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 
and standard deviation 
, the zscore of a measure X is given by:
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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
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 
and standard deviation 
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 
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 
This is 1 subtracted by the pvalue of Z when X = 0.55. So
By the Central Limit Theorem
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.
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
has a pvalue of 0.8897
X = 0.45
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?
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
