The Economic Policy Institute periodically issues reports on workers' wages. Suppose the institute reported that mean wages for male college
graduates were $37.39 per hour and for female college graduates were $27.83 per hour in 2017. Assume the standard deviation for male
graduates is $4.60 and for female graduates it is $4.10. (Round your answers to four decimal places.)
(a) What is the probability that a sample of 60 male graduates will provide a sample mean within $1.00 of the population mean, $37.39?
To solve this problem, we can use the concept of sampling distribution and the Central Limit Theorem.
According to the Central Limit Theorem, as long as the sample size is sufficiently large (typically n > 30), the distribution of sample means will be approximately normal, regardless of the shape of the population distribution.
The mean of the sampling distribution of sample means is equal to the population mean, which is $37.39. The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:
Standard Error = σ / sqrt(n)
where σ is the population standard deviation and n is the sample size.
In this case, the population standard deviation for male graduates is $4.60, and the sample size is 60. Therefore, the standard error is:
Standard Error = 4.60 / sqrt(60) = 0.5937 (approx.)
To find the probability that a sample mean is within $1.00 of the population mean, we need to calculate the z-scores corresponding to the lower and upper limits of the range.
Lower limit: $37.39 - $1.00 = $36.39
Upper limit: $37.39 + $1.00 = $38.39
To calculate the z-scores, we use the formula:
z = (x - μ) / σ
where x is the given value, μ is the population mean, and σ is the standard deviation.
For the lower limit, the z-score is:
z1 = ($36.39 - $37.39) / 0.5937 = -1.6848 (approx.)
For the upper limit, the z-score is:
z2 = ($38.39 - $37.39) / 0.5937 = 1.6848 (approx.)
Next, we need to find the probabilities associated with these z-scores using a standard normal distribution table or a calculator. Since we are interested in the probability that the sample mean is within the specified range, we need to calculate the area under the curve between these two z-scores.
Using a standard normal distribution table or a calculator, the probability corresponding to a z-score of -1.6848 is approximately 0.0447. Similarly, the probability corresponding to a z-score of 1.6848 is also approximately 0.0447.
Finally, to find the probability that a sample of 60 male graduates will provide a sample mean within $1.00 of the population mean, we subtract the probability of being outside this range from 1:
Probability = 1 - (0.0447 + 0.0447) = 0.9106
Therefore, the probability is approximately 0.9106.
To find the probability that a sample of 60 male graduates will provide a sample mean within $1.00 of the population mean ($37.39), we can use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size, the distribution of sample means will be approximately normally distributed.
The formula for calculating the standard error of the mean is:
Standard Error (SE) = standard deviation / √sample size
For male graduates, the standard deviation is given as $4.60, and the sample size is 60.
SE = $4.60 / √60
SE ≈ $0.5945 (rounded to four decimal places)
Since we want the sample mean to be within $1.00 of the population mean, the acceptable range is ($37.39 - $1.00, $37.39 + $1.00), which is ($36.39, $38.39).
We can convert this into z-scores to find the probability using the standard normal distribution.
To calculate the z-scores, we use the formula:
z = (x - μ) / SE
Where:
- x is the value we want to find the probability for (in this case, $37.39)
- μ is the mean of the population (again, $37.39)
- SE is the standard error ($0.5945)
For the lower bound z-score:
z_lower = ($36.39 - $37.39) / $0.5945
For the upper bound z-score:
z_upper = ($38.39 - $37.39) / $0.5945
We can now use these z-scores to find the probability using a standard normal distribution table or a calculator.
P($36.39 ≤ x ≤ $38.39) = P(z_lower ≤ Z ≤ z_upper)
Using the standard normal distribution table or a calculator, you can find the probability associated with the z-scores.
To solve this problem, we need to use the concept of sampling distributions and the Central Limit Theorem. The Central Limit Theorem tells us that for large enough sample sizes, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
Given that we know the population mean ($37.39), the population standard deviation for males ($4.60), and the sample size ($n = 60), we can use the formula for the standard deviation of the sampling distribution of the sample mean:
Standard deviation of the sampling distribution (σx) = population standard deviation (σ) / sqrt(sample size)
In this case, we have:
σ = $4.60
n = 60
So, the standard deviation of the sampling distribution is:
σx = $4.60 / sqrt(60)
To find the probability that the sample mean falls within $1.00 of the population mean, we need to find the z-scores for the lower and upper limits of the range.
Lower limit:
Z1 = (sample mean - population mean) / σx
= ($37.39 - $37.39) / ($4.60 / sqrt(60))
= 0
Upper limit:
Z2 = (sample mean + population mean) / σx
= ($37.39 + $37.39) / ($4.60 / sqrt(60))
= (2 * $37.39) / ($4.60 / sqrt(60))
To find the probability, we need to find the area under the standard normal curve between these two z-scores. We can use a standard normal distribution table or a calculator to find this probability.
Note: Since we are calculating the probability for both the lower and upper limits, we need to find the area in the middle of the two limits. Therefore, we need to find the area from -Z1 to Z1, and then subtract this area from 1 to get the total probability.
Finally, you can use a standard normal distribution table or a calculator to find the probability between -Z1 and Z1. Subtract this probability from 1 to get the final probability.
Note: The standard normal distribution table provides the area to the left of the z-score. To find the area between -Z1 and Z1, you can subtract the area associated with Z1 from 0.5 (which represents the area to the left of the mean) and then double the result.
I hope this explanation helps! Let me know if you have any further questions.