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Given:
- The table of typical hours worked by employees at a company
- salaried employee makes $78,000 per year
- hourly employees get $26 per hour and $39 per hour when they work more than 40 hours.
To Find: Which payment option to recommend to a new employee.
Solution: I would recommend being a salaried employee.
Explanation:
We begin by calculating the typical number of hours worked per week.
Adding up the hours from the table, we have .
The payment for an hourly employee must be calculated as $26 per hour for working till 40 hours, and $39 per hour when they work more than 40 hours.
So, the payment for 47 hours of work per week will be dollars.
As there are 52 weeks in a year, the yearly payment for an hourly emplyee would be . That is, an hourly employee would earn $68276.
On the other hand, we are given that a salaried employee makes $78000 per year which is more money than what an hourly employee makes for the same amount of work.
Therefore, I would recommend a new emplyee to be paid a salary rather than work on an hourly basis.
Answer:
Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.
There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem
Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that .
What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?
For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:
has a pvalue of 0.9495.
This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.
For the average of the sample
What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?
Now, we need to find the standard deviation of the sample before using the zscore formula. That is:
.
Now
has a pvalue of 0.9999.
This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.