Respuesta :
Answer:
a) p = 0.5
b) 11.03% probability that more than 55% of them earned more than $837 per week.
c) 99.29% probability that less than 60% of the sample of 150 employees earned more than $837 per week
d) 77.94% probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week
e) 45% is within 2 standard deviations of the mean, which means it would not be unusual if less than 45% of the sample of 150 employees earned more than $755 per week
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
If X is two or more standard deviations from the mean, it is considered unusual.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In 2016, the median weekly earnings for people employed full-time in the United States was $837.
This means that 50% of employees earn more than $837 and 50% below.
So we use [tex]p = 0.5[/tex]
a) What proportion of full-time employees had weekly earnings of more than $837?
From above, p = 0.5
b) A sample of 150 full-time employees is chosen. What is the probability that more than 55% of them earned more than $837 per week?
n = 150, so [tex]s = \sqrt{\frac{0.5*0.5}{150}} = 0.0408[/tex]
This is 1 subtracted by the pvalue of Z when X = 0.55. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.55 - 0.5}{0.0408}[/tex]
[tex]Z = 1.225[/tex]
[tex]Z = 1.225[/tex] has a pvalue of 0.8897
1 - 0.8897 = 0.1103
11.03% probability that more than 55% of them earned more than $837 per week.
c) What is the probability that less than 60% of the sample of 150 employees earned more than $837 per week?
This is the pvalue of Z when X = 0.6.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.6 - 0.5}{0.0408}[/tex]
[tex]Z = 2.45[/tex]
[tex]Z = 2.45[/tex] has a pvalue of 0.9929
99.29% probability that less than 60% of the sample of 150 employees earned more than $837 per week.
d) What is the probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week?
This is the pvalue of Z when X = 0.55 subtracted by the pvalue of Z when X = 0.45.
X = 0.55
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.55 - 0.5}{0.0408}[/tex]
[tex]Z = 1.225[/tex]
[tex]Z = 1.225[/tex] has a pvalue of 0.8897
X = 0.45
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.45 - 0.5}{0.0408}[/tex]
[tex]Z = -1.225[/tex]
[tex]Z = -1.225[/tex] has a pvalue of 0.1103
0.8897 - 0.1103 = 0.7794
77.94% probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week.
e) Would it be unusual if less than 45% of the sample of 150 employees earned more than $755 per week?
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.45 - 0.5}{0.0408}[/tex]
[tex]Z = -1.225[/tex]
So 45% is within 2 standard deviations of the mean, which means it would not be unusual if less than 45% of the sample of 150 employees earned more than $755 per week
