Answer:
The bonus will be paid on at least 4099 units.
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
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 percentile of this measure.
In this problem, we have that:
The highest 5 percent is the 95th percentile.
Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. This means that
.
If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?
The least units that the bonus will be paid is X when Z has a pvalue of 0.95.
Z has a pvalue of 0.95 between 1.64 and 1.65. So we use ![Z = 1.645](https://tex.z-dn.net/?f=Z%20%3D%201.645)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.645 = \frac{X - 4000}{60}](https://tex.z-dn.net/?f=1.645%20%3D%20%5Cfrac%7BX%20-%204000%7D%7B60%7D)
![X - 4000 = 60*1.645](https://tex.z-dn.net/?f=X%20-%204000%20%3D%2060%2A1.645)
![X = 4098.7](https://tex.z-dn.net/?f=X%20%3D%204098.7)
The number of units is discrete, this means that the bonus will be paid on at least 4099 units.