The Economic Policy Institute periodically issues reports on wages of entry level workers. The Institute reported that entry level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05.

Q.What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

The way I set it up is z=(-.30)/(2.05/square root 120))= -1.60
(.30) /(2.05/square root 120))= 1.60
=.8904 which says not correct?

Please help!

you did it correctly. you just have to take the negative z score of -1.6 and convert it to a percentage that's to the left of that. to 4 decimal places it's .0548

To calculate the probability that a sample of 120 female graduates will provide a sample mean more than $0.30 below the population mean, you need to use the standard normal distribution.

First, calculate the standard error of the sample mean using the formula: standard error = standard deviation / square root of sample size.

For female graduates:
Standard error = 2.05 / √120
Standard error ≈ 0.1874

Next, you need to convert the difference ($0.30) to z-scores using the formula: z = (X - μ) / σ, where X is the difference, μ is the population mean, and σ is the population standard deviation.

For female graduates:
z = (-0.30 - 18.80) / 2.05
z ≈ -9.9756

Now, you need to find the probability of obtaining a z-score greater than -9.9756. Since the standard normal distribution is symmetrical, the probability of obtaining a z-score greater than -9.9756 is essentially 1.

Therefore, the probability that a sample of 120 female graduates will provide a sample mean more than $0.30 below the population mean is approximately 1.

To find the probability that a sample of 120 female graduates will provide a sample mean more than $0.30 below the population mean, we can use the normal distribution.

First, let's calculate the standard error (SE) of the sample mean. The standard error is equal to the population standard deviation divided by the square root of the sample size. In this case, the population standard deviation for female graduates is $2.05, and the sample size is 120.

SE = 2.05 / √120
SE ≈ 0.187

Next, we need to calculate the z-score, which represents the number of standard deviations away from the population mean our desired sample mean is located.

z = (sample mean - population mean) / SE
z = (0.30 - 18.80) / 0.187
z ≈ -95.45

It seems like there is an error in your calculation. Remember to subtract the population mean from the sample mean in the numerator.

Now we have the z-score, we can look up the corresponding probability in the standard normal distribution table. However, the calculated z-score of -95.45 is extremely unlikely and reasonable. It indicates that the sample mean is almost 96 standard deviations below the population mean, which is highly improbable.

Please double-check your calculations and ensure you have correctly entered the values.

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