $538.
So the correct answer is option B.