Respuesta :
Answer:
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 $18.8
Step-by-step explanation:
Answer:
87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68.
91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80.
5.48% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
So, lets see each question:
What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?
The Institute reported that entry level wages for male college graduates were $21.68 per hour, with a standard deviation of 2.30.
So [tex]\mu = 21.68, \sigma = 2.30, n = 50, s = \frac{2.30}{\sqrt{50}} = 0.3253[/tex]
This probability is the pvalue of Z when [tex]X = 21.68 + 0.50 = 22.18[/tex] subtracted by the pvalue of Z when [tex]X = 21.68 - 0.50 = 21.18[/tex]
X = 22.18
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22.18 - 21.68}{0.3253}[/tex]
[tex]Z = 1.54[/tex]
[tex]Z = 1.54[/tex] has a pvalue of 0.9382.
X = 21.18
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{21.18 - 21.68}{0.3253}[/tex]
[tex]Z = -1.54[/tex]
[tex]Z = -1.54[/tex] has a pvalue of 0.0618.
There is a 0.9382 - 0.0618 = 0.8764 = 87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68.
What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?
[tex]\mu = 18.80, \sigma = 2.05, n = 50, s = \frac{2.05}{\sqrt{50}} = 0.29[/tex]
This probability is the pvalue of [tex]Z = \frac{0.50}{0.29} = 1.72[/tex] subtracted by the pvalue of [tex]Z = -\frac{0.50}{0.29} = -1.72[/tex], following the same logic as the question above.
There is a 0.9573 - 0.0427 = 0.9146 = 91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80.
What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?
[tex]\mu = 18.80, \sigma = 2.05, n = 120, s = \frac{2.05}{\sqrt{120}} = 0.1871[/tex]
This is the pvalue of Z when [tex]X = 18.80 - 0.30 = 18.50[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18.50 - 18.80}{0.1871}[/tex]
[tex]Z = - 1.6[/tex]
[tex]Z = - 1.6[/tex] has a pvalue of 0.0548.
So there is a 5.48% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.
