The net sales and number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is 180 million and a standard deviation of 25 million. For the number of employees the mean is 1,500 and the standard deviation is 120. Clair's fabricators had net sales of 170 million and 1,850 employees.
Compare Clair's sales and number of employees with the other fabricators:
What percent of the aluminum fabricators have greater net sales compared with Clair's. Only?
What percent have more employees than Clair's.
What other fabricators?
To compare Clair's sales and number of employees with the other fabricators, we need to calculate the z-scores for Clair's values using the given mean and standard deviation.
For net sales:
Z-score = (Claire's net sales - Mean net sales) / Standard deviation
Z-score = (170 million - 180 million) / 25 million
Z-score = -10 million / 25 million
Z-score = -0.4
For the number of employees:
Z-score = (Claire's number of employees - Mean number of employees) / Standard deviation
Z-score = (1,850 - 1,500) / 120
Z-score = 350 / 120
Z-score = 2.92
Now, we can use the z-scores to find the percent of fabricators that have greater net sales and more employees than Clair's.
1. Percent of fabricators with greater net sales than Clair's:
Looking up the z-score of -0.4 in the z-table, we find the corresponding percentile is 34.13%. This means that approximately 34.13% of fabricators have net sales less than Clair's.
Therefore, the percent of fabricators that have greater net sales compared with Clair's is approximately 100% - 34.13% = 65.87%.
2. Percent of fabricators with more employees than Clair's:
Looking up the z-score of 2.92 in the z-table, we find the corresponding percentile is 99.50%. This means that approximately 99.50% of fabricators have fewer employees than Clair's.
Therefore, the percent of fabricators that have more employees than Clair's is approximately 100% - 99.50% = 0.50%.